Class 12

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 5 Continuity and Differentiability Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Miscellaneous Exercise

Direction (1 – 11) :
Differentiate the following functions with respect to x.
Question 1.
(3x² – 9x + 5)⁹
Solution.
Let y = (3x² – 9x + 5)⁹
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (3x² – 9x + 5)⁹
= 9(3x² – 9x + 5)⁸ – (3x² – 9x + 5) dx
= 9(3x² – 9x + 5)⁸ . ⁸ (6x – 9)
= 9(3x² – 9x + 5)⁸ . 3(2x – 3)
= 27(3x² – 9x + 5)⁸ (2x – 3)

Question 2.
sin³ x + cos⁶ x
Solution.
Let y = sin³ x + cos⁶ x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (sin³ x) + \(\frac{d}{d x}\) (cos⁶ x)
= 3 sin² x \(\frac{d}{d x}\) (sin x) + 6 cos⁵ x . \(\frac{d}{d x}\) (cos x)
= 3 sin² x cos x + 6 cos⁵ x . (- sin x)
= 3 sin x cos x (sin x – 2 cos⁴ x).

Question 3.
(5x)^{3 cos 2x}
Solution.
Let y = (5x)^{3 cos 2x}
Taking logarithm on both sides, we get
log y = 3 cos 2x log 5x
Differentiating on both sides w.r.t. x, we get
\(\frac{1}{y}\) \(\frac{d y}{d x}\) = 3 [log 5x . \(\frac{d}{d x}\) (cos 2x) + cos 2x . \(\frac{d}{d x}\) (log 5x)]

⇒ \(\frac{d y}{d x}\) = 3y [log 5x (- sin 2x) . \(\frac{d}{d x}\) (2x) + cos 2x . \(\frac{1}{5 x}\) . \(\frac{d}{d x}\) (5x)]

⇒ \(\frac{d y}{d x}\) = 3y [- 2 sin 2x log 5x + \(\frac{\cos 2 x}{x}\)]

⇒ \(\frac{d y}{d x}\) = y [\(\frac{3 \cos 2 x}{x}\) – 6 sin 2x log 5x]

∴ \(\frac{d y}{d x}\) = (5x)^{3 cos 2x} [\(\frac{3 \cos 2 x}{x}\) – 6 sin 2x log 5x]

Question 4.
sin^-1(x√x), 0 ≤ x ≤ 1
Solution.
Let y = sin^-1 (x√x)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) sin^-1 (x√x)
= \(\frac{1}{\sqrt{1-(x \sqrt{x})^{2}}} \times \frac{d}{d x}(x \sqrt{x})\)

= \(\frac{1}{\sqrt{1-x^{3}}} \cdot \frac{d}{d x}\left(x^{\frac{3}{2}}\right)\)

= \(\frac{1}{\sqrt{1-x^{3}}} \times \frac{3}{2} \cdot x^{\frac{1}{2}}\)

= \(\frac{3 \sqrt{x}}{2 \sqrt{1-x^{3}}} \Rightarrow \frac{3}{2} \sqrt{\frac{x}{1-x^{3}}}\)

Question 5.
\(\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}\), – 2 < x < 2
Solution.
Let y = \(\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}\)
Differentiating w.r.t. x, we get

Question 6.
cot^-1 \(\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]\), 0 < x < \(\frac{\pi}{2}\)
Solution.

Therefore, Eq. (i) becomes
y = cot^-1 (cot \(\frac{x}{2}\))
⇒ y = \(\frac{x}{2}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{1}{2}\) \(\frac{d}{d x}\) (x)
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2}\).

Question 7.
(log x)^{log x}, x > 1
Solution.
Let y = (log x)^{log x}
Taking logarithm on both sides, we get
log y = log (log x)log x
⇒ log y = log x . log(log x)
Differentiating both sides w.r.t. x, we get
\(\frac{1}{y}\) . \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [log x . log(log x)] y
⇒ \(\frac{1}{y}\) . \(\frac{d y}{d x}\) = log (log x) . \(\frac{d}{d x}\) (log x) + log x . \(\frac{d}{d x}\) [log (log x)]

⇒ \(\frac{d y}{d x}\) = y [log (log x) . \(\frac{1}{x}\) + log x . \(\frac{1}{\log x}\) . \(\frac{d}{d x}\) (log x)]

⇒ \(\frac{d y}{d x}\) = y [\(\frac{1}{x}\) log (log x) + \(\frac{1}{x}\)]

∴ \(\frac{d y}{d x}\) = (log x)^{log x} [\(\frac{1}{x}\) log (log x) + \(\frac{1}{x}\)]

Question 8.
cos(a cos x + b sin x), for some constant a and b.
Solution.
Let y = cos(acosx + bsinx)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) cos(a cos x + b sin x)
\(\frac{d y}{d x}\) = – sin(a cos x + b sin x) . \(\frac{d}{d x}\) (a cos x + b sin x)
= – sin(a cos x + b sin x) . [a (- sin x) + b cos x]
= (a sin x – b cos x) . sin (a cos x + b sin x).

Question 9.
(sin x – cos x)^{(sin x – cos x)}, \(\frac{\pi}{4}<x<\frac{3 \pi}{4}\)
Solution.
Let y = (sin x – cos x) (sin x – cos x)
Taking logarithm on both sides, we get
log y = log (sin x – cos x)[(sin x – cos x)^{(sin x – cos x)}]
⇒ log y = (sin x – cos x) . log(sin x – cos x)
Differentiating on both sides w.r.t. x, we get
\(\frac{1}{y}\) . \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [(sin x – cos x) log(sin x – cos x)]

⇒ \(\frac{1}{y}\) . \(\frac{d y}{d x}\) = log (sin x – cos x) . \(\frac{d}{d x}\) log (sin x – cos x) – (sin x – cos x) + (sin x – cos x) . \(\frac{d}{d x}\) log (sin x – cos x)

⇒ \(\frac{d y}{d x}\) = (sin x – cos x)^{(sin x – cos x)} [(cos x + sin x) . log (sin x – cos x) + (cos x – sin x)]

∴ \(\frac{d y}{d x}\) = (sin x – cos x)^{(sin x – cos x)} (cos x + sin x) [1 + log (sin x – cos x)]

Question 10.
x^x + x^a + a^x + a^a, for some fixed a > 0 and x > 0.
Solution.
Let y = x^x + x^a + a^x + a^a
Also, let x^x = u, x^a = v, a^x = w, and a^a = s
∴ y = u + v + w + s
Differentiating w.r.t. x, we get

\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}+\frac{d s}{d x}\) ……………(i)

Then, u = x^x
⇒ log u = log x^x (Taking log on both sides)
⇒ log u = x logx
Differentiating on both sides w.r.t. x, we get
\(\frac{1}{u}\) \(\frac{d u}{d x}\) = log x . \(\frac{d}{d x}\) (x) + x . \(\frac{d}{d x}\) (log x)
⇒ \(\frac{d u}{d x}\) = u [log x . 1 + x . \(\frac{1}{x}\)]

⇒ \(\frac{d u}{d x}\) = x^x [log x + 1]
= x^x (1 + log x) …………..(ii)

v = x^a
Differentiating w.r.t. x, we get
\(\frac{d v}{d x}\) = \(\frac{d}{d x}\) (x^a)

⇒ \(\frac{d v}{d x}\) = a x^{a – 1} ………….(iii)

w = a^x
⇒ log w = log a^x
⇒ log w = a log x
Differentiating on both sides w.r.t. x, we get
⇒ \(\frac{d w}{d x}\) = log a . \(\frac{d}{d x}\) (x)

⇒ \(\frac{d w}{d x}\) = a^x log a ………………(iv)

s = a^a
Since, a is constant, therefore a^a is also a constant.
∴ \(\frac{d s}{d x}\) = 0 …………..(v)
From Eqs. (i), (ii), (iii), (iv) and (v), we get
\(\frac{d y}{d x}\) = x^x (1 + log x) + a x^{a – 1} + a^x log a + 0
= xx^x (1 + log x) + a x^{a – 1} + a^x log a.

Question 11.
x^{x2 – 3} + (x – 3)^x2, for x > 3.
Solution.
Let y = x^{x2 – 3} + (x – 3)^x2
Also, let u = x^{x2 – 3} and v = (x – 3)^x2
∴ y = u + v
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
Then, u = x^{x2 – 3}
Taking log on both sides, we get
log u = log x^{x2 – 3}
= (x² – 3) log x
Differentiating w.r.t. x, we get
\(\frac{1}{u}\) \(\frac{d u}{d x}\) = 2x log x + (x² – 3) log x × \(\frac{1}{x}\)

∴ \(\frac{d u}{d x}\) = u (2x log x + \(\frac{x^{2}-3}{x}\))
= x^{x2 – 3} (2x log x + \(\frac{x^{2}-3}{x}\)) …………….(ii)
Also, v = (x – 3)^{x2
Differentiating on both sides, we get
\(\frac{1}{v}\) \(\frac{d v}{d x}\) = \(\frac{d v}{d x}\) [2x log (x – 3) + \(\frac{x^{2}-3}{x}\)] ……………..(iii)
From Eqs. (i), (ii) and (iii), we get
\(\frac{d y}{d x}\) = xx2 – 3 \(\left(2 x \log x+\frac{x^{2}-3}{x}\right)+(x-3)^{x^{2}}\left[2 x \log (x-3)+\frac{x^{2}}{x-3}\right]\)}

Question 12.
Find \(\frac{d y}{d x}\), if y = 12 (1 – cos t), x = 10 (t – sin t), \(-\frac{\pi}{2}<t<\frac{\pi}{2}\)
Solution.
Given, y = 12 (1 – cos t) and x = 10 (t – sin t)
Differentiating w.r.t t, we get
\(\frac{d x}{d t}\) = \(\frac{d}{d t}\) [10 (t – sin t)]
= 10 . \(\frac{d}{d t}\) (t – sin t)
= 10 (1 – cos t)

and \(\frac{d y}{d t}\) = \(\frac{d}{d t}\) [12 (1 – cos t)]
= 12 . \(\frac{d}{d t}\) (1 – cos t)
= 12 . [0 – (- sin t)] = 12 sin t

Now, \(\frac{d y}{d x}\) = \(\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{12 \sin t}{10(1-\cos t)}\)

= \(\frac{12 \cdot 2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{10 \cdot 2 \sin ^{2} \frac{t}{2}}=\frac{6}{5} \cot \frac{t}{2}\).

Question 13.
Find \(\frac{d y}{d x}\), if y = sin^-1 x + sin^-1 \(\sqrt{1-x^{2}}\), – 1 ≤ x ≤ 1.
Solution.
Given, y = sin^-1 x + sin^-1 \(\sqrt{1-x^{2}}\)
putting x = sin θ in above eq., we get
y = sin^-1 sin θ + sin^-1 \(\sqrt{1-(\sin \theta)^{2}}\)
y = θ + sin^-1 \(\sqrt{1-(\sin \theta)^{2}}\)
= θ + sin^-1 (cos θ)
= θ + sin^-1 sin (\(\frac{\pi}{2}\) – θ)
= θ + \(\frac{\pi}{2}\) – θ
Differentiating w.r.t x, we get
∴ \(\frac{d y}{d x}\) = 0

Question 14.
If x \(\sqrt{1+y}\) + y \(\sqrt{1+x}\) = 0, for – 1 < x < 1, prove that \(\frac{d y}{d x}=\frac{1}{(1+x)^{2}}\).
Solution.
Given, x \(\sqrt{1+y}\) + y \(\sqrt{1+x}\) = 0
⇒ x \(\sqrt{1+y}\) = – y \(\sqrt{1+x}\)
On squaring bothsides, we get
x² (1 + y) = y² (1 + x)
⇒ x² + x²y = y² + xy²
⇒ x² – y² = xy² – x²y
⇒ x² – y² = xy (y – x)
⇒ (x + y) (x – y) = xy (y – x)
∴ x + y = -xy
⇒ (1 + x) y = – x
⇒ y = \(\frac{-x}{(1+x)}\)
Differentiating on bothsides w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{(1+x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+x)}{(1+x)^{2}}\)

= \(-\frac{(1+x)-x}{(1+x)^{2}}=-\frac{1}{(1+x)^{2}}\)

Hence proved.

Question 15.
If (x – a)² + (y – b)² = c², for some c > 0, prove that \(\frac{\left(1+\left(\begin{array}{l}
d y \\
d x
\end{array}\right)^{2}\right]^{3}}{d^{2} y} \frac{d x^{2}}{}\) is a constant independent of a and b.
Solution.
Given, (x – a)² + (y – b)² = c²
Differentiating on both sides w.r.t. x. we get
\(\frac{d}{d x}\) [(x – a)²] + \(\frac{d}{d x}\) [(y – b)²] = \(\frac{d}{d x}\) (c²)
⇒ 2 (x – a) . \(\frac{d}{d x}\) (x – a) + 2 (y – b) . \(\frac{d}{d x}\) (y – b) = 0
⇒ 2 (x – a) . 1 + 2(y – b) . \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{-(x-a)}{y-b}\) ………………(i)
Again, differentiating w.r.t. x, we get
∴ \(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\left[\frac{-(x-a)}{y-b}\right]\)

= – \(\left[\frac{(y-b) \cdot \frac{d}{d x}(x-a)-(x-a) \cdot \frac{d}{d x}(y-b)}{(y-b)^{2}}\right]\)

Hence proved.

Question 16.
If cos y = x cos (a + y), with cos a ≠ ± 1, prove that \(\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}\).
Solution.
Given, cos y = x cos(a + y)
Differentiating w.r.t. x, we get
\(\frac{d}{d x}\) [cos y] = \(\frac{d}{d x}\) [x cos (a + y)]
⇒ – sin y = cos (a + y) . \(\frac{d}{d x}\) (x) + x . \(\frac{d}{d x}\) [cos (a + y)]
⇒ – sin y \(\frac{d y}{d x}\) = cos (a + y) + x [- sin (a + y)]
⇒ [x sin (a + y) – sin y] \(\frac{d y}{d x}\) = cos (a + y) ……………(i)
∵ cos y = x cos (a + y), x = \(\frac{\cos y}{\cos (a+y)}\)
Then, Eq. (i) reduces to
[\(\frac{\cos y}{\cos (a+y)}\) . sin (a + y) – sin y] \(\frac{d y}{d x}\) = cos (a + y)
⇒ [cos y . sin (a + y)] – sin y . cos (a + y)] . \(\frac{d y}{d x}\) = cos ² (a + y)
⇒ sin (a + y – y) \(\frac{d y}{d x}\) = cos² (a + y)
⇒ \(\frac{d y}{d x}\) = \(\frac{\cos ^{2}(a+y)}{\sin a}\).
Hence proved.

Question 17.
If x = a(cos t + t sin t) and y = a(sin t – t cos t), find \(\frac{d^{2} y}{d x^{2}}\).
Solution.
Given, x = a (cos t + t sin t) and y = a (sin t – t cos t)
Differentiating both Eqs. w.r.t. t, we get
∴ \(\frac{d x}{d t}\) = a \(\frac{d}{d t}\) (cos t + t sin t)

= a [- sin t + sin t . \(\frac{d}{d x}\) (t) + t . \(\frac{d}{d t}\) (sin t)]

= a [- sin t + sin t + t cos t] = at cost
Also, \(\frac{d y}{d t}\) = a \(\frac{d}{d t}\) (sin t – t cost)

= a [cos t – {cos t . \(\frac{d}{d t}\) (t) + t . \(\frac{d}{d t}\) (cost)}]
= a [cos t – {cos t – t sin t}] = at sin t

\(\frac{d y}{d x}\) = \(\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a t \sin t}{a t \cos t}\) = tan t
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (tan t)

= sec² t . \(\frac{d t}{d x}\)

[∵ \(\frac{d x}{d t}\) = at cos t
⇒ \(\frac{d t}{d x}\) = \(\frac{1}{a t \cos t}\)]

= sec² t . \(\frac{1}{a t \cos t}\)

= \(\frac{\sec ^{3} t}{a t}\), 0 < t < \(\frac{\pi}{2}\)

Question 18.
If f(x) = |x|³, show that f”(x) exists for all real x and find it.
Solution.
Here, f(x) = | x |³
When x > 0, f(x) = |x|³ = x³
Differentiating both sides w.r.t. x, we get
\(\frac{d}{d x}\) [f(x)] = \(\frac{d}{d x}\) (x³)
⇒ f'(x) = 3x²
Again, differentiating both sides w.r.t. x, we get
\(\frac{d}{d x}\) [f'(x)] = \(\frac{d}{d x}\) (3x²)
⇒ f”(x) = 6x
When x < 0,
f(x) = |x|³ = – x³
Differentiating both sides w.r.t. x, we get
\(\frac{d}{d x}\) [f(x)] = \(\frac{d}{d x}\) (- x³)
⇒ f'(x) = – 3x²
Again, differentiating both sides w.r.t. x, we get
\(\frac{d}{d x}\) [f'(x)] = \(\frac{d}{d x}\) (- 3x²) = – 6x
f”(x) = – 6x
Hence,f”(x) =

Question 19.
Using mathematical induction, prove that \(\frac{d}{d x}\) (xⁿ) = n x^{n – 1} for all positive integers n.
Solution.
Let P(n): \(\frac{d}{d x}\) (xⁿ) = nx^{n – 1} for all positive integers n
For n = 1,
P(1): \(\frac{d}{d x}\)(x) = 1 = 1 . x^{1 – 1}
∴ P(n) is true for n = 1.
Let P(k) is true for some positive integer k.
i.e., P(k): \(\frac{d}{d x}\) (x^k) = k x^{k – 1}
It has to be proved that P(k +1) is also true.
Consider \(\frac{d}{d x}\) (x^{k – 1}1) = \(\frac{d}{d x}\) (x . x^k)
= x^k . \(\frac{d}{d x}\) (x) + x . \(\frac{d}{d x}\) (x^k) [Applying product rule]
= (k + 1) . x^k
= (k + 1) . x^{(k+1) – 1}
Thus, P(k + 1) is true whenever P(k) is true.
Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n. Hence proved.

Question 20.
Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Solution.
We have, sin(A + B) = sin A cos B + cos A sin B
Differentiating on both sides w.r.t. x, we get
⇒ \(\frac{d}{d x}\) [sin(A + B)] = \(\frac{d}{d x}\) (sin A cos B) + \(\frac{d}{d x}\) (cos A sin B)

⇒ cos (A + B) . \(\frac{d}{d x}\) (A + B) = cos B . \(\frac{d}{d x}\) (sin A) + sin A . \(\frac{d}{d x}\) (cos B) + sin B . \(\frac{d}{d x}\) (cos A) + cos A . \(\frac{d}{d x}\) (sin B)
⇒ cos (A + B) . \(\frac{d}{d x}\) (A + B) = cos B cos A \(\frac{d A}{d x}\) + sin A(- sin B) \(\frac{d B}{d x}\) + sin B (- sin A) . \(\frac{d A}{d x}\) + cos A cos B \(\frac{d B}{d x}\)
cos (A + B) = cos A cos B – sin A sin B.

Question 21.
Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
Solution.
Yes, we have the continuous function f(x) = |x – 1| + |x – 2|, which is continuous at all x ∈ R but differentiable at all x except 1, 2.
Here, f(x) = |x – 1| + |x – 2|
=

i.e., f(x) =
Since, polynomial function is continuous, so it is clear that f is continuous at all except possible at 1, 2.
Now, we have to check the continuity at 1, 2
At x = 1 f(1) = 1,
LHL = \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1^{-}}\) (- 2x + 3)
= – 2 + 3 = 1

RHL = \(\lim _{x \rightarrow 1^{+}}\) f(x) = \(\lim _{x \rightarrow 1^{+}}\) 1 = 1

⇒ f(1) = \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1^{+}}\) f(x)
⇒ f is continuous at x = 1

At x = 2 f(2) = 1
⇒ LHL = \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{-}}\) (1) = 1

RHL = \(\lim _{x \rightarrow 2^{+}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) (2x – 3)
= 2 × 2 – 3 = 1

f(x) = \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) f(x)

So, f is continuous at x = 2. Thus, f is continuous at all x ∈ R. -2, if x < 1

Lf'(2) ≠ Rf'(2)
⇒ f is not differentiale at 2.
Thus, we see that f(x) = |x – 1| + |x – 2| is continuous everywhere and differentiable also at all x ∈ R except at 1, 2.

