Class 11

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.3

Question 1.
For each of the following compound statements first identify the connecting words and then break it into component statements.
(i) All rational numbers are real and all real numbers are not complex.
(ii) Square of an integer is positive or negative.
(iii) The sand heats up quickly in the Sun and does not cool down fast at night.
(iv) x = 2 and x = 3 are the roots of the equation 3x² – x – 10 = 0.
Answer.
(i) Here, the connecting word is ‘and’.
The component statements are as follows.
p : All rational numbers are real.
q : All real numbers are not complex.

(ii) Here, the connecting word is ‘or’.
The component statements are as follows,
p : Square of an integer is positive.
q : Square of an integer is negative.

(iii) Here, the connecting word is ‘and’.
The component statements are as follows.
p : The sand heats up quickly in the sun.
q : The sand does not cool down fast at night.

(iv) Here, the connecting word is ‘and’.
The component statements are as follows.
p : x = 2 is a root of the equation 3x² – x -10 = 0
q : x = 3 is a root of the equation 3x² – x -10 = 0

Question 2.
Identify the quantifier in the following statements and write the negation of the statements.
(i) There exists a number which is equal to its square.
(ii) For every real numbers, x is less than x + 1.
(iii) There exists a capital for every state in India.
Answer.
(i) Quantifier : There exists.
p : There exists a number which is equal to its square
not p : There does not exist a number which is equal to its square.

(ii) Quantifier : For every
p : For every real number x, x is less than x + 1
~p : For every real number x, x is not less than x + 1

(iii) Quantifier : There exists
p : There exists a capital for every state of India.
~ p : There does not exist a capital for every state of India.

Question 3.
Check whether the following pair of statements are negation of each other. Give reasons for the answer.
(i) x + y = y + x is true for every real numbers x andy.
(ii) There exists real number x and y for which x + y = y + x.
Answer.
Let p: x + y = y + x is true for every real numbers x and y.
q : There exists real numbers x and y for which x + y = y + x.
Now ~ p : There exists real numbers x and y for which x + y ≠ y + x. Thus ~ p ≠ q.

Question 4.
State whether the “Or” used in the following statements is exclusive “or” inclusive. Give reasons for your answer.
(i) Sun rises or Moon sets.
(ii) To apply for a driving licence, you should have a ration card or a passport.
(iii) All integers are positive or negative.
Answer.
(i) Here, “or” is exclusive because it is not possible for the Sun to rise and the Moon to set together.
(ii) Here, “or” is inclusive since a person can have both a ration card and a passport to apply for a driving licence.
(iii) Here, “or” is exclusive because all integers cannot be both positive and negative.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Ex 14.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.2

Question 1.
Write the negation of the following statements:
(i) Chennai is the capital of Tamil Nadu.
(ii) √2 is not a complex number.
(iii) All triangles are not equilateral triangle.
(iv) The number 2 is greater than 7.
(v) Every natural number is an integer.
Answer.
The negation of the given statements is as follows:
(i) Chennai is not the capital of Tamil Nadu.
(ii) √2 is a complex number.
(iii) All triangles are equilateral triangles.
(iv) The number 2 is not greater than 7.
(v) Every natural number is not an integer.

Question 2.
Are the following pairs of statements negations of each other?
(i) The number x is not a rational number.
The number x is not an irrational number.
(ii) The number x is a rational number.
The number x is an irrational number.
Answer.
(i) The negation of the first statement:
The number x is “a rational number”.
Which is the same as the second statement.
This is because when a number is statement not rational, it is rational.
Therefore, given statements are negation of each other.

(ii) The negation of the first statement:
The number x is an irrational number.
The second statement, which is the same as the second statement.
Therefore they are negation of each other.

Question 3.
Find the component statements of the following compound statements and check whether they are true or false.
(i) Number 3 is prime or it is odd.
(ii) All integers are positive or negative.
(iii) 100 is divisible by 3,11 and 5.
Answer.
(i) The component statements are as follows.
p : Number 3 is prime.
q : Number 3 is odd.
Both the statements are true.

(ii) The component statements are as follows.
p : All integers are positive.
q : All integers are negative.
Both the statements are false.