Question 22.
If y = \(\begin{array}{ccc}
\boldsymbol{f}(\boldsymbol{x}) & \boldsymbol{g}(\boldsymbol{x}) & \boldsymbol{h}(\boldsymbol{x}) \\
\boldsymbol{l} & \boldsymbol{m} & \boldsymbol{n} \\
\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c}
\end{array}\), prove that \(\frac{d y}{d x}=\left|\begin{array}{ccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
l & m & n \\
a & b & c
\end{array}\right|\).
Solution.

Question 23.
If y = e^{a cos-1 x} ,- 1 ≤ x ≤ 1, show that (1 – x²) \(\frac{d^{2} y}{d x^{2}}\) – x \(\frac{d y}{d x}\) – a² y = 0.
Solution.
Given, y = e^{a cos-1 x}
Taking logarithm on bothsides, we get
⇒ log y = a cos^-1 x log e
⇒ log y = a cos^-1 x [∵ log e = 1]

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 5 Continuity and Differentiability Ex 5.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7

Direction (1 – 10) : Find the second order derivatives of the following functions.

Question 1.
x² + 3x + 2
Solution.
Let y = x² + 3x + 2
Differentiating w.r.t. r, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x²) + \(\frac{d}{d x}\) (3x) + \(\frac{d}{d x}\) (2)
= 2x + 3 + 0
= 2x + 3
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (2x + 3)
= \(\frac{d}{d x}\) (2x) + \(\frac{d}{d x}\) (3)
= 2 + 0 = 2.

Question 2.
x²⁰
Sol.
Let y = x²⁰
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x²⁰)
= 20 x¹⁹
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (20 x¹⁹)
= 20 \(\frac{d}{d x}\) (x¹⁹)
= 20 × 19 × x¹⁸
= 380 × x¹⁸

Question 3.
x . cos x
Solution.
Let y = x cos x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x . cos x)
= cos x . \(\frac{d}{d x}\) (x) + x \(\frac{d}{d x}\) (cos x)
= cos x . 1 + x (- sin x)
= cos x – x sin x
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) [cos x – x sin x]
= \(\frac{d}{d x}\) (cos x ) – \(\frac{d}{d x}\) (x sin x)
= – sin x – [sin x . \(\frac{d}{d x}\) (x) + x . \(\frac{d}{d x}\) (sin x)]
= – sin x – (sin x + x cos x)
= – (x cos x + 2 sin x)

Question 4.
log x
Solution.
Let y = log x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (log x) = \(\frac{1}{x}\)
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (\(\frac{1}{x}\))
= \(\frac{-1}{x^{2}}\)

Question 5.
x³ log x
Solution.
Let y = x³ log x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [x³ log x]
= log x . \(\frac{d}{d x}\) (x³) + x³ . \(\frac{d}{d x}\) (log x)
= log x . 3x² + x³ . \(\frac{1}{x}\)
= log x . 3x² + x²
= x² (1 + 3 log x)
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) [x² (1 + 3 log x)]
= (1 + 3 log x) . \(\frac{d}{d x}\) (x²) + x² \(\frac{d}{d x}\) (1 +3 log x)
= (1 + 3 log x) . 2x + x² . \(\frac{3}{x}\)
= 2x + 6x log x + 3x
= 5x + 6x log x
= x (5 + 6 log x)

Question 6.
e^x sin 5x
Solution.
Let y = e^x sin 5x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (e^x sin 5x)
= sin 5x . \(\frac{d}{d x}\) (e^x) + e^x . cos 5x . \(\frac{d}{d x}\) (5x)
= e^x sin 5x + e^x cos 5x . 5
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) [e^x (sin 5x + 5 cos 5x)]
= (sin 5x + 5 cos 5x) . \(\frac{d}{d x}\) (e^x) + e^x . \(\frac{d}{d x}\) (sin 5x + 5 cos 5x)
= (sin 5x + 5 cos 5x) e^x + e^x [cos 5x . \(\frac{d}{d x}\) (5x) + 5 (- sin 5x) . \(\frac{d}{d x}\) (5x)]
= e^x (sin 5x + 5 cos 5x) + e^x (5 cos 5x – 25 sin 5x)
= e^x (10 cos 5x – 24 sin 5x)
= 2 e^x (5 cos 5x – 12 sin 5x)

Question 7.
e^6x cos 3x
Solution.
Let y = e^6x cos 3x
Differentiating w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (e^6x . cos 3x)

= cos 3x . e^6x \(\frac{d}{d x}\) (6x) + e^6x . (- sin 3x) . \(\frac{d}{d x}\) (3x)

= 6 e^6x cos 3x – 3 e^6x sin 3x …………….(i)
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (6 e^6x cos 3x – 3 e^6x sin 3x)

= 6 . \(\frac{d}{d x}\) (e^6x cos 3x) – 3 \(\frac{d}{d x}\) (e^6x sin 3x)

= 6 . [6e^6x cos 3x – 3 e^6x sin 3x] – 3 . [sin 3x . \(\frac{d}{d x}\) (e^6x) + e^6x \(\frac{d}{d x}\) (sin 3x)]

36e^6x cos 3x – 18 e^6x sin 3x – 3 [sin 3x . e^6x . 6 + e^6x . cos 3x . 3] = 36 e^6x cos 3x – 18 e^6x sin 3x – 18 e^6x sin 3x – 9 e^6x cos 3x
= 27 e^6x cos 3x – 36 e^6x sin 3x
= 9 e^6x (3 cos 3x – 4 sin 3x)

Question 8.
tan^-1 x
Solution.
Let y = tan^-1 x
Differentiating w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (tan^-1 x)

= \(\frac{1}{1+x^{2}}\)

Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (\(\frac{1}{1+x^{2}}\))

= \(\frac{d}{d x}\) (1 + x²)^-1

= (- 1) (1 + x²)^-2 . \(\frac{d}{d x}\) (1 +x²)

= \(\frac{-1}{\left(1+x^{2}\right)^{2}}\) 2x

= \(\frac{-2 x}{\left(1+x^{2}\right)^{2}}\)

Question 9.
log (log x)
Solution.
Lety = log (log x)
Differentiating w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [log (log x)]

= \(\frac{1}{\log x}\) . \(\frac{d}{d x}\) (log x)

= \(\frac{1}{x \log x}\) = (x log x)^-1

Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (x log x)^-1

= (- 1) . (x log x)^-2 . \(\frac{d}{d x}\) (x log x)

= \(\frac{-1}{(x \log x)^{2}} \cdot\left[\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x)\right]\)

= \(\frac{-1}{(x \log x)^{2}} \cdot\left[\log x \cdot 1+x \cdot \frac{1}{x}\right]=\frac{-(1+\log x)}{(x \log x)^{2}}\)

Question 10.
sin (log x)
Solution.
Let y = sin (log x)
Differentiating w.r.t x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [sin (log x)]

= cos log x . \(\frac{d}{d x}\) (log x)

= \(\frac{\cos (\log x)}{x}\)

Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) [latex]\frac{\cos (\log x)}{x}[/latex]

Question 11.
If y = 5 cos x – 3 sin x, prove that \(\frac{d^{2} y}{d x^{2}}\) = 0.
Solution.
Given, y = 5 cos x – 3sin x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (5 cos x) – \(\frac{d}{d x}\) (3 sin x)
= 5 \(\frac{d}{d x}\) (cos x) – 3 \(\frac{d}{d x}\) (sin x)
= 5(- sin x) – 3 cos x
= – (5 sin x + 3 cos x)
Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) [- (5 sin x + 3 cos x)]
= – [5 . \(\frac{d}{d x}\) (sin x) + 3 . \(\frac{d}{d x}\) (cos x)]
= – [5 cos x + 3 (- sin x)]
= – [5 cos x – 3 sin x] = – y
\(\frac{d^{2} y}{d x^{2}}\) + y = 0
∴ Hence proved.

Question 12.
If y = cos^-1 x, find \(\frac{d^{2} y}{d x^{2}}\) in terms of y alone.
Solution.
Given, y = cos^-1 x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (cos^-1 x)

= \(\frac{-1}{\sqrt{1-x^{2}}}=-\left(1-x^{2}\right)^{\frac{-1}{2}}\)

Again, differentiating w.r.t. x, we get

Question 13.
If y = 3 cos (log x) + 4 sin (log x), show that x²y₂ + xy₁ + y = 0.
Solution.
Given, y = 3 cos (log x) + 4 sin (log x), show that x²y₂ + xy₁ + y = 0.
Differentiating w.r.t. x, we get
\(\frac{d x}{d y}\) = y₁
= 3 . \(\frac{d}{d x}\) [cos (log x)] + 4 . \(\frac{d}{d x}\) [sin (log x)]

= 3 . [- sin(log x) . \(\frac{d}{d x}\) (log x)] + 4 . [cos (log x) . \(\frac{d}{d x}\) (log x)]

= – sin (log x) – 7 cos (log x) + 4 cos (log x) – 3 sin (log x) + 3 cos (log x) + 4 sin (log x)
= 0
Hence proved.

Question 14.
If y = A e^mx + B e^nx, show that \(\frac{d^{2} y}{d x^{2}}\) – (m + n) \(\frac{d y}{d x}\) + mny = 0.
Solution.
\(\frac{d y}{d x}\) = A . \(\frac{d}{d x}\) (e^mx) + B . \(\frac{d}{d x}\) (e^nx)

= A . e^mx . \(\frac{d}{d x}\) (mx) + B . e^nx . \(\frac{d}{d x}\) (nx)

= Ame^mx + Bne^nx

Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = \(\frac{d}{d x}\) (Ame^mx + Bne^nx)

= Am . \(\frac{d}{d x}\) (e^mx) + Bn . \(\frac{d}{d x}\) (e^nx)

Am . e^mx . \(\frac{d}{d x}\) (mx) + Bne^nx \(\frac{d}{d x}\) (nx)

= Am² e^mx + Bn² e^nx

∴ \(\frac{d^{2} y}{d x^{2}}\) – (m + n) \(\frac{d y}{d x}\) + mny

= Am² e^mx + Bn² e^nx – (m + n) . (Ame^mx + Bne^nx) + mn (Ae^mx + Be^nx)

= Am² e^mx + Bn² e^nx – Am² e^mx – Bmne^nx – Amne^nx – Bn² e^nx + Amne^mx + Bmne^nx = 0

Hence proved.

Question 15.
If y = 500 e^7x + 600 e^-7x, show that \(\frac{d^{2} y}{d x^{2}}\) = 49y.
Solution.
Given, y = 500 e^7x + 600 e^-7x
Differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}\) = 500 . \(\frac{d}{d x}\) (e^7x) + 600 . \(\frac{d}{d x}\) (e^-7x)

= 500 . e^7x \(\frac{d}{d x}\) (7x) + 600 . \(\frac{d}{d x}\) . (- 7x)

Again, differentiating w.r.t. x, we get

\(\frac{d^{2} y}{d x^{2}}\) = 3500 . \(\frac{d}{d x}\) (e^7x) – 4200 \(\frac{d}{d x}\) (e^-7x)

= 3500 . e^7x \(\frac{d}{d x}\) (7x) – 4200 . e^-7x \(\frac{d}{d x}\) (- 7x)

= 7 × 3500 . e^7x + 7 × 4200 . e^-7x
= 49 × 500 e^7x + 49 × 600 ^-7x
= 49 (500 e^7x + 600 e^-7x) = 49 y
Hence proved.

Question 16.
If e^y (x + 1) = 1, show that \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\).
Solution.
Given, e^y (x + 1) = 1

e^y = \(\frac{1}{x+1}\)

Taking logarithm on both sides, we get
y = log \(\frac{1}{x+1}\)
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}\) = (x + 1) \(\frac{d}{d x}\) \(\frac{1}{x+1}\)

= (x + 1) . \(\frac{-1}{(x+1)^{2}}=\frac{-1}{x+1}\)

Again, differentiating w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}=-\frac{d}{d x}\left(\frac{1}{x+1}\right)\) = \(-\left[\frac{-1}{(x+1)^{2}}\right]=\frac{1}{(x+1)^{2}}\)

⇒ \(\frac{d^{2} y}{d x^{2}}=\left(\frac{-1}{x+1}\right)^{2}\)

⇒ \(\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\)
Hence proved.

Question 17.
If y = (tan^-1 x)², show that (x² + 1)² y₂ + 2x (x² + 1) y₁ = 2.
Solution.
Given, y = (tan x)²
Differentiating w.r.t. x, we get
y₁ = 2 tan^-1 x \(\frac{d}{d x}\) (tan^-1 x)
⇒ y₁ = 2 tan^-1 . \(\frac{1}{1+x^{2}}\)

⇒ (1 + x²) y₁ = 2 tan^-1 x
Again, differentiating w.r.t. x, we get

(1 + x²) \(\frac{d y_{1}}{d x}\) + y₁ \(\frac{d}{d x}\) (1 + x²) = \(\frac{2}{1+x^{2}}\)

⇒ (1 + x²) y₂ + y₁ (0 + 2x) = \(\frac{2}{1+x^{2}}\)
[∵ \(\frac{d}{d x}\) (y₁) = y₁]

⇒ (1 + x²)² y₂ + 2x (1 + x²) y₁ = 2

Hence proved.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 6 Application of Derivatives Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1

Question 1.
Find the rate of change of the area of a circle with respect to its radius r when
(a) r = 3 cm
(b) r = 4 cm
Solution.
The area of a circle (A) with radius (r) is given by
A = πr²
Now, the rate of change of the area with respect to its radius is given by
\(\frac{d A}{d r}\) = \(\frac{d}{d r}\) (πr²) = 2πr

(a) When, r = 3 cm
\(\frac{d A}{d r}\) = 2π(3) = 6π
Hence, the area of the circle is changing at the rate of 6π cm²/s when its radius is 3 cm.

(b) When r = 4 cm,
\(\frac{d A}{d r}\) = 2π(4) = 8π
Hence, the area of the circle is changing at the rate of 8π cm²/s when its radius is 4 cm.

Question 2.
The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution.
Let x be the length of a side (edge), V be the volume and S be the surface area of the cube.
Then, V = x³ and S = 6x² where, x is a function of time t.
It is given that \(\frac{d V}{d t}\) = 8 cm³ /s
Then, y using the chain rule, we have

Thus, when x = 12 cm,
\(\frac{d S}{d t}\) = \(\frac{32}{12}\) cm²/s

= \(\frac{8}{3}\) cm²/s

Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \(\frac{8}{3}\) cm²/s.

Question 3.
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Solution.
The area of circle (A) with radius (r) is given by A = πr²
Now, the rate of change of area (A) with respect to time (t) is given by

\(\frac{d A}{d t}=\frac{d}{d t}\left(\pi r^{2}\right) \cdot \frac{d r}{d t}=2 \pi r \frac{d r}{d t}\)

It i s given that,
\(\frac{d r}{d t}\) = 3 cm/s
∴ \(\frac{d A}{d t}\) = 2πr (3) = 6πr
Thus, when r = 10 cm, dA
\(\frac{d A}{d t}\) = 6π(10) = 60π cm²/s dt
Hence, the rate at which the area of the circle is increasing when the radius is 10 cm, is 60π cm²/s.

Question 4.
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Solution.
Let x be the length of a side and V be the volume of the cube.
Then, V = x³
∴ \(\frac{d V}{d t}\) = 3x² . \(\frac{d x}{d t}\)
It is given that,
\(\frac{d x}{d t}\) = 3 cm/sec
⇒ \(\frac{d V}{d t}\) = 3x² (3) = 9x²
Thus, when x = 10 cm,
\(\frac{d V}{d t}\) = 9 (10)² = 900 cm³/s
Hence, the volume of the cube is increasing at the rate of 900 cm³/s when the edge is 10 cm long.

Question 5.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Solution.
The area of a circle (A) with radius (r) is given by A = πr².
Therefore, the rate of change of area (A) with respect to time (t) is given by

\(\frac{d A}{d t}=\frac{d}{d t}\left(\pi r^{2}\right)=\frac{d}{d r}\left(\pi r^{2}\right) \frac{d r}{d t}=2 \pi r \frac{d r}{d t}\)

It is given that \(\frac{d r}{d t}\) = 5 cm/s dt
Thus, when r = 8 cm,
\(\frac{d A}{d t}\) = 2π (8) (5) = 80 π dt
Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm²/s.

Question 6.
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
Solution.
The circumference of a circle (C) with radius (r) is given by C = 2nr
Therefore, the rate of change of circumference (C) with respect to time (t) is given by
\(\frac{d C}{d t}=\frac{d C}{d r} \cdot \frac{d r}{d t}\)

= \(\frac{d}{d r}(2 \pi r) \frac{d r}{d t}=2 \pi \cdot \frac{d r}{d r}\)

It is given that,
\(\frac{d r}{d t}\) = 0.7 cm/s
Hence, the rate of increase of the circumference is 2π (0.7) = 1.4π cm/s.

Question 7.
The length x of a rectangle is decreasing at the rate of 5 cm/min and the width y is increasing at the rate of 4 cm/min. When x = 8 cm and y = 6 cm, find the rates of change of
(a) the perimeter, and
(b) the area of the rectangle.
Solution.
Since, the length (x) is decreasing at the rate of 5 cm/min and the width (y) is increasing at the rate of 4 cm/min, we have
\(\frac{d x}{d t}\) = – 5 cm/min and

\(\frac{d y}{d t}\) = 4 cm/min

(a) The perimeter (P) of a rectangle is given by P = 2 (x + y)
∴ \(\frac{d P}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)\)
= 2 (- 5 + 4) = – 2 cm/min
Hence, the perimeter is decreasing at the rate of 2 cm/min.

(b) The area (A) of a rectangle is given by A = x × y
∴ \(\frac{d A}{d t}=\frac{d x}{d t} \cdot y+x \cdot \frac{d y}{d t}\)
= – 5y + 4x
When x = 8 cm and y = 6 cm,
\(\frac{d A}{d t}\) = – 5 × 6 +4 × 8
= 2 cm²/min
Hence, the area of the rectangle is increasing at the rate of 2 cm²/min.

Question 8.
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic cm of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Solution.
The volume of a sphere (V) with radius (r) is given by

V = \(\frac{4}{3}\) πr³
∴ Rate of change of volume (V) with respect to time (r) is given by
\(\frac{d V}{d t}=\frac{d V}{d r} \cdot \frac{d r}{d t}\)

= \(\frac{d}{d r}\left(\frac{4}{3} \pi r^{3}\right) \cdot \frac{d r}{d t}=4 \pi r^{2} \cdot \frac{d r}{d t}\)

It is given that \(\frac{d V}{d t}{/latex] = 900 cm³/s
∴ 900 = 4πr² . [latex]\frac{d r}{d t}\)

⇒ \(\frac{d r}{d t}\) = \(\frac{900}{4 \pi r^{2}}=\frac{225}{\pi r^{2}}\)
Therefore, when radius = 15 cm,
\(\frac{d r}{d t}=\frac{225}{\pi(15)^{2}}=\frac{1}{\pi}\).

Question 9.
A balloon, which always remains spherical, has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Solution.
The volume of a sphere (V) with radius (r) is gwen by V = \(\frac{4}{3}\) πr³.
Rate of change of volume (V) with respect to its radius (r) is given by
\(\frac{d V}{d r}=\frac{d}{d r}\left(\frac{4}{3} \pi r^{3}\right)\)

= \(\frac{4}{3}\) π (3r²) = 4πr²
Therefore, when radius = 10 cm,
\(\frac{d V}{d r}{/latex] = 4π(10)² = 400π
Hence, the volume of the balloon is increasing at the rate of 400 π cm³/s.
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm, is [latex]\frac{1}{\pi}\) cm/s.

Question 10.
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
Solution.
Let y m be the height of the wall at which the ladder touches.
Also, let the foot of the ladder be x away from the wall.
Then, by Pythagoras theorem, we have
x² + y² = 25 [∵ Length of the ladder = 5 m]
⇒ y = \(\sqrt{25-x^{2}}\)
Then, the rate of change of height (y) with respect to time (t) is given by

\(\frac{d y}{d t}=\frac{-x}{\sqrt{25-x^{2}}} \cdot \frac{d x}{d t}\)

It is given that \(\frac{d x}{d t}\) = 2 cm/s dt

∴ \(\frac{d y}{d t}=\frac{-2 x}{\sqrt{25-x^{2}}}\)

Now, when x = 4 m, we have

∴ \(\frac{d y}{d t}=\frac{-2 \times 4}{\sqrt{25-4^{2}}}=-\frac{8}{3}\)

Hence, the height of the ladder on the wall is decreasing at the rate of \(\frac{8}{3}\) cm/s.

Question 11.
A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y – coordinate is changing 8 times as fast as the x – coordinate.
Solution.
The equation of the curve is given as 6y = x³ + 2

The rate of change of the position of the particle with respect to time (t) is given by

\(6 \frac{d y}{d t}=3 x^{2} \frac{d x}{d t}+0\)

⇒ \(2 \frac{d y}{d t}=x^{2} \frac{d x}{d t}\)

When the y – coordinate of the particle changes 8 times as fast as the x-coordinate i.e., \(\left(\frac{d y}{d t}=8 \frac{d x}{d t}\right)\), we have

Hence, the points required on the curve are (4, 11) and (- 4, \(-\frac{31}{3}\)).

Question 12.
The radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/s; At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution.
The air bubble is in the shape of a sphere.
Now, the volume of an air bubble (V) with radius (r) is given by

V = \(\frac{4}{3}\) πr³.

The rate of change of volume (V) with respect to time (t) is given by
\(\frac{d V}{d t}=\frac{4}{3} \pi \frac{d}{d r}\left(r^{3}\right) \cdot \frac{d r}{d t}\)

= \(\frac{4}{3} \pi\left(3 r^{2}\right) \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t}\)

It is given that \(\frac{d r}{d t}\) = \(\frac{1}{2}\) cm/s
Therefore, when r = 1 cm,
\(\frac{d V}{d t}\) = 4π(1)² (\(\frac{1}{2}\))
= 2π cm³/s
Hence, the rate at which the volume of the bubble increases, is 2π cm³/s.

Question 13.
A balloon, which always remains spherical, has a variable diameter \(\frac{3}{2}\) (2x + 1). Find the rate of change of its volume with respect to x.
Solution.
The volume of a sphere (V) with radius (r) is given by
V = \(\frac{4}{3}\) πr³
It is given that, Diameter = \(\frac{3}{2}\) (2x + 1)
⇒ r = \(\frac{3}{4}\) (2x + 1)
∴ V = \(\frac{4}{3} \pi\left(\frac{3}{4}\right)^{3}(2 x+1)^{3}\)

= \(\frac{9}{16} \pi(2 x+1)^{3}\)

Hence, the rate of change of volume with respect to x is

\(\frac{d V}{d x}=\frac{9}{16} \pi \frac{d}{d x}(2 x+1)\)

= \(\frac{9}{16} \pi \times 3(2 x+1)^{2} \times 2\)

= \(\frac{27}{8} \) π (2x + 1)²

Question 14.
Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Solution.
The volume of a cone (V) with radius (r) and height (h) is given by V = \(\frac{1}{3}\) πr²h
It is given that, h = \(\frac{1}{6}\) r
⇒ r = 6h
⇒ V = \(\frac{1}{3}\) π(6h)² h
= 12 πh³
The rate of change of volume with respect to time (t) is given by
\(\frac{d V}{d t}\) = 12π \(\frac{d}{d h}\) (h³) . \(\frac{d h}{d t}\)

= 12π (3h²) \(\frac{d h}{d t}\)

= 36πh² \(\frac{d h}{d t}\)

It is also given that \(\frac{d V}{d t}\) = 12 cm³/ s
Therefore, when h = 4 cm, we have
12 = 36π(4)² \(\frac{d h}{d t}\)
Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of\(\frac{1}{48 \pi}\) cm/s.

Question 15.
The total cost C(x) in rupees associated with the production of x units of an item is given by
C(x) = 0.007 x³ – 0.003 x² + 15x + 4000.
Find the marginal cost when 17 units are produced.
Solution.
Marginal cost is the rate of change of total cost with respect to output.

Marginal cost (MC) = \(\frac{d C}{d x}\)
= 0.007 (3x²) – 0.003 (2x) + 15
= 0.021 x² – 0.006x + 15
When x = 17, then MC = 0.021 (17²) – 0.006 (17) +15
= 0.021 (289) – 0.006 (17) +15
= 6.069 – 0.102 + 15 = 20.967
Hence, when 17 units are produced, the marginal cost is ₹ 20.967.

Question 16.
The total revenue in rupees received from the sale of x units of a product is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.
Solution.
Marginal revenue is the rate of change of total revenue with respect to the number of units sold.
Marginal Revenue (MR) = \(\frac{d R}{d x}\)
= 13(2x) + 26
= 26x + 26

When x = 7, then MR = 26(7) + 26
= 182 + 26 = 208
Hence, the required marginal revenue is ₹ 208.

Direction (17 – 18):
Choose the correct answer.

Question 17.
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π
(B) 12π
(C) 8π
(D) 11π
Solution.
The area of a circle (A) with radius (r) is given by
A = πr²
Therefore, the rate of change of the area with respect to its radius r is
\(\frac{d A}{d r}\) = \(\frac{d}{d r}\) (πr²) = 2πr
∴ When r = 6 cm
⇒ \(\frac{d A}{d r}\) = 2π × 6 = 12π cm²/s
Hence, the required rate of change of the area of a circle is 12π cm²/s.
The correct answer is (B).

Question 18.
The total revenue in rupees received from the sale of x units of a product is given by R(x) = 3x² + 36x + 5. The marginal revenue, when x = 15, is
(A) 116
(B) 96
(C) 90
(D) 126
Solution.
Marginal revenue is the rate of change of total revenue with respect to the number of units sold.
∴ Marginal Revenue (MR) = \(\frac{d R}{d x}\)
= 3(2x) + 36 = 6x + 36
∴ When x = 15, then MR = 6(15) + 36 = 90 + 36 = 126
Hence, the required marginal revenue is ₹ 126.
The correct answer is (D).

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 5 Continuity and Differentiability Ex 5.8 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8

Question 1.
Verify Rolle’s theorem for the function f(x) = x² + 2x – 8, x ∈ [- 4, 2].
Solution
The given function, f(x) = x² + 2x – 8, being a polynomial function, is continuous in [- 4, 2] and is differentiable in (- 4, 2)
f(- 4) = (- 4)²+ 2 × (- 4) – 8
= 16 – 8 – 8 = 0

f(2) = (2)² + 2 × 2 – 8
= 4 + 4 – 8 = 0
∴ f(- 4) = f(2) = 0
⇒ The value of f(x) at – 4 and 2 coincides.
Rolle’s theorem states that there is a point c ∈ (- 4, 2) such that f'(c) = 0
f(x) = x² +2x – 8
⇒ f(x) = 2x + 2
⇒ f(c) = 0
⇒ 2c + 2 = 0
⇒ c = – 1, where c = – 1 ∈ (- 4, 2)
Hence, Rolle’s theorem is verified for the given function.

Question 2.
Examine if Rolle’s theorem is applicable to any of the following functions. Can you say something about the converse of Rolle’s theorem from these examples?
(i) f(x) = [x] for x ∈ [5, 9]
(ii) f(x) = [x] for x ∈ [- 2, 2]
(iii) f(x) = x² – 1 for x ∈ [1, 2]
Solution.
(i) In the interval [5, 9], f(x) = [x] is neither continuous nor derivable at x = 6, 7, 8.
Hence, Rolle’s theorem is not applicable.

(ii) f(x) = [x] is not continuous and derivable at – 1, 0, 1.
Hence, Rolle’s theorem is not applicable.

(iii) f(x) = (x² – 1), f(1) = 1 – 1 = 0, f(2) = 2² – 1 = 3
f(a) ≠ f(b)
Though it is continuous and derivable in the interval [1, 2].
Rolle’s theorem is not applicable.

In case of converse if f(c) = 0, c ∈ [a, b], then conditions of Rolle’s theorem are not true.

(i) f(x) = [x] is the greatest integer less than or equal to x.
∴ f'(x) = 0, But f is neither continuous nor differentiable in the interval [5, 9]
(ii) Here also, though f'(x) = 0, but f is neither continuous nor differentiable in the interval [- 2, 2]
(iii) f(x) = x² – 1, f'(x) = 2x.
Here f'(x) is not zero in the [1, 2]
So, f(2) ≠ f'(2).

Question 3.
If f: [- 5, 5] → R is a differentiable function and if f'(x) does not vanish anywhere, then prove that f(- 5) ≠ f(5).
Solution.
It is given that f : [- 5, 5] → R is a differentiable function.
Since, every differentiable function is a continuous function, therefore we get
(a) f is continuous on [- 5, 5].
(b) f is differentiable on (- 5, 5).
Therefore, by the Mean Value Theorem, there exists c ∈ (- 5, 5) such that
f'(c) = \(\frac{f(5)-f(-5)}{5-(-5)}\)
⇒ 10f'(c) = f(5) – f(- 5)

It is also given that f'(x) does not vanish anywhere.
∴ f'(c) ≠ 0
⇒ 10 f'(c) ≠ 0
⇒ f(5) – f(- 5) ≠ 0
⇒ f(5) ≠ f(- 5)
Hence proved.

Question 4.
Verify Mean Value Theorem, if f(x) = x² – 4x – 3 in the interval [o, 6], where a = 1 and 6=4
Solution.
The given function is f(x) = x² – 4x – 3, x e [1, 4] which is a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x – 4.
f(1) = 1² – 4 × 1 – 3 = – 6,
f(4) = 4² – 4 × 4 – 3 = – 3

∴ \(\frac{f(b)-f(a)}{b-a}=\frac{f(4)-f(1)}{4-1}=\frac{-3-(-6)}{3}=\frac{3}{3}\) = 1

Mean Value Theorem states that there is a point c ∈ (1, 4) such that f'(c) = 1.
∴ f'(c) = 1
⇒ 2c – 4 = 1
⇒ c = \(\frac{5}{2}\)
where c = \(\frac{5}{2}\) ∈ (1, 4)
Hence, Mean Value Theorem is verified for the given function.

Question 5.
Verify mean value Theorem, if f(x) = x³ – 5x² – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f'(c) = 0.
Solution.
The given function is f(x) = x³ – 5x² – 3x, x ∈(1, 3) which is a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3x² – 10x – 3.
f(1) = 1³ – 5 × 1² – 3 × 1 = – 7,
f(3) = 3³ – 5 × 3² – 3 × 3 = – 27 .

∴ \(\frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(1)}{3-1}=\frac{-27-(-7)}{3-1}\) = – 10

Mean Value Theorem states that there exists a point c e (1, 3) such that f'(c) = – 10.
∴ f'(c) = – 10
⇒ 3c² – 10c – 3 = – 10
⇒ 3c² – 10c+ 7 = 0
⇒ 3c² – 3c – 7c + 7 = 0
⇒ 3c (c – 1) – 7(c – 1) = 0
⇒ (c – 1) (3c – 7) = 0
⇒ c = 1, \(\frac{7}{3}\) where c = \(\frac{7}{3}\) ∈ (1, 3)

Hence, Mean Value Theorem is verified for the given function and c = \(\frac{7}{3}\) ∈ (1, 3) is the only point for which f'(c) = 0.

Question 6.
Examine the applicability of Mean Value Theorem for the given functions.
(i) f(x) = [x] for x ∈ [5, 9]
(ii) f(x) = [x] for x ∈ [- 2, 2]
(iii) f(x) = x² – 1 for x ∈ [1, 2]
Solution.
(i) f(x) = [x] for x ∈ [5, 9]
f(x) = [x] in the interval [5,9] is neither continuous, nor differentiable.
∴ Mean Value Theorem is not applicable.

(ii) f(x) = [x], for x ∈ [- 2, 2]
Again f(x) = [x] in the interval [- 2,2] is neither continuous, nor differentiable.
Hence, Mean Value Theorem is not applicable.

(iii) f(x) = x² – 1 for x ∈ [1, 2]
It is a polynomial. Therefore it is continuous in the interval [1, 2] and differentiable in the interval (1,2)
∴ f(x) = 2x,
f(1) = 1 – 1 = 0,
f(2) = 4 – 1 = 3
∴ f'(c) = 2c
By Mean Value Theorem f'(c) = \(\frac{f(b)-f(a)}{b-a}\)

2c = \(\frac{3-0}{2-1}=\frac{3}{1}\)

⇒ c = \(\frac{3}{2}\), which belongs to (1, 3)
Hence, mean value theorem is applicable.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 5 Continuity and Differentiability Ex 5.6 Textook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6

Direction (1 – 10) : If x and y are connected parametrically by the equations, given in the following questions without eliminating the parameter, find

Question 1.
x = 2at², y = at⁴
Solution.
The given equations are x = 2at² and y = at⁴
On differentiating w.r.t. t, we get
\(\frac{d x}{d t}\) = \(\frac{d}{d t}\) (2at²)
= 2a . \(\frac{d}{d t}\) (t²)
= 2a . 2t = 4at

and \(\frac{d y}{d t}\) = \(\frac{d}{d t}\) (at⁴)
= a . \(\frac{d}{d t}\) (t⁴)
= a . 4 . t³
= 4at³

∴ \(\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{4 a t^{3}}{4 a t}\) = t²

(∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\))

Question 2.
x = a cos θ, y = b cos θ
Solution.
The given equations are x = a cos θ and y = b cos θ
On differentiating w.r.t. θ, we get
\(\frac{d x}{d θ}\) = \(\frac{d}{d θ}\) (a cos θ)
= a (- sin θ) = – a sin θ
and \(\frac{d y}{d θ}\) = \(\frac{d}{d θ}\) (cos θ)
= b (- sin θ) = – sin θ
∴ \(\frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{-b \sin \theta}{-a \sin \theta}=\frac{b}{a}\)

Question 3.
x = sin t, y = cos 2t
Solution.
The given equations are x = sin t and y = cos 2t
On differentiating w.r.t. t, we get
\(\frac{d x}{d t}\) = \(\frac{d}{d t}\) (sin t) = cos t
and \(\frac{d y}{d t}\) = \(\frac{d}{d t}\) (cos 2t)
= – sin (2t) . \(\frac{d}{d t}\) (2t) = – 2 sin 2t
∴ \(\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}\)

= \(\frac{-2 \sin 2 t}{\cos t}=\frac{-2 \cdot 2 \sin t \cos t}{\cos t}\)
= – 4 sin t.

Question 4.
x = 4t, y = \(\frac{4}{t}\)
Sol.
The given equations are x = 4t and y = \(\frac{4}{t}\)
On differentiating w.r.t. t, we get

Question 5.
x = cos θ – cos 2θ, y = sin θ – sin 2θ
Solution.
The given equations are x = cos θ – cos 2θ, y = sin θ – sin 2θ
On differentiating w.r.t θ, we get
\(\frac{d x}{d θ}\) = \(\frac{d}{d θ}\) (cos θ – cos 2θ)

= \(\frac{d}{d θ}\) (cos θ) – \(\frac{d}{d θ}\) (cos 2θ)
= – sin θ – (- sin 2θ)
= 2 sin 2θ – sin θ

and \(\frac{d y}{d θ}\) = \(\frac{d}{d θ}\) (sin θ – sin 2θ)

= \(\frac{d}{d θ}\) (sin θ) – \(\frac{d}{d θ}\) (sin 2θ)
= cos θ – 2 cos 2θ
∴ \(\frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}\)

Question 6.
x = a (θ – sin θ), and y = a(1 + cos θ)
Solution.
The given equations are x = a (θ – sin θ), and y = a(1 + cos θ)
On differentiating w.r.t x, we get
\(\frac{d x}{d θ}\) = a [\(\frac{d}{d θ}\) (θ) – \(\frac{d}{d θ}\) (sin θ)]
= a (1 – cos θ)

and \(\frac{d y}{d θ}\) = a[\(\frac{d}{d θ}\) (1) + \(\frac{d}{d θ}\) (cos \(\frac{d}{d θ}\))]
= a[0 + (- sin θ)]
– a sin θ
∴ \(\frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{-a \sin \theta}{a(1-\cos \theta)}\)

= \(\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}=\frac{-\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}=-\cot \frac{\theta}{2}\)

Question 7.
x = \(\frac{\sin ^{3} t}{\sqrt{\cos ^{2} t}}\), y = \(\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}\)
Solution.
The given equations are x = \(\frac{\sin ^{3} t}{\sqrt{\cos ^{2} t}}\), y = \(\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}\)
On differentiating w.r.t t, we get

Question 8.
x = a(cos t + log tan \(\frac{t}{2}\)), y = a sin t
Solution.
The given equations are x = a (cos t + log tan \(\frac{t}{2}\)) and y = a sin t
On differentiating w.r.t t, we get

Question 9.
x = a sec θ, y = b tan θ
Solution.
The given equations are x = a sec θ and y = b tan θ
On differentiating w.r.t to θ, we get
\(\frac{d x}{d θ}\) = a . \(\frac{d}{d θ}\) (sec θ)
= a sec θ tan θ

and \(\frac{d y}{d θ}\) = b . \(\frac{d}{d θ}\) (tan θ)
= b sec² θ

∴ \(\frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{b \sec ^{2} \theta}{a \sec \theta \tan \theta}\)

= \(\frac{b}{a}\) sec θ cot θ

= \(\frac{b \cos \theta}{a \cos \theta \sin \theta}=\frac{b}{a} \times \frac{1}{\sin \theta}=\frac{b}{a} \ {cosec} \theta\)

Question 10.
x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)
Solution.
The given equations are x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)
On differentiating w.r.t to θ, we get

\(\frac{d x}{d θ}\) = a [\(\frac{d}{d θ}\) (cos θ ) + \(\frac{d x}{d θ}\) (θ sin θ)]

= a [- sin θ + θ \(\frac{d}{d θ}\) (sin θ) + sin θ \(\frac{d}{d θ}\) (θ)]

= a [- sin θ + θ cos θ + sin θ] = a θ cos θ

and \(\frac{d y}{d θ}\) = a [\(\frac{d}{d θ}\) (sin θ) – \(\frac{d}{d θ}\) (θ cos θ)]

= a [cos θ – {θ latex]\frac{d}{d θ}[/latex] (cos θ) + cos θ . latex]\frac{d}{d θ}[/latex] (θ)}]

= a [cos θ + θ sin θ – cos θ] = aθ sin θ

∴ \(\frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{a \theta \sin \theta}{a \theta \cos \theta}\) = tan θ.

Question 11.
If x = \(\sqrt{a^{\sin ^{-2} t}}\), y = \(\sqrt{a^{\cos ^{-1} t}}\), show that \(\frac{d y}{d x}=-\frac{y}{x}\).
Solution.
The given equations are x = \(\sqrt{a^{\sin ^{-2} t}}\) and y = \(\sqrt{a^{\cos ^{-1} t}}\)

⇒ x = \(\sqrt{a^{\sin ^{-2} t}}\) and y = \(\sqrt{a^{\cos ^{-1} t}}\)

Consider x = \(a^{\frac{1}{2} \sin ^{-1} t}\)
Taking logarithm on bothsides, we get
log x = \(\frac{1}{2}\) sin^-1 t log a

⇒ \(\frac{1}{x} \cdot \frac{d x}{d t}=\frac{1}{2} \log a \cdot \frac{d}{d t}\left(\sin ^{-1} t\right)\)

⇒ \(\frac{d x}{d t}=\frac{x}{2} \log a \cdot \frac{1}{\sqrt{1-t^{2}}}\)

⇒ \(\frac{d x}{d t}=\frac{x \log a}{2 \sqrt{1-t^{2}}}\)

Then, consider y = \(a^{\frac{1}{2} \cos ^{-1} t}\)

Taking logarithm on bothsides, we get

Hence proved.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

Find the value of the following :

Question 1.
cos^-1 (cos \(\frac{13 \pi}{6}\)).
Solution.
We know that cos^-1(cos x) = x if x ∈ [0, π], which is the principal value of cos^-1 x.
Here, \(\frac{13 \pi}{6}\) ∉ [0, π].
Now, cos^-1 (cos \(\frac{13 \pi}{6}\)) can be written as
cos^-1 (cos \(\frac{13 \pi}{6}\)) = cos^-1 [cos (2π + \(\frac{pi}{6}\))]
= cos^-1 [cos (\(\frac{pi}{6}\))],
where \(\frac{pi}{6}\) ∈ [0, π]
[∵ cos(2π + x) = cos x]
∴ cos^-1 (cos \(\frac{13 \pi}{6}\)) = cos^-1 [cos (\(\frac{pi}{6}\))]
= \(\frac{pi}{6}\).