(iii) The component statements are as follows.
p : 100 is divisible by 3.
q : 100 is divisible by 11.
r : 100 is divisible by 5.
Here, the statements, p and q, are false and statement r is true.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 14 Mathematical Reasoning Ex 14.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 14 Mathematical Reasoning Ex 14.1

Question 1.
Which of the following sentences are statements? Give reasons for your answer.
(i) There are 35 days in a month.
(ii) Mathematics is difficult.
(iii) The sum of 5 and 7 is greater than 10.
(iv) The square of a number is an even number.
(v) The side of a quadrilateral have equal length.
(vi) Answer this question.
(vii) The product of (-1) and 8 is 8.
(viii) The sum of all interior angles of a triangle is 180°.
(ix) Today is a windy day.
(x) All real numbers are complex numbers.
Answer.
(i) This sentence is incorrect because the maximum number of days in a month is 31. Hence, it is a statement.
(ii) This sentence is subjective in the sense that for some people, mathematics can be easy and for some others, it can be difficult. Hence, it is not a statement.
(iii) The sum of 5 and 7 is 12, which is greater than 10. Therefore, this sentence is always correct. Hence, it is a statement.
(iv) This sentence is sometimes correct and sometimes incorrect. For example, the square of 2 is an even number. However, the square of 3 is an odd number. Hence, it is not a statement.
(v) This sentence is sometimes correct and sometimes incorrect. For example, squares and rhombus have sides of equal lengths. However, trapezium and rectangles have sides of unequal lengths. Hence, it is not a statement.
(vi) It is an order. Therefore, it is not a statement.
(vii) The product of (- 1) and 8 is (- 8). Therefore, the given sentence is incorrect. Hence, it is a statement.
(viii) This sentence is correct and hence, it is a statement.
(ix) The day that is being referred to is not evident from the sentence. Hence, it is not a statement.
(x) All real numbers can be expressed as a + ib. Therefore, the given sentence is always correct. Hence, it is a statement.

Question 2.
Give three examples of sentences which are not statements. Give reasons for the answers.
Answer.
The three examples of sentences, which are not statements, are as follows:
(i) He is a doctor.
It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement.
(ii) Geometry is difficult.
This is not a statement because for some people, geometry can be easy and for some others, it can be difficult.
(iii) Where is she going?
This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is not a statement.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 13 Limits and Derivatives Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Miscellaneous Exercise

Question 1.
Find the derivative of the following functions from first principle:
(i) – x
(ii) (- x)^{– 1}
(iii) sin (x + 1)
(iv) cos (x – \(\frac{\pi}{8}\))
Answer.
(i) Let f(x) = – x. Accordingly, f(x + h) = – (x + h)
By first prinicple,
f’(x) = \(\lim -{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{-(x+h)-(-x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{-x-h+x}{h}=\lim -{h \rightarrow 0} \frac{-h}{h}\)

= \(\lim -{h \rightarrow 0}\) (- 1) = – 1.

(ii) Let f(x) = (- x)^{– 1} = \(=\frac{1}{-x}=\frac{-1}{x}\)
Accordingly, f(x + h) = =\frac{1}{-x}=\frac{-1}{x}
By first prinicple,

(iii) Let f(x) = sin (x +1).
Accordingly, f(x + h) = sin (x + h + 1)
By first prinicple,
f’(x) = \(\lim -{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{1}{h}\) [sin (x + h + 1) – sin (x + 1)]

= \(\lim -{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{x+h+1+x+1}{2}\right) \sin \left(\frac{x+h+1-x-1}{2}\right)\right]\)

(iv) Let f(x) = cos (x – \(\frac{\pi}{8}\))
By using first principle of derivative,
We have,
f'(x) = \(\lim -{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

Question 2.
Find the derivative of the following functions (it is to be understood that a, b, e, d, p, q, r and s are fixed non-zero constants and m and n are integers): (x + a)
Answer.
Let f(x) = x + a.
Accordingly, f(x + h) = x + h + a
By first prinicple,
f’(x) = \(\lim -{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{x+h+a-x-a}{h}=\lim -{h \rightarrow 0}\left(\frac{h}{h}\right)\)

= \(\lim -{h \rightarrow 0}\) (1) = 1.