Question 2.
tan^-1 (tan \(\frac{7 \pi}{6}\))
Solution.
We know that tan^-1(tan x) = x if x ∈ (\(-\frac{\pi}{2}, \frac{\pi}{2}\)) which is the principal value of cos^-1 x.
Here, \(\frac{7 \pi}{6}\) ∉ (\(-\frac{\pi}{2}, \frac{\pi}{2}\))
Now, tan^-1 (tan \(\frac{7 \pi}{6}\)) can be written as
tan^-1 (tan \(\frac{7 \pi}{6}\)) = tan^-1 (tan (π + \(\frac{\pi}{6}\)))
= tan^-1 [tan (\(\frac{\pi}{6}\))]
where \(\frac{pi}{6}\) ∈ (\(-\frac{\pi}{2}, \frac{\pi}{2}\))
[∵ tan(π + x) = tan x]
∴ tan^-1 (tan \(\frac{7 \pi}{6}\)) = tan^-1 [tan (\(\frac{\pi}{6}\))]
= \(\frac{pi}{6}\)

Prove that

Question 3.
2 sin^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{24}{7}\).
Solution.
Let sin^-1 \(\frac{3}{5}\) = x.
Then, sin x = \(\frac{3}{5}\)
⇒ cos x = \(\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
∴ tan x = \(\frac{3 / 5}{4 / 5}=\frac{3}{4}\)
∴ x = tan^-1 \(\frac{3}{4}\)
⇒ sin^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{3}{4}\)
Now, we have
L.H.S = 2 sin^-1 \(\frac{3}{5}\) = 2 tan^-1 \(\frac{3}{4}\)
= tan^-1 \(\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)\)

[∵ 2 tan^-1 x = tan^-1 \(\frac{2 x}{1-x^{2}}\)]

= tan^-1 \(\left(\frac{\frac{3}{2}}{\frac{16-9}{16}}\right)\)

= tan^-1 \(\left(\frac{3}{2} \times \frac{16}{7}\right)\)

= tan^-1 \(\frac{24}{7}\)

= R.H.S.
Hence proved.

Question 4.
sin^-1 \(\frac{8}{17}\) + sin^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{77}{36}\)
solution.
Let sin^-1 \(\frac{8}{17}\) = x.
Then, sin x = \(\frac{8}{17}\)
⇒ cos x = \(\sqrt{1-\left(\frac{8}{17}\right)^{2}}=\sqrt{\frac{225}{289}}=\frac{15}{17}\)
∴ tan x = \(\frac{8 / 17}{15 / 17}=\frac{8}{15}\)
⇒ x = tan^-1 \(\frac{8}{15}\)
∴ sin^-1 \(\frac{8}{17}\) = tan^-1 \(\frac{8}{15}\) …………..(i)

Now, let sin^-1 \(\frac{3}{5}\) = y.
Then, sin y = \(\frac{3}{5}\)
⇒ cos y = \(\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
∴ tan y = \(\frac{3 / 5}{4 / 5}=\frac{3}{4}\)
⇒ y = tan^-1 \(\frac{3}{4}\)
∴ sin^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{3}{4}\) …………..(ii)

Now, we have
L.H.S = sin^-1 \(\frac{8}{17}\) + sin^-1 \(\frac{3}{5}\)
[Using Eqs. (i) and (ii)]
= tan^-1 \(\frac{8}{15}\) + tan^-1 \(\frac{3}{4}\)
= tan^-1 \(\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}\)
= tan^-1 \(\left(\frac{32+45}{60-24}\right)\)
[tan^-1 x + tan^-1 y = tan^-1 \(\left(\frac{x+y}{1-x y}\right)\)]
= tan^-1 \(\frac{77}{36}\)
= R.H.S
Hence proved.

Question 5.
cos^-1 \(\frac{4}{5}\) + cos^-1 \(\frac{12}{13}\) = cos^-1 \(\frac{33}{65}\)
Solution.
Let cos^-1 \(\frac{4}{5}\) = x
Then, cos x = \(\frac{4}{5}\)
⇒ sin x = \(\sqrt{1-\left(\frac{4}{5}\right)^{2}}=\sqrt{\frac{9}{25}}=\frac{3}{5}\)
∴ tan x = \(\frac{3 / 5}{4 / 5}=\frac{3}{4}\) …………(i)

⇒ x = tan^-1 \(\frac{3}{4}\)
Now, let cos^-1 \(\frac{12}{13}\) = y.
Then cos y = \(\frac{12}{13}\)
⇒ sin y = \(\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\frac{5}{13}\)
∴ tan y = \(\frac{5 / 13}{12 / 13}=\frac{5}{12}\)
⇒ y = tan^-1 \(\frac{5}{12}\)
∴ cos^-1 \(\frac{12}{13}\) = tan^-1 \(\frac{5}{12}\) ……………(ii)

Let cos^-1 \(\frac{33}{65}\) = z.
Then, cos z = \(\frac{33}{65}\)
⇒ sin z = \(\sqrt{1-\left(\frac{33}{65}\right)^{2}}=\sqrt{\frac{3136}{4225}}=\frac{56}{65}\)
∴ tan z = \(\frac{56 / 65}{33 / 65}=\frac{56}{33}\)
⇒ z = tan^-1 \(\frac{56}{33}\)
∴ cos^-1 \(\frac{33}{65}\) = tan^-1 \(\frac{56}{33}\) …………..(iii)

Now, we have
L.H.S = cos^-1 \(\frac{4}{5}\) + cos^-1 \(\frac{12}{13}\)
= \(\frac{3 / 5}{4 / 5}=\frac{3}{4}\) + tan^-1 \(\frac{5}{12}\)
[∵ Usin Eqs. (i) and (ii)]
= tan^-1 \(\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}\)
[∵ tan^-1 x + tan^-1 y = tan^-1 \(\left(\frac{x+y}{1-x y}\right)\)]
= tan^-1 \(\frac{36+20}{48-15}\)
= tan^-1 \(\frac{56}{33}\)
= cos^-1 \(\frac{33}{65}\) = R.H.S
Hence proved.

Question 6.
cos^-1 \(\frac{12}{13}\) + sin^-1 \(\frac{3}{5}\) = sin^-1 \(\frac{56}{65}\)
Solution.
Let sin^-1 \(\frac{3}{5}\) = x.
Then, sin x = \(\frac{3}{5}\)
⇒ cos x= \(\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
∴ tan x = \(\frac{3 / 5}{4 / 5}=\frac{3}{4}\)
⇒ x = tan^-1 \(\frac{3}{4}\)
∴ sin^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{3}{4}\) …………(i)

Now, let cos^-1 \(\frac{12}{13}\) = y.
Then, cos y = \(\frac{12}{13}\)
⇒ sin y = \(\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\frac{5}{13}\)
∴ tan y = \(\frac{5 / 13}{12 / 13}=\frac{5}{12}\)
⇒ y = tan^-1 \(\frac{5}{12}\)
∴ cos^-1 \(\frac{12}{13}\) = tan^-1 \(\frac{5}{12}\) ………………(ii)

Let sin^-1 \(\frac{56}{65}\) = z.
Then, sin z = \(\frac{56}{65}\)
⇒ cos z = \(\sqrt{1-\left(\frac{56}{65}\right)^{2}}=\sqrt{\frac{1089}{4225}}=\frac{33}{65}\)
∴ tan z = \(\frac{56 / 65}{33 / 65}=\frac{56}{33}\)
⇒ z = tan^-1 \(\frac{56}{33}\)
∴ sin^-1 \(\frac{56}{65}\) = tan^-1 \(\frac{56}{33}\) …………….(iii)

Now, we have
L.H.S. = cos^-1 \(\frac{12}{13}\) + sin^-1 \(\frac{3}{5}\)
= tan^-1 \(\frac{3}{4}\) + tan^-1 \(\frac{5}{12}\)
[Using Eqs. (i) and (ii)]
= tan^-1 \(\frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12} \cdot \frac{3}{4}}\)

[∵ tan^-1 x + tan^-1 y = tan^-1 \(\left(\frac{x+y}{1-x y}\right)\)]
= tan^-1 \(\frac{20+36}{48-15}\)
= tan^-1 \(\frac{56}{33}\)
= sin^-1 \(\frac{56}{65}\)

Question 7.
tan^-1 \(\frac{63}{16}\) = sin^-1 \(\frac{5}{13}\) + cos ^-1 \(\frac{3}{5}\)
Solution.
Let sin^-1 \(\frac{5}{13}\) = x
Then, sin x = \(\frac{5}{13}\)
⇒ cos x = \(\sqrt{1-\left(\frac{5}{13}\right)^{2}}=\sqrt{\frac{144}{169}}=\frac{12}{13}\)
∴ tan x = \(\frac{5 / 13}{12 / 13}=\frac{5}{12}\)
⇒ x = tan^-1 \(\frac{5}{12}\)
∴ sin^-1 \(\frac{5}{13}\) = tan^-1 \(\frac{5}{12}\) ………….(i)

Let cos^-1 \(\frac{3}{5}\) = y.
Then, cos y = \(\frac{3}{5}\)
⇒ sin y = \(\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
∴ tan y = \(\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
⇒ y = tan^-1 \(\frac{4}{5}\)
∴ cos^-1 \(\frac{3}{5}\) = tan^-1 \(\frac{4}{5}\) ………………(ii)

Using Eqs. (i) and (ii), we have
R.H.S = sin^-1 \(\frac{5}{13}\) + cos^-1 \(\frac{3}{5}\)
= tan^-1 \(\frac{5}{12}\) + tan^-1 \(\frac{4}{5}\)
[∵ tan^-1 x + tan^-1 y = tan^-1 \(\left(\frac{x+y}{1-x y}\right)\)]
= tan^-1 \(\left(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}\right)\)

= tan^-1 \(\left(\frac{15+48}{36-20}\right)\)

= tan^-1 \(\frac{63}{16}\) = L.H.S

Question 8.
tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{3}\) + tan^-1 \(\frac{1}{8}\) = \(\frac{\pi}{4}\)
Solution.
tan^-1 \(\frac{1}{5}\) + tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{3}\) + tan^-1 \(\frac{1}{8}\)

Question 9.
tan^-1 √x = \(\frac{1}{2}\) cos^-1 \(\left(\frac{1-x}{1+x}\right)\), x ∈ [0, 1]
Solution.
Let x = tan² θ.
Then, √x = tan θ
⇒ θ = tan^-1 √x
∴ \(\frac{1-x}{1+x}=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\) = cos 2θ
[∵ cos 2θ = \(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\)]
Now, we have
R.H.S = \(\frac{1}{2}\) cos^-1 \(\left(\frac{1-x}{1+x}\right)\)
= \(\frac{1}{2}\) cos^-1 2θ = θ
= tan^-1 √x = L.H.S
Hence Proved.

Question 10.
cot^-1 \(\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}\), x ∈ (0, \(\frac{\pi}{4}\))
Solution.
Consider, \(\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)\)

= \(\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x})^{2}-(\sqrt{1-\sin x})^{2}}\) (By rationalising)

= \(\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1+\sin x)(1-\sin x)}}{1+\sin x-1+\sin x}\)

= \(\frac{2\left(1+\sqrt{\left.1-\sin ^{2} x\right)}\right.}{2 \sin x}=\frac{1+\cos x}{\sin x}=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\)

= cot \(\frac{x}{2}\)

L.H.S = cot^-1 \(\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)\)

= cot^-1 (cot \(\frac{x}{2}\))
= \(\frac{x}{2}\) = R.H.S
Hence proved.

Question 11.
tan^-1 \(\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\) = \(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\), \(-\frac{1}{\sqrt{2}}\) ≤ x ≤ 1.
[Hint: put x = cos 2θ]
Solution.
Put x = cos 2θ, so that θ = \(\frac{1}{2}\) cos^-1 x.
Then, we have

Question 12.
\(\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\)
Solution.
L.H.S = \(\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}\)

= \(\frac{9}{4}\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{3}\right)\)

= \(=\frac{9}{4}\left(\cos ^{-1} \frac{1}{3}\right)\) [∵ sin^-1 x + cos^-1 x = \(\frac{\pi}{2}\)]

= \(\frac{9}{4}\left(\sin ^{-1} \sqrt{1-\left(\frac{1}{3}\right)^{2}}\right)\)
[∵ cos^-1 x = sin^-1 \(\sqrt{1-x^{2}}\)]

= \(\frac{9}{4} \sin ^{-1} \sqrt{\frac{8}{9}}\)

= \(\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\)

= R.H.S
Hence proved.

Direction (13 – 17): Solve the following equations.

Question 13.
2 tan^-1 (cos x) = tan^-1 (2cosec x)
Solution.
We have, 2 tan^-1 (cos x) = tan^-1 (2 cosec x)
⇒ tan^-1 \(\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)\) = tan^-1 (2 cosec x)

[∵ 2 tan^-1 x = tan^-1 \(\frac{(2 x)}{1-x^{2}}\)]

⇒ \(\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)\) = 2 cosec x

⇒ \(\frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x}\)

⇒cos x = sin x

⇒ tan x = 1 = tan \(\frac{\pi}{4}\).

Question 14.
tan^-1 \(\frac{1-x}{1+x}\) = \(\frac{1}{2}\) tan^-1 x, (x > 0)
Solution.
We have, sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
⇒ – 2 sin^-1 x = \(\frac{\pi}{2}\) – sin^-1 (1 – x)
⇒ – 2 sin^-1 x = cos^-1 (1 – x) ………….(i)
Let sin^-1 x = θ
⇒ sin θ = x
⇒ cos θ = \(\sqrt{1-x^{2}}\)
∴ θ = cos^-1 \(\sqrt{1-x^{2}}\)
∴ sin^-1 x = cos^-1 \(\sqrt{1-x^{2}}\)
Therefore, from Eq. (i), we have
– 2 cos^-1 (\(\sqrt{1-x^{2}}\) ) = cos^-1 (1 – x)
Put x = sin y. Then, we have
– 2 cos^-1 (\(\)) = cos^-1 (1 – sin y)
⇒ – 2 cos^-1 (cos y) = cos^-1 (1 – sin y)
⇒ – 2y = cos^-1 (1 – sin y)
⇒ 1 – sin y = cos(- 2y) = cos 2y
⇒ 1 – sin y = 1 – 2 sin2 y
⇒ 2 sin² y – sin y = 0
⇒ sin y(2 sin y – 1) = 0
sin y = 0 or \(\frac{1}{2}\)
∴ x = 0 or x = \(\frac{1}{2}\)
But, when x = \(\frac{1}{2}\), it can be observed that
We have, tan^-1 \(\frac{1-x}{1+x}\) = \(\frac{1}{2}\) tan^-1 x
⇒ tan^-1 1 – tan^-1 x = \(\frac{1}{2}\) tan^-1 x
[∵ tan^-1 x – tan^-1 y = tan^-1 \(\frac{(x-y)}{1+x y}\)]
[∵ tan^-1 (1) = \(\frac{\pi}{4}\)]
⇒ \(\frac{\pi}{4}\) = \(\frac{3}{2}\) tan^-1 x
⇒ tan^-1 x = \(\frac{\pi}{6}\)
⇒ x = tan \(\frac{\pi}{6}\)
∴ x = \(\frac{1}{\sqrt{3}}\).

Question 15.
sin(tan^-1 x), |x| < 1 is equal to
(A) \(\frac{x}{\sqrt{1-x^{2}}}\)

(B) \(\frac{1}{\sqrt{1-x^{2}}}\)

(C) \(\frac{1}{\sqrt{1+x^{2}}}\)

(D) \(\frac{x}{\sqrt{1+x^{2}}}\)
Solution.
Let tan^-1 x = y.
Then, tan y = x
⇒ sin y = \(\frac{x}{\sqrt{1+x^{2}}}\)
∴ y = sin^-1 (\(\frac{x}{\sqrt{1+x^{2}}}\))
⇒ tan^-1 x = sin^-1 (\(\frac{x}{\sqrt{1+x^{2}}}\))
Now, sin(tan^-1 x) = sin(sin^-1 (\(\frac{x}{\sqrt{1+x^{2}}}\)))
= \(\frac{x}{\sqrt{1+x^{2}}}\)
The correct answer is (D).

Question 16.
sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\), then x is equal to
(A) 0, \(\frac{1}{2}\)
(B) 1, \(\frac{1}{2}\)
(C) 0
(D) \(\frac{1}{2}\)
Solution.
Given, sin^-1 (1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
putting \(\frac{\pi}{2}\) = sin^-1 (1 – x) + cos^-1 (1 – x)
or sin^-1 (1 – x) – 2 sin^-1 (1 – x) = sin^-1 (1 – x) + cos^-1 (1 – x)
⇒ – 2 sin^-1 x = cos^-1 (1 – x)
Let sin^-1 x = α
∴ sin α = x
∴ – 2 sin^-1 x = – 2 α = cos^-1 (1 – x)
or cos 2α = 1 – x [∵ cos(- θ) = cos θ]
∴ 1 – 2 sin² α = (1 – x)
Putting sin α = x
⇒ 1 – 2x² = 1 – x
or 2x² – x = 0
x(2x – 1) = 0
∴ x = 0, \(\frac{1}{2}\)
But x = \(\frac{1}{2}\) does not satisfy the equation.
∴ x = 0
Hence, the correct answer is (C).

Question 17.
tan^-1 \(\left(\frac{x}{y}\right)\) – tan^-1 \(\frac{x-y}{x+y}\) is equal to
(A) \(\frac{\pi}{2}\)

(B) \(\frac{\pi}{3}\)

(C) \(\frac{\pi}{4}\)

(D) \(\frac{3 \pi}{4}\)
Solution.
We have tan^-1 \(\left(\frac{x}{y}\right)\) – tan^-1 \(\frac{x-y}{x+y}\)

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 5 Continuity and Differentiability Ex 5.5 Textook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5

DirectIon (1 – 11): Differentiate the following functions w.r.t. x:

Question 1.
cos x . cos 2x . cos 3x
Solution.
Let y = cos x . cos 2x . cos 3x
Taking logarithm on othsides, we get
log y = log(cos x . cos 2x . cos 3x)
⇒ log y = log (cos x) + log (cos 2x) + log (cos 3x)
On differentiating bothsides w.r.t x, we get
\(\frac{1}{y}\) . \(\frac{d y}{d x}\) = \(\frac{1}{\cos x}\) \(\frac{d}{d x}\) (cos x) + \(\frac{1}{\cos 2 x}\) . \(\frac{d}{d x}\) (cos 2x) + \(\frac{1}{\cos 3 x}\) . \(\frac{d}{d x}\) (cos 3x)

⇒ \(\frac{d y}{d x}\) = y \(\left[-\frac{\sin x}{\cos x}-\frac{\sin 2 x}{\cos 2 x} \cdot \frac{d}{d x}(2 x)-\frac{\sin 3 x}{\cos 3 x} \cdot \frac{d}{d x}(3 x)\right]\)

∴ \(\frac{d y}{d x}\) = – cos x . cos 2x . cos 3x [tan x + 2 tan 2x + 3 tan 3x]

Question 2.
\(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)

Solution.
Let y = \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)

Taking logarithm on othsides, we get
log y = log \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)

⇒ log y = \(\frac{1}{2}\) log \(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)

⇒ log y = \(\frac{1}{2}\) – [log {(x – 1) (x – 2)} – log {(x – 3) (x – 4) (x – 5)}]

⇒ log y = \(\frac{1}{2}\) [log (x – 1) + log (x – 2) – log (x – 3) – log (x – 4) – log (x – 5)]

On differentiating bothsides w.r.t x, we get

Question 3.
(log x)^{cos x}
Solution.
Let y = (log x)^{cos x}
On differentiating bothsides w.r.t x, we get

\(\frac{1}{y}\) . \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (cos x) × log(log x) + cos x × \(\frac{d}{d x}\) [log (log x)]

⇒ \(\frac{1}{y}\) . \(\frac{d y}{d x}\) = – sin x log (log x) + cos x × \(\frac{1}{\log x}\) . \(\frac{d}{d x}\) (log x)

⇒ \(\frac{d y}{d x}\) = y [- sin x log (log x) + \(\frac{\cos x}{\log x} \times \frac{1}{x}\)]

∴ \(\frac{d y}{d x}\) = (log x)^{cos x} [\(\frac{\cos x}{x \log x}\) – sin x log (log x)]

Question 4.
x^x – 2^{sin x}
Solution.
Let y = x^x – 2^{sin x}
Also, let x^x = u and 2^{sin x} = v
∴ y = u – v
⇒ \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) – \(\frac{d v}{d x}\)
Now, u = x^x
Taking lagarithm on bothsides, we get
log u = x log x
On differentiating bothsides w.r.t x, we get
\(\frac{1}{u}\) . \(\frac{d u}{d x}\) = [\(\frac{d}{d x}\) (x) × log x + x × \(\frac{d}{d x}\) (log x)]

⇒ \(\frac{d u}{d x}\) = u [1 × log x + x × \(\frac{1}{x}\)]

⇒ \(\frac{d u}{d x}\) = x^x (log x + 1)

⇒ \(\frac{d u}{d x}\) = x^x (1 + log x)
And, v = 2^{sin x}
Taking lagarithm on bothsides, we get
log v = sin x log 2
On differentiating bothsides w.r.t x, we get
\(\frac{1}{v}\) . \(\frac{d v}{d x}\) = log 2 . [\(\frac{d}{d x}\) (sin x)

⇒ \(\frac{d v}{d x}\) = v log 2 cos x

⇒ \(\frac{d v}{d x}\) = 2^{sin x} cos x log 2

⇒ \(\frac{d y}{d x}\) = x^x (1 + log x) – 2^{sin x} cos x log 2.

Question 5.
(x + 3)² . (x + 4)³ . (x + 5)⁴
Solution.
Let y = (x + 3)² . (x + 4)³ . (x + 5)⁴
Taking lagarithm on bothsides, we get
log y = log (x + 3)² + log (x + 4)³ + log (x + 5)⁴
log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)
On differentiating bothsides w.r.t x, we get
\(\frac{1}{y}\) . \(\frac{d y}{d x}\) = 2 . \(\frac{1}{x+3}\) . \(\frac{d}{d x}\) (x + 3) + 3 . \(\frac{1}{x+4}\) . \(\frac{d}{d x}\) (x + 4) + 4 . \(\frac{1}{x +5}\) \(\frac{d}{d x}\) (x + 5)

⇒ \(\frac{d y}{d x}\) = y \(\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]\)

⇒ \(\frac{d y}{d x}\) = (x + 3)² . (x + 4)³ . (x + 5)⁴ \(\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]\)

⇒ \(\frac{d y}{d x}\) = (x + 3)² + (x + 4)³ + (x + 5)⁴ . \(\left[\frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\right]\)

⇒ \(\frac{d y}{d x}\) = (x + 3) (x + 4)² (x + 5)³ . [2 (x² + 9x + 20) + 3 (x² + 8x + 15 + 4 (x² + 7x + 12)]

⇒ \(\frac{d y}{d x}\) = (x + 3) (x + 4)² (x + 5)³ . [2x² + 18x + 40 + 3x² + 24x + 45 + 4x² + 28x + 48]

∴ \(\frac{d y}{d x}\) = (x + 3) (x + 4)² (x + 5)³ (9x² + 70x + 133)

Question 6.
\(\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}\)
Solution.
Let y = \(\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}\)

Also, let u = \(\left(x+\frac{1}{x}\right)^{x}\) and v = \(x^{\left(x+\frac{1}{x}\right)}\)

∴ y = u + v
⇒ \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) + \(\frac{d v}{d x}\)
Then, u = \(\left(x+\frac{1}{x}\right)^{x}\)
⇒ log u = log \(\left(x+\frac{1}{x}\right)^{x}\)
⇒ log u = x log (x + \(\frac{1}{x}\)) (Taking log on bothsides)
On differentiating bothsides w.r.t x, we get

\(\frac{d y}{d x}\) = v \(\left(\frac{-\log x+x+1}{x^{2}}\right)\)

\(\frac{d y}{d x}\) = \(x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^{2}}\right)\) ………….(iii)

Therefore from eqs. (i), (ii) and (iii), we get

\(\frac{d y}{d x}\) = \(\left(x+\frac{1}{x}\right)^{x}\left[\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left(\frac{x+1-\log x}{x^{2}}\right)\)

Question 7.
(log x)^x + x^{log x}
Solution.
Let y = (log x)^x + x^{log x}
Also, let u = (log x)^x and v = x^{log x}
∴ y = u + v
⇒ \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) + \(\frac{d v}{d x}\) ………….(i)
Then, u = (log x)^x
⇒ log u = log [(log x)^x]
(Taking log on bothsides)
⇒ log u = x log (log x)
On differentiating bothsides w.r.t x, we get

Again, v = x^{log x}
⇒ log v = log (x^{log x}) (Taking log on bothsides)
⇒ log v = log x log x
= (log x)²
On differentiating bothsides w.r.t x, we get
\(\frac{1}{v}\) \(\frac{d v}{d x}\) = \(\frac{d}{d x}\) [(log x)²]

⇒ \(\frac{1}{v}\) \(\frac{d v}{d x}\) = 2 (log x) . \(\frac{d}{d x}\) (log x)

⇒ \(\frac{d v}{d x}\) = 2v (log x) . \(\frac{1}{x}\)

⇒ \(\frac{d v}{d x}\) = 2x^{log x} \(\frac{\log x}{x}\)

⇒ \(\frac{d v}{d x}\) = 2x^{log x – 1} . log x
Therefore, from Eqs. (i), (ii) and (iii), we get

\(\frac{d y}{d x}\) = (log x)^{log x – 1} [1 + log x . log (log x)] + 2x^{log x – 1} . log x

Question 8.
(sin x)^x + sin^-1 √x
Solution.
Let y = (sin x)^x + sin^-1 √x
Also, let u = (sin x)^x and v = sin^-1 √x
⇒ y = u + v
⇒ \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) + \(\frac{d v}{d x}\) ………….(i)
Then, u = (sin x)^x
⇒ log u = log (sin x)^x (Taking log on bothsides)
log u = x log (sin x)
On differentiating bothsides w.r.t x, we get

Question 9.
x^{sin x} + (sin x)^{cos x}
Solution.
Let y = x^{sin x} + (sin x)^{cos x}
also, let u = x^{sin x} and v = (sin x)^{cos x}
⇒ y = u + v
⇒ \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) + \(\frac{d v}{d x}\)
Then, u = x^{sin x}
⇒ log u = log (x^{sin x}) (Taking log on bothsides)
⇒ log u = sin x log x
On differentiating bothsides w.r.t x, we get
\(\frac{1}{u}\) . \(\frac{d u}{d x}\) = \(\frac{d}{d x}\) (sin x) . log x + sin x . \(\frac{d}{d x}\) (log x)

⇒ \(\frac{d u}{d x}\) = u [cos x log x + sin x . \(\frac{1}{x}\)]

⇒ \(\frac{d u}{d x}\) = x^{sin x} [cos x log x + \(\frac{\sin x}{x}\)] …………..(ii)
Again, v = (sin x)^{cos x}
⇒ log v = log (sin x)^{cos x}
⇒ log v = cos x log (sin x)
On differentiating bothsides w.r.t x, we get
\(\frac{1}{v}\) . \(\frac{d v}{d x}\) = \(\frac{d}{d x}\) (cos x) × log(sin x) + cos x × \(\frac{d}{d x}\) [log(sin x)]

⇒ \(\frac{d v}{d x}\) = v[- sin x . log(sin x) + cos x . \(\frac{1}{\sin x} \cdot \frac{d}{d x}\) (sin x)]

⇒ \(\frac{d v}{d x}\) = (sin x)^{cos x} [- sin x log sin x + \(\frac{\cos x}{\sin x}\) cos x]

⇒ \(\frac{d v}{d x}\) = (sin x)^{cos x} [- sin x log . sin x + cot x cos x]

⇒ \(\frac{d v}{d x}\) = (sin x)^{cos x} [cot x cos x – sin x log sin x] ……………(iii)
Therefore, from Eqs. (i), (ii) and (iii), we get
\(\frac{d v}{d x}\) = x^{sin x} [cos x log x + \(\frac{\sin x}{x}\)] + (sin x)^{cos x} [cot x cos x – sin x log sin x]

Question 10.
x^{x cos x} + \(\frac{x^{2}+1}{x^{2}-1}\)
Solution.
Let y = x^{x cos x} + \(\frac{x^{2}+1}{x^{2}-1}\)
Also, let u = x^{cos x} and v = \(\frac{x^{2}+1}{x^{2}-1}\)
⇒ y = u + v
⇒ \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) + \(\frac{d v}{d x}\) …………..(i)
Then, u = x^{x cos x}
⇒ log u = log (x)^{x cos x} (Taking log on bothsides)
⇒ log u = x cos x log x
On differentiating bothsides w.r.t x, we get
\(\frac{1}{u}\) . \(\frac{d u}{d x}\) = \(\frac{d}{d x}\) (x) . cos x . log x + x . \(\frac{d}{d x}\) (cos x) . log x + x cos x . \(\frac{d}{d x}\) (log x)

⇒ \(\frac{d u}{d x}\) = u [1 . cos x . log x + x . (- sin x) log x + x cos x . \(\frac{1}{x}\)]

⇒ \(\frac{d u}{d x}\) = x^{x cos x} (cos x log x – x sin x log x + cos x)

\(\frac{d u}{d x}\) = x^{x cos x} [cos x (1 + log x) – x sin x log x] …………..(ii)

Again, v = \(\frac{x^{2}+1}{x^{2}-1}\)

⇒ log v = log(x² + 1) – log (x – 1) (Taking log on both sides)
On differentiating both sides w.r.t x, we get

Question 11.
(x cos x)^x + (x sin x)^{\(\frac{1}{x}\)}
Solution.
Let y = (x cos x)^x + (x sin x)^{\(\frac{1}{x}\)}
Also, let u = (x cos x)^x and v = (x sin x)^{\(\frac{1}{x}\)}
⇒ y = u + v
⇒ \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) + \(\frac{d v}{d x}\)
Then, u = (x cos x)^x
⇒ log u = log (x cos x)^x (Taking log on both sides)
⇒ log u = x log (x cos x)
⇒ log u = x[log x + log cos x]
⇒ log u = x log x + x log cos x
On differentiating both sides w.r.t x, we get

⇒ \(\frac{d u}{d x}\) = (x cos x)^x [(1 + log x) + (log cos x – x tan x)]
⇒ \(\frac{d u}{d x}\) = (x cos x)^x [1 – x tan x + (log x + log cos x)]
⇒ \(\frac{d u}{d x}\) = (x cos x)^x [1 – x tan x + log (x cos x)] …………..(ii)
Again, v = (x sin x)^{\(\frac{1}{x}\)}
⇒ log v = log (x sin x)^{\(\frac{1}{x}\)} (Taking log on both sides)
⇒ log v = \(\frac{1}{x}\) log (x sin x)
⇒ log v = \(\frac{1}{x}\) (log x + log sin x)
⇒ log v = \(\frac{1}{x}\) log x + \(\frac{1}{x}\) log sin x
On differentiating both sides w.r.t x, we get

DIrectIon (12 – 15) : Find of the following functions:

Question 12.
x^y + y^x = 1
Solution.
Given, x^y + y^x = 1
Let x^y = u and y^x = v
Then, the function becomes u + v = 1
∴ \(\frac{d u}{d x}\) + \(\frac{d v}{d x}\) = 0
Now, u = x^y
⇒ log u = log(x^y) (Taking log on both sides)
⇒ log u = y log x
On differentiating both sides w.r.t. x, we get
\(\frac{1}{u}\) . \(\frac{d u}{d x}\) = log x \(\frac{d y}{d x}\) + y . \(\frac{d}{d x}\) (log x)

⇒ \(\frac{d u}{d x}\) = x^y [log x \(\frac{d y}{d x}\) + y . \(\frac{1}{x}\)]

⇒ \(\frac{d u}{d x}\) = x^y (log x \(\frac{d y}{d x}\) + \(\frac{y}{x}\)) …………(ii)

Again, v = y^x
⇒ log v = log y^x
⇒ log v = x log y
On differentiating both sides w.r.t. x, we get

Question 13.
y^x = x^y
Solution.
Given, y^x = x^y
Taking logarithm on both sides , we get
log y^x = log x^y
⇒ x log y = y log x
On differentiating both sides w.r.t. x, we get

Question 14.
(cos x)^y = (cos y)^x
Solution.
Given, (cos x)^y = (cos y)^x
Taking logarithm on both sides , we get
⇒ log (cos x)^y = log (cos y)^x
⇒ y log (cos x) = x log (cos y)
On differentiating both sides w.r.t. x, we get
log cos x . \(\frac{d y}{d x}\) + y . \(\frac{d}{d x}\) (log cos x) = log cos y . \(\frac{d}{d x}\) (x) + x . \(\frac{d}{d x}\) (log cos y)

⇒ log cos x . \(\frac{d y}{d x}\) + \(\frac{y}{\cos x}\) (- sin x) = log cos y + \(\frac{x}{\cos y}\) (- sin y) . \(\frac{d y}{d x}\)

⇒ log cos x \(\frac{d y}{d x}\) – y tan x = log cos y – x tan y \(\frac{d y}{d x}\)

⇒ (log cos x + x tan y) \(\frac{d y}{d x}\) = y tan x + log cos y

∴ \(\frac{d y}{d x}=\frac{y \tan x+\log \cos y}{x \tan y+\log \cos x}\).

Question 15.
xy = e^{(x – y)}
Solution.
Given, xy = e^{(x – y)}
Taking logarithm on bothsides, we get
⇒ log (xy) = log (e^{(x – y)})
⇒ log x + log y = (x – y) log e
⇒ log x + log y = (x – y) × 1
⇒ log x + log y = x – y
On differentiating both sides w.r.t. x, we get
\(\frac{d}{d x}\) (log x) + \(\frac{d}{d x}\) (log y) = \(\frac{d}{d x}\) (x) – \(\frac{d y}{d x}\)

⇒ \(\frac{1}{x}\) + \(\frac{1}{y}\) . \(\frac{d y}{d x}\) = 1 – \(\frac{d y}{d x}\)

⇒ (1 + \(\frac{1}{y}\)) \(\frac{d y}{d x}\) = 1 – \(\frac{1}{x}\)

⇒ \(\left(\frac{y+1}{y}\right) \frac{d y}{d x}=\frac{x-1}{x}\)

⇒ \(\frac{d y}{d x}=\frac{y(x-1)}{x(y+1)}\)

Question 16.
Find the derivative of the function given by f(x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸) and hence find f(1).
Solution.
Given f(x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸)
Taking logarithm on both sides, we get
log f(x) = log (1 + x) + log (1 + x²) + log(1 + x⁴) + log (1 + x⁸)
On differentiating both sides w.r.t. x, we get
\(\) . \(\frac{d}{d x}\) [f(x)] = \(\frac{d}{d x}\) log (1 + x) + \(\frac{d}{d x}\) log (1 + x²) + log (1 + x⁴) + log (1 + x⁸)

Question 17.
Differentiate(x² – 5x + 8) (x³ + 7x +9) in three ways mentioned below:
(i) By using product rule.
(ii) By expanding the product to obtain a single polynomial.
(iii) By logarithmic differentiation.
Do they all give the same answer?
Solution.
(i) Let y = (x² – 5x + 8) (x³ + 7x + 9)
Let x² – 5x + 8 = u and x³ + 7x + 9 = v
∴ y = uv
⇒ \(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) . v + u . \(\frac{d v}{d x}\)

⇒ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x² – 5x + 8) . (x³ + 7x + 9) + (x² – 5x + 8) . \(\frac{d}{d x}\) (x³ + 7x + 9)

⇒ \(\frac{d y}{d x}\) = (2x – 5) (x³ + 7x + 9) + (x² – 5x + 8) (3x² + 7)

⇒ \(\frac{d y}{d x}\) = 2x (x³ + 7x + 9) – 5 (x³ + 7x + 9) + x² (3x² + 7) – 5x (3x² + 7) + 8 (3x² + 7)

⇒ \(\frac{d y}{d x}\) = (2x⁴ + 14x² + 18x) – 5x³ – 35x – 45 + (3x⁴ + 7x²) – 15x³ – 35x + 24x² + 56

∴ \(\frac{d y}{d x}\) = 5x⁴ – 20x³ + 45x² – 52x + 11

(ii) y = (x² – 5x + 8) (x³ + 7x + 9)
= x² (x³ +7x + 9) – 5x (x³ + 7x + 9) + 8(x³ + 7x + 9)
= x⁵ + 7x³ + 9x² – 5x⁴ – 35x² – 45x + 8x³ 56x + 72
= x⁵ – 5x⁴ + 15x³ – 26x² + 11x + 72
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (x⁵ – 5x⁴ + 15x³ – 26x² + 11x + 72)

= \(\frac{d}{d x}\) (x⁵ – \(\frac{d}{d x}\) (5x⁴) + 15 \(\frac{d}{d x}\) (x³ – 26 \(\frac{d}{d x}\) (x²) + 11 \(\frac{d}{d x}\) (x) + \(\frac{d}{d x}\) (72)

= 5x⁴ – 5 × 4x³ + 15 × 3x² – 26 × 2x + 11 × 1 + 0
= 5x⁴ – 20x³ + 45x² – 52x + 11

(iii) y = (x² – 5x + 8) (x³ + 7x + 9)
Taking logarithm on both sides, we get
log y = log (x² – 5x + 8) + log (x³ + 7x + 9)
On differentiating both sides w.r.t x, we get
\(\frac{1}{y}\) . \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) log (x² – 5x + 8) + \(\frac{d}{d x}\) log (x³ + 7x + 9)

⇒ \(\frac{d y}{d x}\) = 2x (x³ + 7x + 9) -5 (x³ + 7x + 9) + 3x² (x² – 5x + 8) + 7 (x² – 5x + 8)

⇒ \(\frac{d y}{d x}\) = (2x⁴ + 14x² + 18x) – 5x³ – 35x – 45 + (3x⁴ – 15x³ + 24x²) + (7x² – 35x + 56)

⇒ \(\frac{d y}{d x}\) = 5x⁴ – 20x³ + 45x² – 52x + 11

From the above three observations, it can be concluded that all the results of \(\frac{d y}{d x}\) are same.

Question 18.
If u, v and w are functions of x, then show that
\(\frac{d}{d x}\) (u . v . w) = \(\frac{d u}{d x}\) v . w + u . \(\frac{d v}{d x}\) . w + u . v . \(\frac{d w}{d x}\)
in two ways-first by repeated application of product rule, second by logarithmic differentiation.
Solution.
Let y = u . v . w = u(v . w)
By applying product rule, we get
\(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) (v . w) + u . \(\frac{d}{d x}\) (v . w)

\(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) v . w + u [\(\frac{d v}{d x}\) . w + v . \(\frac{d w}{d x}\)] (Again, applying product rule)

\(\frac{d y}{d x}\) = \(\frac{d u}{d x}\) . v . w + u . \(\frac{d v}{d x}\) . w + u . v . \(\frac{d w}{d x}\)

By taking logarithm on both sides of the equation y = u u w, we get
log y = log u + log u + log w
On differentiating both sides w.r.t. x, we get

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 7 The p-Block Elements Textbook Exercise Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 7 The p-Block Elements

Very Short Answer Type Questions

Question 1.
What is the difference between the nature of n-bonds present in H₃PO₃ and HNO₃ molecules?
Answer:
In H₃PO₃, there is pπ-dπ bond whereas in HNOs there is pπ-pπ bond.

Question 2.
Complete the following equations:
(i) PCl₃ + H₂O →
(ii) XeF₂ + PF₅ →

Answer:
(i) PCl₃ + 3H₂O → H₃PO₃ + 3HCl
(ii) XeF₂ + PF₅ → [XeF]⁺ [PF₆]^–

Question 3.
Which allotrope of sulphur is thermally stable at room temperature?
Answer:
Rhombic sulphur

Question 4.
O—O bond has lower bond dissociation enthalpy than S—S bond. Why?
Answer:
Due to smaller size the lone pairs of electrons on the O atom repel the bond pair of O—O bond to a greater extent as compared to the lone pairs of electrons on S atom in S—S bond. Consequently O—O bond has lower bond dissociation enthaltpy than S—S bond.

Question 5.
How would you account for the following:
(i) H₂S is more acidic than H₂O.
(ii) Both O₂ and F₂ stabilise higher oxidation states hut the ability of oxygen to stabilise the higher oxidation state exceeds that of fluorine.
Answer:
(i) This is because bond dissociation enthalpy of H—S bond is lower than that of H—O bond.
(ii) This is due to tendency of oxygen to form multiple bonds with metal atom.

Question 6.
Why solid PCl₅ is ionic in nature?
Answer:
Because in solid state, PCl₅ exists as [PCl₄]⁺[PCl₆]^– and conducts electricity on melting.

Question 7.
On adding NaOH to ammonium sulphate, a colourless gas with pungent odour is evolved which form a blue coloured complex with Cu²⁺ ion. Identify the gas.
Answer:
Ammonia (NH₃).

Question 8.
N₂O₅ is more acidic thanNaO₃. Why?
Answer:
N₂O₅ is the anhydride of nitric acid, forms the stable acid with water as follows :

While N₂O₃ is the anhydride of nitrous acid, HNO₂. It dissolves in water to form the unstable acid as follows :

Hence, N₂O₅ is more acidic thanN₂O₃.

Question 9.
Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling is diamagnetic?
Answer:
In gaseous state, NO₂ exists as monomer which has one unpaired electron but in solid state, it dimerises to NO₂ so no unpaired electron is left hence, the solid formed is diamagnetic.

Question 10.
In the preparation of H₂SO₄ by contact process, why is SO₃ not absorbed directly in water to form H₂SO₄ ?
Answer:
Acid fog is formed, which is difficult to condense.

Short Answer Type Questions

Question 1.
Account for the following:
(i) Bi(V) is a stronger oxidising agent than Sb(V).
(ii) N—N single bond is weaker than P—P single bond.
Answer:
(i) Due to inert pair effect +3 oxidation state of Bi is more stable than its +5 oxidation state while +5 oxidation state of Sb is more stable than its +3 oxidation state. Therefore, Bi (V) can accept a pair of electrons to form more stable Bi (III) more easily than Sb (V). Hence, Bi (V) is a stronger oxidising agent than Sb (V).

(ii) N—N single bond is weaker than P—P single bond due to large interelectronic repulsion between the lone pairs of electrons present on the N atoms of N—N bond having small bond length.

Question 2.
Account for the following:
(i) PCl₅ is more covalent than PCl₃.
(ii) Iron on reaction with HCl forms FeCl₂ and not FeCl₃.
(iii) The two O—O bond lengths in the ozone molecule are equal.
Answer:
(i) The oxidation state of central atom, i.e., phosphorus is +5 in PCl₅ whereas it is +3 in PCl₃. Higher the positive oxidation of central atom, more will be its polarising power which, in turn, increases the covalent character of bond formed between the central atom and the atoms surrounding it.

(ii) Iron reacts with HCl to form FeCl₂ and H₂.
Fe + 2HCl → FeCl₂ + H₂
H₂ thus produced prevents the oxidation of FeCl₂ to FeCl₃.

(iii) Ozone is a resonance hybrid of the following two main structures :

As a result of resonance, the two O—O bond lengths in O₃ are equal.

Question 3.
How would you account for the following:
(i) The electron gain enthalpy with negative sign is less for oxygen than that for sulphur.
(ii) Fluorine never acts as the central atom in polyatomic interhalogen compounds.
Answer:
(i) This is due to smaller size of oxygen the electron cloud is distributed over a small region of space, making electron density high which repels the incoming electrons.

(ii) Fluorine never acts as the central atom in polyatomic interhalogen compounds since it is the most electronegative element of the group.

Question 4.
PCl₅ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH₃ solution. Write the reactions involved to explain what happens.
Answer:
PCl₅ on reaction with finely divided silver produced silver halide.
PCl₅ + 2Ag → 2AgCl + PCl₃
AgCl on further reaction with aqueous ammonia solution produces a soluble complex of [Ag(NH₃)₂]⁺Cl^–.

Question 5.
Account for the following:
(i) Sulphur in vapour state exhibits paramagnetism.
(ii) H₃PO₂ is a stronger reducing agent than H₃PO₂.
Answer:
(i) In vapour form, sulphur partly exists as S₂ molecules which have two unpaired electrons in the antibonding n molecular orbitals like 02 molecule and hence, exhibits paramagnetism.

(ii) Acids which contain P—H bonds, have reducing character. Since, H₃PO₂ contains two P—H bonds while H₃PO₃ contains only one P—H bond, therefore H₃PO₂ is a stronger reducing agent than H₃PO₃.

Question 6.
What happens when:
(i) ortho phosphorus acid is heated?
(ii) XeF₆ undergoes complete hydrolysis?
Answer:
(i) On heating, ortho phosphorus acid disproportionates to give orthophosphoric acid and phosphine gas.

(ii) When XeF₆ undergoes complete hydrolysis, it forms XeO₃.
XeF₆ + 3H₂O → 6HF + XeO₃

Long Answer Type Questions

Question 1.
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(b) Draw the structures of the following:
(i) ClF₃
(ii) XeF₄
Answer:
(a) (i) As the size of halogen atom increases from F to I, the bond dissociation enthalpy of H—X bond decreases from H—F to H—I. Due to this, acidic character increases from HF to HI.

(ii) Because of small size and high electronegativity oxygen forms pπ-pπ multiple bonds and exists as a diatomic, O₂ molecule. The molecules are held together by weak van der Waal forces. Sulphur on the other hand due to its higher tendency for catenation and lower tendency for pπ-pπ multiple bond formation, forms octa-atomic, S₈ molecule. Because of bigger size of S₈ molecule than O₂ molecule the force of attraction holding the S₈ molecules together are much stronger than O₂ molecules. Hence, there is large difference between the melting and boiling points of oxygen and sulphur.

(iii) Nitrogen with n = 2, has s and p-orbitals only. It does not have d-orbitals to expand its covalency beyond four. Due to this, it does not form pentahalide.

(b)

Question 2.
(a) Give reasons for the following:
(i) Bond enthalpy of F₂ is lower than that of Cl₂.
(ii) PH₃ has lower boiling point than NH₃.
(b) Draw the structures of the following molecules :
(i) BrF₃
(ii)BrF₅
(iii) (HPO₃)₃
Answer:
(a) (i) Bond dissociation enthalpy decreases as the bond distance increases from F₂ to I₂ because of the corresponding increase in the size of the atom as we move from F to I. The F—F bond dissociation enthalpy is, however, smaller than that of Cl—Cl and even smaller than that of Br—Br. This is because F atom is very small and hence the three lone pairs of electrons on each F atom repel the bond pair holding the F-atoms in F2 molecule resulting lower bond enthalpy than Cl₂.

(ii) Unlike NH₃, PH₃ molecules are not associated through hydrogen bonding in liquid state. That is why the boiling point of PH₃ is lower than NH₃.

Question 3.
(a) Complete the following chemical reaction equations:
(i) AgCl(s) + NH₃ (aqr) →
(ii) P₄(s) + NaOH(aq) + H₂O(l) →
(b) Explain the following observations :
(i) H₂S is less acidic than H₂Te.
(ii) Fluorine is a stronger oxidising agent than chlorine.
(iii) Noble gases are the least reactive elements.
Answer:
(a) (i) AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ Cl^–
(ii) P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂

(b) (i) This is because bond dissociation enthalpy of H—Te bond is less than H—S as the size of Te is larger than S.
(ii) Fluorine is a stronger oxidising agent than chlorine due to low dissociation enthalpy of F—F bond and high hydration enthalpy of F^– ions.
(iii) Noble gases are the least reactive elements due to fully filled outermost shells, high ionisation enthalpy and positive electron gain enthalpy.

Question 4.
(a) Arrange the following in the order of property indicated against each set:
(i) F₂, Cl₂, Br₂, I₂ (increasing bond dissociation enthalpy)
(ii) H₂O, H₂S, H₂Se, H₂Te (increasing acidic character)
(b) A colourless gas ‘A’ with a pungent odour is highly soluble in water and its aqueous solution is weakly basic. As a weak base it precipitates the hydroxides of many metals from their salt solution. Gas ‘A* finds application in detection of metal ions. It gives a deep blue colouration with copper ions. Identify the gas A’ and write the chemical equations involved in the following :
(i) Gas ‘A’ with copper ions
(ii) Solution of gas ‘A’ with ZnSO₄ solution.
Answer:
(a) (i) I₂ < F₂ < Br₂ < Cl₂
(ii) H₂O < H₂S < H₂Se < H₂Te

(b) The gas ‘A’ is ammonia (NH₃).
(i) Cu²⁺(aq) + 4 NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺(aq)
(ii) ZnSO₄(aq) + 2 NH₄OH(aq) → Zn(OH)₂(s) + (NH₄) ₂ SO₄(aq)

Question 5.
Answer the following questions
(a) Write the formula of the neutral molecule which is isoelectronic with ClO^–.
(b) Draw the shape of H₂S₂O₇.
(c) Nitric acid forms an oxide of nitrogen on reaction with P₄. Write the formula of the stable molecule formed when this oxide undergoes dimerisation.
(d) Bleaching action of chlorine is permanent. Justify.
(e) Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in + 3 oxidation state.
Answer:
(a) ClF

(c) N₂O₄
(d) Bleaching action of chlorine is permanent due to oxidation.
Cl + H₂O → 2HCl + [O]
(d) 3HNO₂ → HNO₃ + H₂O + 2NO

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 5 Continuity and Differentiability Ex 5.4 Textook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4

Question 1.
\(\frac{e^{x}}{\sin x}\)

Solution.
Let y = \(\frac{e^{x}}{\sin x}\)

By using the quotient rule, we get

\(\frac{d y}{d x}\) = \(\frac{\sin x \frac{d}{d x}\left(e^{x}\right)-e^{x} \frac{d}{d x}(\sin x)}{\sin ^{2} x}\)

= \(\frac{\sin x \cdot\left(e^{x}\right)-e^{x} \cdot(\cos x)}{\sin ^{2} x}\)

= \(\frac{e^{x}(\sin x-\cos x)}{\sin ^{2} x}\), x ≠ nπ, n ∈ Z.

Question 2.
e^sin-1 x.
Solution.
Let y = e^sin-1 x
By using the chain rule, we get

Question 3.
e^x3
Solution.
Let y = e^x3
By using the chain rule, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (e^x3)

= e^x3 . \(\frac{d}{d x}\) (x³)

= e^x3 . 3x²
= 3x² e^x3

Question 4.
sin(tan^-1 e^-x
Solution.
Let y = sin(tan^-1 e^-x
By using the chain rule, we get

Question 5.
log (cos e^x)
Solution.
Let y = log (cos e^x)
By using the chain rule, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [log (cos e^x)]

= \(\frac{1}{\cos e^{x}}\) . \(\frac{d}{d x}\) (cos e^x)

= \(\frac{1}{\cos e^{x}}\) . (- sin e^x) . \(\frac{d}{d x}\) (e^x)

= \(\frac{-\sin e^{x}}{\cos e^{x}}\) . e^x

= – e^x tan e^x, e^x ≠ (2n + 1) \(\frac{\pi}{2}\), n ∈ N.

Question 6.
\(e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}\)
Solution.
\(e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}\)
By using the chain rule, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (\(e^{x}+e^{x^{2}}+\ldots+e^{x^{5}}\))

= \(\frac{d}{d x}\) (e^x ) + \(\frac{d}{d x}\) (e^x2) + \(\frac{d}{d x}\) (e^x3) + \(\frac{d}{d x}\) (e^x4) + \(\frac{d}{d x}\) (e^x5)

= e^{x + [ex2 . \(\frac{d}{d x}\) (x2)] + [ ex3 . \(\frac{d}{d x}\) (x3)] + [ex4 . \(\frac{d}{d x}\) (x4)] + [ex5 . \(\frac{d}{d x}\) (x5)]}

= e^x + (e^x2 × 2x) + (e^x3 × 3x²) + (e^x4 × 4x³) + (e^x5 × 5x⁴)

= e^x + 2x e^x2 + 3x² e^x3 + 4x³ e^x4 + 5x⁴ e^x5

Question 7.
\(\sqrt{e^{\sqrt{x}}}\), x > 0
Solution.
Let y = \(\sqrt{e^{\sqrt{x}}}\)
Then, y² = e^√x
On differentiating w.r.t x, we get
y² = e^√x
⇒ 2y \(\frac{d y}{d x}\) = e^√x (√x)

⇒ 2y \(\frac{d y}{d x}\) = e^√x \(\frac{1}{2} \cdot \frac{1}{\sqrt{x}}\)

⇒ \(\frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 y \sqrt{x}}\)

⇒ \(\frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{e^{\sqrt{x}}} \sqrt{x}}\)

⇒ \(\frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}\), x > 0

Question 8.
log (log x), x > 1
Solution.
Let y = log (log x)
By using the chain rule, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [log (log x)]
= \(\frac{1}{\log x} \cdot \frac{d}{d x}\) (log x)

= \(\frac{1}{\log x} \cdot \frac{1}{x}=\frac{1}{x \log x}\), x > 1.

Question 9.
\(\frac{\cos x}{\log x}\), x > 0
Solution.
Let y = \(\frac{\cos x}{\log x}\)
By using quotient rule, we get
\(\frac{d y}{d x}=\frac{\frac{d}{d x}(\cos x) \times \log x-\cos x \times \frac{d}{d x}(\log x)}{(\log x)^{2}}\)

= \(\frac{-\sin x \log x-\cos x \times \frac{1}{x}}{(\log x)^{2}}\)

= \(\frac{-[x \log x \sin x+\cos x]}{x(\log x)^{2}}\), x > 0.

Question 10.
cos (log x + e^x), x > 0
Solution.
Let y = cos (log x + e^x)
By using chain rule, we get
\(\frac{d y}{d x}\) = – sin (log x + e^x) + \(\frac{d}{d x}\) (log x + e^x)
= – sin (log x + e^x) . [\(\frac{d}{d x}\) (log x) + \(\frac{d}{d x}\) (e^x)]
= – sin (log x + e^x) . (\(\frac{1}{x}\) + e^x)
= – (\(\frac{1}{x}\) + e^x) sin (log x + e^x).

Punjab State Board PSEB 12th Class Physical Education Book Solutions ਬੈਡਮਿੰਟਨ (Badminton) Game Rules.

ਬੈਡਮਿੰਟਨ (Badminton) Game Rules – PSEB 12th Class Physical Education

ਬੈਡਮਿੰਟਨ ਦਾ ਇਤਿਹਾਸ
(History of Badminton)

ਬੈਡਮਿੰਟਨ ਸ਼ਬਦ ਦੀ ਉਤਪੱਤੀ, ਗਲਾਉਸੇਸਟਰ ਸ਼ਾਇਰ (ਇੰਗਲੈਂਡ) ਵਿਚ ਇਕ ਐਸਟੇਟ, ‘ਬੈਡਮਿੰਟਨ’ ਸ਼ਹਿਰ ਦੇ ਨਾਂ ਤੋਂ ਹੋਇਆ ਹੈ । 1873 ਵਿਚ, ਪਹਿਲੇ ਬੈਡਮਿੰਟਨ ਕਲੱਬ ਦਾ ਆਰੰਭ ਇੰਗਲੈਂਡ ਵਿਚ ਹੋਇਆ ਸੀ । ਜਦਕਿ, ਇਸ ਤਰ੍ਹਾਂ ਦਾ ਵਿਸ਼ਵਾਸ ਕੀਤਾ ਜਾਂਦਾ ਹੈ ਕਿ ਇਸ ਤਰ੍ਹਾਂ ਦਾ ‘ਬੈਟੇਡੋਕ’ ਨਾਮ ਦਾ ਇਕ ਖੇਡ, ਆਧੁਨਿਕ ਯੁੱਗ ਤੋਂ ਪਹਿਲਾਂ-ਚੀਨ ਦਾ ਇਕ ਭਾਗ ਸੀ ।

ਭਾਰਤ ਵਿਚ ਫਿਰ ਇਸ ਖੇਡ ਦਾ ਵਿਕਾਸ ਸੈਨਿਕ ਅਧਿਕਾਰੀਆਂ ਦੁਆਰਾ ਕੀਤਾ ਗਿਆ ਅਤੇ ਪੁਨੇ ਸ਼ਹਿਰ ਦੇ ਨਾਮ ‘ਤੇ ਇਸ ਨੂੰ ਪੁਨਾ ਕਿਹਾ ਗਿਆ। ਇੰਗਲੈਂਡ ਦੀ ਬੈਡਮਿੰਟਨ ਐਸੋਸੀਏਸ਼ਨ ਦਾ ਗਠਨ 1893 ਵਿਚ ਹੋਇਆ ਸੀ । ਅੰਤਰ ਰਾਸ਼ਟਰੀ ਬੈਡਮਿੰਟਨ ਫੈਡਰੇਸ਼ਨ ਦਾ ਗਠਨ 1934 ਵਿਚ ਹੋਇਆ | ਭਾਰਤ ਵਿਚ ਇਹ ਖੇਡ ਦੁਸਰੇ ਵਿਸ਼ਵ ਯੁੱਧ ਦੇ ਬਾਅਦ ਲੋਕਪ੍ਰਿਯ ਹੋਇਆ | ਭਾਰਤ ਦੀ ਬੈਡਮਿੰਟਨ ਐਸੋਸੀਏਸ਼ਨ ਦਾ ਗਠਨ 1935 ਵਿਚ ਹੋਇਆ । ਜਦਕਿ, ਪਹਿਲਾਂ ਨੈਸ਼ਨਲ ਬੈਡਮਿੰਟਨ ਚੈਂਪੀਅਨਸ਼ਿਪ 1936 ਵਿਚ ਹੋਈ । ਬੈਡਮਿੰਟਨ 1972 ਮਿਊਨਿਖ ਉਲੰਪਿਕ ਅਤੇ 1988 ਸੀਓਲ ਉਲੰਪਿਕ ਵਿਚ ਇਕ ਪ੍ਰਦਰਸ਼ਨੀ ਖੇਡ ਦੇ ਰੂਪ ਵਿਚ ਉਨ੍ਹਾਂ ਦਾ ਹਿੱਸਾ ਰਹੀ । ਬਾਰਸੀਲੋਨਾ ਦੇ 1992 ਉਲੰਪਿਕ ਖੇਡਾਂ ਵਿਚ ਇਹ ਇਕ ਪਦਕ ਖੇਡ ਬਣੀ ।

ਯਾਦ ਰੱਖਣ ਯੋਗ ਗੱਲਾਂ
(Tips to Remember)

1. ਡਬਲਜ਼ (Doubles) ਦੇ ਲਈ ਬੈਡਮਿੰਟਨ ਕੋਰਟ ਦਾ ਆਕਾਰ : 13.40 × 6.10 ਮੀ. ਜਾਂ 44’× 20 ਫੁੱਟ
2. ਸਿੰਗਲਜ਼ (Singles) ਦੇ ਲਈ ਬੈਡਮਿੰਟਨ ਕੋਰਟ ਦਾ ਆਕਾਰ : 13.40 × 5.18 ਮੀ. ਜਾਂ 44′ × 17′ ਫੁੱਟ
3. ਨੈੱਟ ਦੀ ਚੌੜਾਈ : 760 ਮਿ.ਮੀ. (76 ਸੈਂ. ਮੀ.).
4. ਵਿਚਕਾਰ ਤੋਂ ਨੈੱਟ ਦੀ ਉੱਚਾਈ : 1.524 ਮੀ.
5. ਪੋਸਟਸ (Posts) ਤੋਂ ਨੈੱਟ ਦੀ ਉੱਚਾਈ : 1.550 ਮੀ.
6. ਕੋਰਟ ਦਾ ਆਕਾਰ : ਆਇਤਾਕਾਰ ।
7. ਰੈਕਟ ਦਾ ਨਾਪ : ਲੰਬਾਈ 680 ਮਿ.ਮੀ. × ਚੌੜਾਈ 230 ਮਿ.ਮੀ.
8. ਸ਼ਟਲ ਦਾ ਭਾਰ : 4,73 ਗ੍ਰਾਮ ਤੋਂ 5.50 ਗ੍ਰਾਮ
9. ਸ਼ਟਲ ਦੇ ਪੱਖਾਂ ਦੀ ਸੰਖਿਆ : 14 ਤੋਂ 16
10. ਪੰਖਾਂ ਦੀ ਲੰਬਾਈ : 62 ਮਿ.ਮੀ. ਤੋਂ 70 ਮਿ.ਮੀ.
11. ਬੈਂਕ ਗੈਲਰੀ ਦੀ ਚੌੜਾਈ : 2′ – ” ( 76 ਮਿ.ਮੀ.)
12. ਸਾਈਡ ਗੈਲਰੀ ਦੀ ਚੌੜਾਈ : 1′ – 6″ ( 46 ” “)
13. ਮੱਧ ਤੋਂ ਸ਼ਾਰਟ ਸਰਵਿਸ ਦੀਆਂ ਰੇਖਾਵਾਂ : 6′ – 6″ (1.98 ਮੀ.)
14. ਅਧਿਕਾਰੀਆਂ ਦੀ ਸੰਖਿਆ : ਅੰਪਾਇਰ – 1, ਸਰਵਿਸ ਅੰਪਾਇਰ – 1, ਰੈਫ਼ਰੀ – 1, ਲਾਈਨਮੈਨ – 10.

ਬੈਡਮਿੰਟਨ ਕੋਰਟ

ਨਿਯਮ ਅਤੇ ਵਿਨਿਯਮ (Rules and Regulations)-
1. ਟਾਸ (Toss) – ਖੇਡ ਸ਼ੁਰੂ ਹੋਣ ਤੋਂ ਪਹਿਲਾਂ ਦੋਵੇਂ ਟੀਮਾਂ ਵੱਲੋਂ ਟਾਸ ਕੀਤਾ ਜਾਵੇਗਾ ਅਤੇ ਵਿਜੇਤਾ ਟੀਮ ਦੇ ਕੋਲ ਪਹਿਲਾਂ ਸਰਵ (Serve) ਕਰਨ ਅਤੇ ਰਸੀਵ (Receive) ਕਰਨ ਦਾ ਵਿਕਲਪ ਹੁੰਦਾ ਹੈ ।

2. ਸਕੋਰਿੰਗ (Scoring) –

• ਇਕ ਮੈਚ ਵਿਚ ਤਿੰਨ ਗੇਮਾਂ ਦੇ ਸਰਵੋਤਮ ਸ਼ਾਮਿਲ ਹੁੰਦੇ ਹਨ ।
• ਜਿਹੜੀ ਸਾਈਡ ਪਹਿਲੇ 21 ਪੁਆਇੰਟ ਸਕੋਰ ਕਰਦੀ ਹੈ, ਉਸ ਦੇ ਦੁਆਰਾ ਗੇਮ ਜਿੱਤੀ ਜਾਣੀ ਹੈ ।
• ਰੋਮ ਜਿੱਤਣ ਵਾਲੀ ਸਾਈਡ ਹੀ ਅਗਲੀ ਗੇਮ ਵਿਚ ਪਹਿਲੇ ਸਰਵ (Serve) ਕਰੇਗੀ ।
• ਰੈਲੀ (Rally) ਜਿੱਤਣ ਵਾਲੀ ਸਾਈਡ ਵਿਚ ਇਕ ਪੁਆਇੰਟ ਜੋੜਿਆ ਜਾਵੇਗਾ ।

3. ਇੰਡਸ ਵਿਚ ਪਰਿਵਰਤਨ (Change of Ends) – 11 ਪੁਆਇੰਟ ਦੇ ਬਾਅਦ ਪਹਿਲੀ ਗੇਮ ਅਤੇ ਤੀਸਰੀ ਗੇਮ ਦੇ ਅੰਤ ਵਿਚ ਇੰਡਸ ਵਿਚ ਪਰਿਵਰਤਨ ਹੁੰਦਾ ਹੈ ।

4. ਸਰਵਿਸ ਦੇ ਨਿਯਮ (Service Rules)-

• ਜੇਕਰ ਇਕ ਵਾਰ ਸਰਵਰ (Server) ਅਤੇ ਰਸੀਵਰ (Receiver) ਤਿਆਰ ਹੋਵੇ ਤਾਂ ਸਰਵ ਕਰਨ ਵਿਚ ਗ਼ੈਰ| ਜ਼ਰੂਰੀ ਦੇਰੀ ਕਰਨ ਦੀ ਪ੍ਰਵਾਨਗੀ ਨਹੀਂ ਹੁੰਦੀ ।
• ਸਰਵਿਸ ਦੇ ਦੌਰਾਨ ਸਰਵਰ ਅਤੇ ਰਸੀਵਰ ਕੋਰਟ ਦੇ ਸਾਹਮਣੇ ਤਿਰਛੇ ਖੜੇ ਹੋਣਗੇ ।
• ਸਰਵਰ ਅਤੇ ਰਸੀਵਰ ਦੇ ਦੋਵਾਂ ਪੈਰਾਂ ਦੇ ਕੁਝ ਭਾਗ ਸਥਿਰ ਅਵਸਥਾ ਵਿਚ ਕੋਰਟ ਦੇ ਧਰਾਤਲ ਦੇ ਸੰਪਰਕ ਵਿਚ ਹੋਣੇ ਚਾਹੀਦੇ ਹਨ ।
• ਸਰਵਿਸ ਦੇ ਦੌਰਾਨ ਸਰਵਰ ਦਾ ਰੈਕੇਟ ਪਹਿਲੇ ਸ਼ਟਲ ਦੇ ਆਧਾਰ ਨੂੰ ਹਿੱਟ ਕਰੇਗਾ ।
• ਸਰਵਿਸ ਦੇ ਦੌਰਾਨ ਸ਼ਟਲ ਪਹਿਲੇ ਕਮਰ ਦੀ ਪੱਧਰ ਦੇ ਹੇਠਾਂ ਹੋਵੇਗੀ ।
• ਡਬਲ ਵਿਚ, ਸਾਥੀ (Partner) ਆਪਣੇ ਕੋਰਟ ਦੇ ਅੰਦਰ ਕੋਈ ਵੀ ਪੋਜੀਸ਼ਨ ਲੈ ਸਕਦਾ ਹੈ ।
• ਸਰਵ ਕਰਦੇ ਹੋਏ ਜੇਕਰ ਸਰਵਰ ਸ਼ਟਲ ਮਿਸ ਕਰਦਾ ਹੈ, ਤਾਂ ਇਸਨੂੰ ਗਲਤੀ ਮੰਨਿਆ ਜਾਵੇਗਾ ।

5. ਗਲਤੀ (Fault)-

• ਜੇਕਰ ਸਰਵਿਸ ਦਾ ਸਹੀ ਸਰਵ ਨਹੀਂ ਹੈ, ਤਾਂ ਇਸਨੂੰ ਗਲਤੀ ਮੰਨਿਆ ਜਾਵੇਗਾ ।
• ਜੇਕਰ ਸ਼ਟਲ ਨੈੱਟ ਨੂੰ ਪਾਰ ਕਰਨ ਵਿਚ ਅਸਫਲ ਰਹਿੰਦੀ ਹੈ ਜਾਂ ਨੈੱਟ ਵਿਚੋਂ ਜਾਂ ਹੇਠਾਂ ਤੋਂ ਪਾਰ ਕਰਦੀ ਹੈ ।
• ਜੇਕਰ ਇਸ ਨੂੰ ਵਿਅਕਤੀ, ਖਿਡਾਰੀ ਜਾਂ ਕਿਸੇ ਹੋਰ ਪਦਾਰਥ ਨਾਲ ਸੰਪਰਸ਼ ਕੀਤਾ ਜਾਂਦਾ ਹੈ ।
• ਜੇਕਰ ਕੋਈ ਖਿਡਾਰੀ ਰੈਕੇਟ ਦੇ ਨਾਲ ਸੈੱਟ ਵਿਚੋਂ ਵਿਰੋਧੀ ਦੇ ਕੋਰਟ ’ਤੇ ਹਮਲਾ ਕਰਦਾ ਹੈ ਤਾਂ ਚੀਕਾਂ ਮਾਰਨ ਜਾਂ ਇਸ਼ਾਰਿਆਂ ਦੁਆਰਾ ਰੁਕਾਵਟ ਪੈਦਾ ਕਰਦਾ ਹੈ ।
• ਜੇਕਰ ਇਕ ਹੀ ਖਿਡਾਰੀ ਜਾਂ ਸਾਈਡ ਦੁਆਰਾ ਅਨੁਕੂਮ ਵਿਚ ਸ਼ਟਲ ਨੂੰ ਦੋ ਵਾਰ ਹਿੱਟ ਕੀਤਾ ਜਾਂਦਾ ਹੈ ।

ਲੈਂਟ ਨਿਯਮ (Let Rule)-
ਅੰਪਾਇਰ ਦੁਆਰਾ ‘ਲੈਟ’ (Let) ਸ਼ਬਦ ਦਾ ਪ੍ਰਯੋਗ ਖੇਡਾਂ ਨੂੰ ਰੋਕਣ ਦੇ ਲਈ ਕੀਤਾ ਜਾਂਦਾ ਹੈ ।

1. ਜੇਕਰ ਸ਼ਟਲ ਨੈੱਟ ਵਿਚ ਫਸ ਜਾਂਦੀ ਹੈ ਜਾਂ ਨੈੱਟ ‘ਤੇ ਲਟਕੀ ਰਹਿੰਦੀ ਹੈ, ਤਾਂ ਸਰਵਿਸ ਦੇ ਇਲਾਵਾ ਇਸਨੂੰ ਨੈੱਟ ਕਿਹਾ ਜਾਵੇਗਾ ।
2. ਜੇਕਰ ਸਰਵਿਸ ਦੇ ਦੌਰਾਨ, ਰਸੀਵਰ ਅਤੇ ਸਰਵਰ ਨਾਲ-ਨਾਲ ਗਲਤੀ ਕਰਦੇ ਹਨ, ਤਾਂ ਇਸਨੂੰ “ਲੈਂਟ’ ਕਿਹਾ ਜਾਵੇਗਾ ।
3. ਜੇਕਰ ਖੇਡ ਦੇ ਦੌਰਾਨ, ਸਰਵਿਸ ਕੋਰਟ ਵਿਚ ਗਲਤੀ ਹੁੰਦੀ ਹੈ, ਤਾਂ ਇਸਨੂੰ “ਲੈਂਟ ਕਿਹਾ ਜਾਵੇਗਾ |
4. ਜੇਕਰ ਖੇਡ ਦੇ ਦੌਰਾਨ ਸ਼ਟਲ ਪੂਰੀ ਤਰ੍ਹਾਂ ਨਾਲ ਟੁੱਟ ਜਾਂਦੀ ਹੈ, ਤਾਂ ਇਸਨੂੰ ਲੈਂਟ ਕਿਹਾ ਜਾਵੇਗਾ ।

ਮਹੱਤਵਪੂਰਨ ਸ਼ਬਦਾਵਲੀ
(Important Terminology)

1. ਲੌਂਟ (Let) ਅਚਾਨਕ ਵਾਪਰਨ ਵਾਲੀਆਂ (Unfareseen) ਪ੍ਰਸਥਿਤੀਆਂ ਦੇ ਪਰਿਣਾਮਸਵਰੂਪ ਗੇਮ ਨੂੰ ਰੋਕਣ ਦੇ ਲਈ ਅੰਪਾਇਰ ਦੁਆਰਾ ਉਪਯੁਕਤ ਸ਼ਬਦ ਹੈ ।
2. ਰੈਲੀ (Rally) – ਗੇਮ ਵਿਚ ਸ਼ਟਲ ਦੇ ਰੁਕ ਜਾਣ ਤਕ, ਸਰਵਿਸ ਤੋਂ ਆਰੰਭ ਕਰਕੇ ਇਕ ਜਾਂ ਵਧੇਰੇ ਸਟੋਕਸ ਦਾ ਕੁਮ ।
3. ਸਰਵ (Serve) – ਹਰੇਕ ਰੈਲੀ ਦੇ ਸ਼ੁਰੂ ਵਿਚ, ਸ਼ਟਲ ਕਾਕ ਨੂੰ ਖੇਡ ਵਿਚ ਲਿਆਉਣ ਦੇ ਲਈ ਪ੍ਰਯੋਗ ਕੀਤਾ ਗਿਆ ਸਟਰੋਕ ।
4. ਵੁੱਡ ਸ਼ਾਟ (Wood Shot) – ਇਕ ਨਿਯਮ ਅਨੁਸਾਰ ਖੇਡਿਆ ਗਿਆ ਸ਼ਾਟ, ਜਿਸ ਵਿਚ ਸ਼ਟਲ ਰੈਕਟ ਦੇ ਫਰੇਮ ਨੂੰ ਸਪੱਰਸ਼ ਕਰਦੀ ਹੈ ।
5. ਗਲਤੀ (Fault) – ਸਰਵਿਸ, ਰਸੀਵਿੰਗ ਜਾਂ ਖੇਡ ਦੇ ਦੌਰਾਨ ਖੇਡਣ ਦੇ ਨਿਯਮਾਂ ਦਾ ਉਲੰਘਣ ।
6. ਸ਼ਾਰਟ ਸਰਵਿਸ ਲਾਈਨ (Short Service line)-ਨੈੱਟ ਤੋਂ 1.98 ਮੀ. ਦੀ ਦੂਰੀ ਤੇ ਇਕ ਰੇਖਾ, ਜਿਸ ਵਿਚ ਨਿਯਮ ਅਨੁਸਾਰ ਸਰਵ ਕਰਨ ਦੇ ਲਈ ਸਰਵ ਨੂੰ ਕਰਾਸ ਜ਼ਰੂਰ ਕਰਨਾ ਚਾਹੀਦਾ ਹੈ ।
7. ਡਿਉਸ (Deuce) – ਜਦੋਂ ਸਕੋਰ 20-20 ਤਕ ਪਹੁੰਚਦਾ ਹੈ, ਤਾਂ ਇਸ ਸ਼ਬਦ ਦਾ ਪ੍ਰਯੋਗ ਕੀਤਾ ਜਾਂਦਾ ਹੈ । ਡਿਉਸ ਦੇ ਮਾਮਲੇ ਵਿਚ ਗੇਮ ਜਿੱਤਣ ਦੇ ਲਈ 2 ਪੁਆਇੰਟ ਦੀ ਲੀਡ (Lead) ਲੈਣੀ ਜ਼ਰੂਰੀ ਹੁੰਦੀ ਹੈ ।
8. ਸਮੈਸ਼ (Smash) – ਇਹ ਸਿਵਰ ਦੇ ਉਪਰ ਜ਼ੋਰ ਨਾਲ ਮਾਰਿਆ ਗਿਆ ਸਰੋਕ ਹੁੰਦਾ ਹੈ, ਜਿਸ ਵਿਚ ਸ਼ਟਲ ਤੇਜ਼ੀ ਨਾਲ ਹੇਠਾਂ ਡਿੱਗਦੀ ਹੈ ।

ਖੇਡ ਦੇ ਮੈਦਾਨ/ਕੋਰਟ ਜਾਂ ਉਪਕਰਨਾਂ ਦੇ ਮਾਪ
(Dimensions of Play Field/Court or Equipments)

1. ਕੋਰਟ (Court) – ਸਿੰਗਲਜ਼ ਅਤੇ ਡਬਲਜ਼ ਦੋਵਾਂ ਦੇ ਲਈ ਬੈਡਮਿੰਟਨ ਦੇ ਕੋਰਟ ਦਾ ਆਕਾਰ ਆਇਤਾਕਾਰ ਹੁੰਦਾ ਹੈ, ਜਿਸਦੀ ਲੰਬਾਈ 13.4 ਮੀਟਰ (44 ਫੁੱਟ) ਹੁੰਦੀ ਹੈ । ਡਬਲਜ਼ ਦੇ ਲਈ ਕੋਰਟ ਦੀ ਚੌੜਾਈ 6.1 ਮੀਟਰ (20 ਫੁੱਟ) ਹੁੰਦੀ ਹੈ ਅਤੇ ਸਿੰਗਲਜ਼ ਦੇ ਲਈ ਇਹ ਘੱਟ ਹੋ ਕੇ 5.18 ਮੀਟਰ (17 ਫੁੱਟ) ਰਹਿ ਜਾਂਦੀ ਹੈ ।

2. ਪੋਸਟਸ (Posts) – ਪੋਸ਼ਟਸ ਕੋਰਟ ਦੇ ਧਰਾਤਲ ਤੋਂ 1.55 ਮੀ. ਉੱਚੇ ਹੁੰਦੇ ਹਨ | ਭਾਵੇਂ ਸਿੰਗਲਜ਼ ਜਾਂ ਡਬਲਜ਼ ਖੇਡਿਆ ਜਾਵੇ, ਪੋਸਟਸ ਨੂੰ ਡਬਲਜ਼ ਸਾਈਡ ਰੇਖਾਵਾਂ ‘ਤੇ ਲਗਾਇਆ ਜਾਂਦਾ ਹੈ ।

3. ਨੈੱਟ (Net) – ਨੈੱਟ ਵਧੀਆ ਰੱਸੀ ਜਾਂ ਕੇਬਲ (Cable) ਦੀ ਬਣੀ ਹੋਵੇਗੀ । ਇਹ ਗੂੜ੍ਹੇ ਰੰਗ ਦੀ ਹੋਣੀ ਚਾਹੀਦੀ ਹੈ ਜਿਸ ਦੀ ਜਾਲੀ 3/4″ ਤੋਂ 1” ਹੋਵੇਗੀ । ਨੈੱਟ ਦੀ ਚੌੜਾਈ 2’-6″ ਹੋਵੇਗੀ । ਨੈੱਟ ਦੀ ਉੱਚਾਈ ਮੱਧ ਤੋਂ ਜ਼ਮੀਨ ਤੋਂ 5 ਫੁੱਟ ਅਤੇ ਪੋਸਟਸ ਤੇ 5 ਫੁੱਟ 1 ਇੰਚ ਹੋਵੇਗੀ ।

4. ਰੈਕਟ (Racket) – ਰੈਕਟ ਦਾ ਫਰੇਮ ਰੈਕਟ ਦੀ ਸਹਾਇਤਾ ਨਾਲ ਜੁੜਿਆ ਹੋਵੇਗਾ । ਰੈਕਟ ਦੇ ਮੁੱਖ ਤੌਰ ਤੇ ਤਿੰਨ ਭਾਗ ਹੁੰਦੇ ਹਨ, ਸਿਰ, ਰੋਫਟ ਅਤੇ ਗਰਦਨ । ਸਿਰ ਲੰਬਾਈ ਵਿਚ 280 ਮਿ.ਮੀ. ਤੋਂ ਅਤੇ ਚੌੜਾਈ ਵਿਚ 220 ਮਿ.ਮੀ. ਤੋਂ ਅਧਿਕ ਨਹੀਂ ਹੋਵੇਗਾ | ਰੈਕਟ ਦੀ ਕੁੱਲ ਲੰਬਾਈ 680 ਮਿ.ਮੀ. ਅਤੇ ਚੌੜਾਈ 230 ਮਿ. ਮੀ. ਹੋਣੀ ਚਾਹੀਦੀ ਹੈ ।

5. ਸ਼ਟਲ (Shuttle) – ਸ਼ਟਲ ਆਮ ਜਾਂ ਸਿੰਥੈਟਿਕ ਪਦਾਰਥ ਦੀ ਬਣੀ ਹੋਵੇ । ਸ਼ਟਲ ਦਾ ਆਧਾਰ ਜਾਂ ਕਾਕ ਦਾ ਘੇਰਾ 25 ਤੋਂ 28 ਮਿ.ਮੀ. ਦੇ ਵਿਚ ਹੋਣਾ ਚਾਹੀਦਾ ਹੈ । ਖੰਭਾਂ (ਪੰਖਾਂ ਦੀ ਕੁੱਲ ਲੰਬਾਈ 62 ਤੋਂ 70 ਮਿ.ਮੀ. ਦੇ ਵਿਚ ਹੋਵੇਗੀ । ਸ਼ਟਲ ਦਾ ਭਾਰ 4.74 ਤੋਂ 5.50 ਗ੍ਰਾਮ ਹੋਣਾ ਚਾਹੀਦਾ ਹੈ ।

ਬੁਨਿਆਦੀ ਗੁਣ (Fundamental Skills)-
1. ਰੈਕਟ ਨੂੰ ਪਕੜਨਾ (Holding the Racket) – ਇਸ ਖੇਡ ਵਿਚ ਸਭ ਤੋਂ ਮਹੱਤਵਪੂਰਨ ਅਤੇ ਬੁਨਿਆਦੀ ਕਾਰੀਗਰੀ ਰੈਕਟ ਨੂੰ ਪਕੜਨਾ ਹੁੰਦਾ ਹੈ । ਰੈਕਟ ਨੂੰ ਪਕੜਦੇ ਸਮੇਂ ਕਲਾਈ ਕਠੋਰ (Stiff) ਨਹੀਂ ਹੋਣੀ ਚਾਹੀਦੀ । ਰੈਕੇਟ ਨੂੰ ਪਕੜਣ ਦੀਆਂ ਮੁੱਖ ਤੌਰ ‘ਤੇ ਦੋ ਸ਼ੈਲੀਆਂ ਹੁੰਦੀਆਂ ਹਨ-ਫਰਾਇੰਗ ਪੈਂਨ ਪਕੜ ਅਤੇ ਬੈਕ ਹੈੱਡ ਪਕੜ ।

2. ਸਰਵਿਸ (Service) – ਹਰੇਕ ਰੈਲੀ (Rally) ਦੇ ਸ਼ੁਰੂ ਵਿਚ ਸ਼ਟਲ ਨੂੰ ਖੇਡ ਵਿਚ ਲਿਆਉਣ ਵਾਲੇ ਸਟ੍ਰੋਕ ਨੂੰ ਸਰਵਿਸ ਕਹਿੰਦੇ ਹਨ । ਮੁੱਖ ਤੌਰ ‘ਤੇ ਦੋ ਤਰ੍ਹਾਂ ਦੀ ਸਰਵਿਸ ਹੁੰਦੀ ਹੈ-ਹਾਈ ਸਰਵਿਸ (High Service) ਅਤੇ ਲੋ ਸਰਵਿਸ (Low Service) । ਹਾਈ ਸਰਵਿਸ ਵਿਚ ਖਿਡਾਰੀ ਸ਼ਟਲ ਨੂੰ ਕੋਰਟ ਦੇ ਪਿੱਛੇ ਗਹਿਰਾਈ ਵਿਚ ਕਰਨ ਦੀ ਕੋਸ਼ਿਸ਼ ਕਰਦਾ ਹੈ । ਦੂਸਰੇ ਪਾਸੇ, ਲੋ ਸਰਵਿਸ ਵਿਚ ਸਿਰਫ ਨੈੱਟ ਨੂੰ ਪਾਰ ਕਰਨਾ ਹੁੰਦਾ ਹੈ ਅਤੇ ਸ਼ਾਰਟ ਸਰਵਿਸ ਲਾਈਨ ਤੋਂ ਕੁਝ ਇੰਚ ਦੂਰ ਰੱਖਣਾ ਹੁੰਦਾ ਹੈ ।

3. ਸਟ੍ਰੋਕ (Stroke) – ਸ਼ਟਲ ਅਤੇ ਰੈਕਟ ਦੇ ਵਿਚ ਸੰਪਰਕ ਨੂੰ ਸਕ ਕਹਿੰਦੇ ਹਨ । ਵਿਭਿੰਨ ਸਕਾਂ ਨੂੰ ਮੁੱਖ ਤੌਰ ‘ਤੇ ਤਿੰਨ ਵਰਗਾਂ ਵਿਚ ਵੰਡਿਆ ਜਾ ਸਕਦਾ ਹੈ : (ੳ) ਫਾਰਹੁੱਡ ਸਟੋਕ (Forhead Stroke) (ਅ ਬੈਕ ਹੈੱਡ ਸਟ੍ਰੋਕ (Backhand Stroke) (ਈ ਔਵਰ ਹੈਡ ਸਟਰੋਕ (Overhead Stroke) ।

• ਫਾਰਹੁੱਡ ਸਟੂਕ Forehead Stroke) – ਇਸ ਦਾ ਪ੍ਰਯੋਗ ਖੇਡ ਵਿਚ ਅਕਸਰ ਹੁੰਦਾ ਹੈ । ਇਸ ਸਟ੍ਰੋਕ ਨੂੰ ਤਦ ਖੇਡਿਆ ਜਾਂਦਾ ਹੈ, ਜਦ ਸ਼ਟਲ ਰਸੀਵਰ ਦੇ ਅੱਗੇ ਡਿੱਗਦੀ ਹੈ । ਵਿਰੋਧੀ ਦੇ ਕੋਰਟ ਦੇ ਕਿਸੇ ਵੀ ਪੁਆਇੰਟ ਤੇ ਸ਼ਟਲ ਨੂੰ ਮਾਰਨਾ ਆਸਾਨ ਹੁੰਦਾ ਹੈ ।
• ਬੈਕ ਹੈੱਡ ਸਟਰੋਕ (Backhand Stroke) – ਇਹ ਸਧਾਰਨਤਾ ਮੁਸ਼ਕਿਲ ਸ਼ਾਟ ਹੁੰਦਾ ਹੈ ਕਿਉਂਕਿ ਸ਼ਟਲ ਖਿਡਾਰੀ ਦੇ ਉਸ ਪਾਸੇ ਡਿੱਗਦੀ ਹੈ, ਜਿਸ ਪਾਸੇ ਗਤੀਸ਼ੀਲਤਾ ਨਹੀਂ ਹੁੰਦੀ । ਇਸ ਸ਼ਾਟ ਨੂੰ ਜ਼ੋਰਦਾਰ ਢੰਗ ਨਾਲ ਵਾਪਿਸ ਦੇਣਾ ਮੁਸ਼ਕਿਲ ਹੁੰਦਾ ਹੈ ।
• ਝਾਪ ਸ਼ਾਟ (Drop Shot) – ਪ ਸ਼ਾਟਸ ਕਮਜ਼ੋਰ ਬੈਡਮਿੰਟਨ ਸ਼ਾਟਸ ਹੁੰਦੇ ਹਨ, ਜਿਸ ਵਿਚ ਇਕ ਪੁਆਇੰਟ ਜਿੱਤਣ ਦੇ ਲਈ ਚਕਮਾ ਦੇਣ ਵਾਲੇ ਢੰਗ ਨਾਲ ਖੇਡਿਆ ਜਾਂਦਾ ਹੈ । ਇਸ ਸ਼ਾਟ ਦਾ ਉਦੇਸ਼ ਹੁੰਦਾ ਹੈ ਕਿ ਵਿਰੋਧੀ ਨੂੰ ਕਮਜ਼ੋਰ ਵਾਪਸੀ ਦੇ ਲਈ ਮਜ਼ਬੂਰ ਕੀਤਾ ਜਾਵੇ ।
• ਲੋਬ ਸ਼ਾਟ (Lob Shot) – ਇਸ ਵਿਚ ਸ਼ਟਲ ਨੂੰ ਵਿਰੋਧੀ ਦੀ ਆਧਾਰ ਲਾਈਨ ਤੇ ਉੱਚੇ ਅਤੇ ਡੂੰਘੇ ਵਿਚ ਹਿੱਟ ਕੀਤਾ ਜਾਂਦਾ ਹੈ । ਇਸ ਵਿਚ ਹਾਈ ਸਰਵ ਨੂੰ ਪੂਰਨ ਵਿਸਤਾਰ ਨਾਲ ਖੇਡਿਆ ਜਾਂਦਾ ਹੈ ।
• ਸਮੈਸ਼ (Smash-ਇਹ ਔਵਰ ਹੈੱਡ ਤੇ ਹਮਲਾ ਕਰਨ ਵਾਲਾ ਸਟ੍ਰੋਕ ਹੁੰਦਾ ਹੈ । ਜੋ ਸ਼ਟਲ ਨੂੰ ਵਿਰੋਧੀ ਦੇ ਕੋਰਟ ਵਿਚ ਤੇਜ਼ੀ ਨਾਲ ਹੇਠਾਂ ਦੇ ਵੱਲ ਡਿੱਗਣ ਦੇ ਲਈ ਮਜ਼ਬੂਰ ਕਰ ਦਿੰਦੀ ਹੈ ।
• ਹੇਅਰ ਪਿੰਨ ਸ਼ਾਟ (Hair Pin Shot) – ਅਸੀਂ ਸ਼ਾਟ ਵਿਚ, ਸ਼ਟਲ ਨੂੰ ਨੈੱਟ ਦੇ ਬਿਲਕੁਲ ਕੋਲੋਂ ਤੇਜ਼ ਗਤੀ ਨਾਲ ਵਾਪਿਸ ਭੇਜਿਆ ਜਾਂਦਾ ਹੈ । ਸ਼ਟਲ ਦੀ ਗਤੀਵਿਧੀ ਇਕ ਹੇਅਰ ਪਿਨ ਦੀ ਤਰ੍ਹਾਂ ਹੁੰਦੀ ਹੈ । ਜਿਹੜੀ ਨੈੱਟ ਦੇ ਦੂਸਰੇ ਪਾਸੇ ਬਹੁਤ ਨੇੜੇ ਹੀ ਡਿੱਗਦੀ ਹੈ ।

ਮਹੱਤਵਪੂਰਨ ਟੂਰਨਾਮੈਂਟਸ
(Important Tournaments)

ਅੰਤਰ-ਰਾਸ਼ਟਰੀ ਪੱਧਰ (International Levely)-

1. ਥਾਮਸ ਕੱਪ
2. ਵਿਸ਼ਵ ਕੱਪ
3. ਵਿਲਸ ਕੱਪ
4. ਚਾਈਨਾ ਕੱਪ
5. ਉਬੇਰ ਕੱਪ
6. ਸ਼ਾਹ ਜੀ ਕੁਰੈਸ਼ੀ ਕੱਪ
7. ਉਲੰਪਿਕ ਖੇਡ ।
8. ਕਾਮਨਵੈਲਥ ਖੇਡ
9. ਏਸ਼ਿਆਈ ਖੇਡ
10. ਐਲਬਾ ਵਿਸ਼ਵ ਕੱਪ
11. ਆਲ ਇੰਗਲੈਂਡ ਚੈਂਪੀਅਨਸ਼ਿਪ
12. ਯੋਨੈਕਸ ਕੱਪ । ਰਾਸ਼ਟਰੀ ਪੱਧਰ (National Level)-
13. ਸੀਨੀਅਰ ਨੈਸ਼ਨਲ ਚੈਂਪੀਅਨਸ਼ਿਪ
14. ਅੱਗਰਵਾਲ ਕੱਪ
15. ਅੰਮ੍ਰਿਤ ਦੀਵਾਨ ਕੱਪ
16. ਆਲ ਇੰਡੀਆ ਇੰਟਰਵਰਸਿਟੀ ਚੈਂਪੀਅਨਸ਼ਿਪ

ਸਪਰੋਟਸ ਅਵਾਰਡ
(Sports Award)

ਅਰਜੁਨ ਪੁਰਸਕਾਰ ਜੇਤੂਆਂ ਦੀ ਸੂਚੀ (List of Arjuna Award Winners) – ਨੰਦ ਐੱਮ. ਨਾਟੇਕਰ ( 1961), ਮੀਨਾ ਸ਼ਾਹ (1962), ਦਿਨੇਸ਼ ਖੰਨਾ ( 1965), ਸੁਰੇਸ਼ ਗੋਇਲ (1967), ਦੀਪੂ ਘੋਸ਼ ( 1969), ਡੀ. ਵੀ. ਤਾਬੇ (1970), ਸ਼ੋਭਾ ਮੂਰਥੀ (1971), ਪ੍ਰਕਾਸ਼ ਪਾਦੂਕੋਨੋ (1972), ਰਮਨ ਘੋਸ਼ (1974), ਦੇਵੇਂਦਰ ਆਹੂਜਾ (1975), ਅਮੀ ਘੀਆ (1976), ਕੰਵਲ ਠਾਕੁਰ ਸਿੰਘ (1977-78), ਸੈੱਯਦ ਮੋਦੀ (1980-81), ਪਾਰਥੇ ਗਾਂਗੁਲੀ, ਕੁਮਾਰੀ ਮਧੂਮੀਤਾ ਗੋਸਵਾਮੀ (1981), ਰਾਜੀਵ ਬੱਗਾ (1991), ਅਰਪਣਾ ਪੋਪਟ (1999), ਵੀ.ਗੋਪੀਚੰਦ (2000), ਜਾਰਜ ਥਾਮਸ (2001), ਪਾਰੂਲ ਡੀ ਪਰਮਾਰ (2009), ਸੇਨਾ ਨੇਹਵਾਲ (2009), ਜਵਾਲਾ ਗੁੱਟਾ (2011) ।
ਦਰੋਣਾਚਾਰੀਆ ਅਵਾਰਡ-ਪੀ. ਗੋਪੀ ਚੰਦ (2009) ।

ਬੈਡਮਿੰਟਨ
(Badminton)

ਅਰਜੁਨ ਅਵਾਰਡ ਵਿਜੇਤਾ-ਨੰਦੂ ਨਾਟੇਕਰ (1961), ਮੀਨਾ ਸ਼ਾਹ (1962), ਦੀਨੇਸ਼ ਖੰਨਾ (1965), ਸੁਰੇਸ਼ ਗੋਇਲ (1967), ਦੀਪੂ ਘੋਸ਼ (1969), ਡੀ.ਵੀ. ਟਮਬੇ (1970), ਐੱਸ. ਊਰਥੀ (1971), ਪ੍ਰਕਾਸ਼ ਪਾਦੂਕੋਨ (1972), ਰਮਨ ਘੋਸ਼ (1974), ਦਵਿੰਦਰ ਅਹੂਜਾ (1975), ਅਮੀ ਘੀਆ (1976), ਕੇ.ਟੀ. ਸਿੰਘ (1977-78), ਸੱਯਦ ਮੋਦੀ (1980-81), ਪੀ. ਗਾਂਗੁਲੀ, ਮਧੁਮਿਤਾ ਬਿਸ਼ਟ (1982), ਰਾਜੀਵ ਬੱਗਾ (1991), ਪੁਲੇਲਾ ਗੋਪੀਚੰਦ (2000), ਜਾਰਜ ਥੋਮਸ (2000-01), ਮਦਾਸੂ ਸ੍ਰੀਨਿਵਾਸ ਰਾਓ (ਫਿਜ਼ਿਕਲੀ ਚੈਲੇਂਜਡ) (2003), ਅਭਿੰਨ ਸ਼ਾਮ ਗੁਪਤਾ (2004), ਅਪਰਨਾ ਪੋਪਟ (2005), ਚੇਤਨ ਆਨੰਦ (2006), ਰੋਹੀਤ ਭਾਕਰ (ਫਿਜ਼ਿਕਲੀ ਚੈਲੇਂਜਡ) (2006), ਅਨੂਪ ਸ਼੍ਰੀਧਰ (2008), ਸਾਇਨਾ ਨੇਹਵਾਲ (2009), ਅਸ਼ਵਨੀ ਪੋਨਅੱਪਾ, ਪਾਰੂਪਲੀਕੇਸ਼ਅਪ (2012), ਪੀ.ਵੀ. ਸਿੰਧੂ (2013), ਵੀ. ਦੀਜੂ (2014), ਕੇ. ਸੀਰੀਕਨਾਥ (2015)
ਦਰੋਣਾਚਾਰੀਆ ਅਵਾਰਡ ਵਿਜੇਤਾ-ਐੱਸ. ਐੱਮ. ਆਰਿਫ (2000), ਪੁਲੇਲਾ ਗੋਪੀਚੰਦ (2009) ।

PSEB 12th Class Physical Education Practical ਬੈਡਮਿੰਟਨ (Badminton)

ਪ੍ਰਸ਼ਨ 1.
ਅੰਤਰ-ਰਾਸ਼ਟਰੀ ਬੈਡਮਿੰਟਨ ਫੈਡਰੇਸ਼ਨ-ਕਦੋਂ ਸ਼ੁਰੂ ਹੋਇਆ ਸੀ ?
ਉੱਤਰ-
ਸਾਲ 1934 ਵਿਚ ।

ਪ੍ਰਸ਼ਨ 2.
ਉਲੰਪਿਕ ਖੇਡਾਂ ਵਿਚ ਬੈਡਮਿੰਟਨ ਨੂੰ ਇਕ ਪਦਕ ਖੇਡ ਦੇ ਰੂਪ ਵਿਚ ਕਦੋਂ ਸਵੀਕਾਰ ਕੀਤਾ ਗਿਆ ਸੀ ?
ਉੱਤਰ-
1992 ਉਲੰਪਿਕ ਖੇਡ ਬਰਸੀਲੋਨਾ ਵਿਚ ਇਹ ਇਕ ਪਦਕ ਖੇਡ ਬਣਿਆ ।

ਪ੍ਰਸ਼ਨ 3.
ਡਬਲਜ਼ ਦੇ ਲਈ ਬੈਡਮਿੰਟਨ ਕੋਰਟ ਦੇ ਪੜਾਅ ਕੀ ਹਨ ?
ਉੱਤਰ-
13.00 × 6.10 ਮੀ. ਜਾਂ 4 × 20 ਫੁੱਟ ।

ਪ੍ਰਸ਼ਨ 4.
ਨੈੱਟ ਦੀ ਚੌੜਾਈ ਕਿੰਨੀ ਹੁੰਦੀ ਹੈ ?
ਉੱਤਰ-
760 ਕਿ.ਮੀ. (76 ਸੈਂ.ਮੀ.)

ਪ੍ਰਸ਼ਨ 5.
ਇਕ ਸ਼ਟਲ ਵਿਚ ਪੰਖਾਂ ਦੀ ਸੰਖਿਆ ਦੱਸੋ ।
ਉੱਤਰ-
14 ਤੋਂ 16 ਪੰਖ ।

ਪ੍ਰਸ਼ਨ 6.
ਬੈਡਮਿੰਟਨ ਮੈਚ ਦੇ ਲਈ ਕਿੰਨੇ ਅਧਿਕਾਰੀਆਂ ਦੀ ਜ਼ਰੂਰਤ ਹੁੰਦੀ ਹੈ ?
ਉੱਤਰ-
1. ਅੰਪਾਇਰ, ਸਰਵਿਸ ਅੰਪਾਇਰ, 11 ਰੈਫ਼ਰੀ ਅਤੇ 10 ਲਾਈਨਮੈਨ ।

ਪ੍ਰਸ਼ਨ 7.
ਬੈਡਮਿੰਟਨ ਵਿਚ ਟੱਸ ਦੇ ਬਾਰੇ ਵਿਚ ਤੁਸੀਂ ਕੀ ਜਾਣਦੇ ਹੋ ?
ਉੱਤਰ-
ਖੇਡ ਦੇ ਸ਼ੁਰੂ ਹੋਣ ਤੋਂ ਪਹਿਲਾਂ ਟਾਸ ਕੀਤਾ ਜਾਂਦਾ ਹੈ ਅਤੇ ਜਿੱਤੇ ਪਾਸੇ ਦੇ ਕੋਲ ਪਹਿਲੇ ਸਰਵ ਕਰਨ ਜਾਂ ਰਸੀਵ ਕਰਨ ਦਾ ਵਿਕਲਪ ਹੁੰਦਾ ਹੈ ।

ਪ੍ਰਸ਼ਨ 8.
ਮੱਧ ਤੋਂ ਸ਼ਾਰਟ ਸਰਵਿਸ ਲਾਈਨ ਦੀ ਦੂਰੀ ਕਿੰਨੀ ਹੁੰਦੀ ਹੈ ?
ਉੱਤਰ-
6′-6″ (1.98 ਮੀ.) ।

ਪ੍ਰਸ਼ਨ 9.
ਡਿਊਸ (Deuce-ਸ਼ਬਦ ਤੋਂ ਤੁਸੀਂ ਕੀ ਸਮਝਦੇ ਹੋ ? ਉੱਤਰ-ਇਸ ਸ਼ਬਦ ਦਾ ਪ੍ਰਯੋਗ ਤਦ ਕੀਤਾ ਜਾਂਦਾ ਹੈ, ਜਦੋਂ ਸਕੋਰ 20-20 ਤਕ ਪਹੁੰਚਦਾ ਹੈ ।ਡਿਊਸ ਦੇ ਮਾਮਲੇ ਵਿਚ ਗੇਮ ਜਿੱਤਣ ਦੇ ਲਈ ਅਤੇ 2 ਪੁਆਇੰਟ ਸਕੋਰ ਕਰਨੇ ਜ਼ਰੂਰੀ ਹੁੰਦੇ ਹਨ ।

ਪ੍ਰਸ਼ਨ 10.
ਪੋਸਟਸ (Posts) ਦੀ ਉੱਚਾਈ ਕਿੰਨੀ ਹੁੰਦੀ ਹੈ ? ਉੱਤਰ-ਪੋਸਟਸ ਕੋਰਟ ਦੇ ਧਰਾਤਲ ਤੋਂ 1.55 ਮੀ. ਉੱਚੇ ਹੁੰਦੇ ਹਨ ।

ਪ੍ਰਸ਼ਨ 11.
ਬੈਡਮਿੰਟਨ ਵਿਚ ਮਹੱਤਵਪੂਰਨ ਅੰਤਰਰਾਸ਼ਟਰੀ ਪੱਧਰ ਦੇ ਟੂਰਨਾਮੈਂਟਸ ਦੇ ਨਾਂ ਦੱਸੋ ।
ਉੱਤਰ-
ਥਾਮਸ ਕੱਪ, ਵਿਸ਼ਵ ਕੱਪ, ਵਿਲਸ ਵਿਸ਼ਵ ਕੱਪ, ਚਾਈਨਾ ਕੱਪ, ਉਬੇਰ ਕੱਪ, ਉਲੰਪਿਕ ਖੇਡ, ਏਸ਼ਿਆਈ ਖੇਡ, ਕਾਮਨਵੈਲਥ ਖੇਡ, ਆਲ ਇੰਡੀਆ ਚੈਂਪੀਅਨਸ਼ਿਪ, ਯੋਮੈਕਸ ਕੱਪ ।