Question 3.
(Px + q) (\(\frac{r}{x}\) + s).
Answer.
Let f(x) = (Px + q) (\(\frac{r}{x}\) + s)
By Leibnitz product rule,
f’(x) = (px + q) (\(\frac{r}{x}\) + s) + (\(\frac{r}{x}\) + s) (px + q)’
= (px + q)(rx^{– 1} + s)’ + (\(\frac{r}{x}\) + s) (p)
= (px + q) (- rx^{– 2}) + (\(\frac{r}{x}\) + s) p
= (px + q) \(\left(\frac{-r}{x^{2}}\right)\) + (\(\frac{r}{x}\) + s) p
= \(\frac{-p r}{x}-\frac{q r}{x^{2}}+\frac{p r}{x}\) + ps
= ps – \(\frac{q r}{x^{2}}\).

Question 4.
(ax + b) (cx + d)².
Answer.
Let f(x) = (ax + b) (cx + d)²
By Leibnitz product rule,
f(x) = (ax + b) \(\frac{d}{d x}\) (cx + d)² + (cx + d)² \(\frac{d}{d x}\) (ax + b)
= (ax + b) \(\frac{d}{d x}\) (c²x² + 2cdx + d²) + (cx + d)² \(\frac{d}{d x}\) (ax + b)
= (ax + b) [\(\frac{d}{d x}\) (c²x²) + \(\frac{d}{d x}\) (2cdx) + \(\frac{d}{d x}\) d²] + (cx + d)² [\(\frac{d}{d x}\) ax + \(\frac{d}{d x}\) b]
= (ax + b) (2c²x + 2cd) + (cx + d²) a
= 2c (ax + b) (cx+ d) + a (cx + d)².

Question 5.
\(\frac{a x+b}{c x+d}\)
Answer.

Question 6.
\(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\)
Answer.

Question 7.
\(\frac{1}{a x^{2}+b x+c}\)
Answer.
Let f(x) = \(\frac{1}{a x^{2}+b x+c}\)
By quotient rule,
f'(x) = \(\frac{\left(a x^{2}+b x+c\right) \frac{d}{d x}(1)-\frac{d}{d x}\left(a x^{2}+b x+c\right)}{\left(a x^{2}+b x+c\right)^{2}}\)

= \(\frac{\left(a x^{2}+b x+c\right)(0)-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}\)

= \(\frac{-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}\)

Question 8.
\(\frac{a x+b}{p x^{2}+q x+r}\)
Answer.

Question 9.
\(\frac{p x^{2}+q x+r}{a x+b}\)
Answer.

Question 10.
\(\frac{a}{x^{4}}-\frac{b}{x^{2}}\) + cos x
Answer.

Question 11.
4√x – 2
Answer.
Let f(x) = 4√x – 2
f'(x) = \(\frac{d}{d x}\) (4√x – 2)
= \(\frac{d}{d x}\) (4√x) – \(\frac{d}{d x}\) (2)
= 4 \(\frac{d}{d x}\) (x^{\(\frac{1}{2}\)}) – 0
= 4 \(\left(\frac{1}{2} x^{\frac{1}{2}-1}\right)\)
= \(\left(2 x^{-\frac{1}{2}}\right)=\frac{2}{\sqrt{x}}\)

Question 12.
(ax + b)ⁿ
Answer.
Let f(x) = (ax + b)ⁿ
Accordingly, f(x + h) = {a(x + h) + b}ⁿ
= (ax + ah + b)ⁿ
By first principle,
f'(x) = \(\lim -{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{(a x+a h+b)^{n}-(a x+b)^{n}}{h}\)

Question 13.
(ax + b)ⁿ (cx + d)^m.
Answer.
Let f(x) = (ax + b)ⁿ (cx + d)^m
By Leibnitz product rule,

= \(\frac{m c(c x+d)^{m}}{(c x+d)}\)

= mc (cx + d)^{m – 1}
\(\frac{d}{d x}\) (cx + d)^m = mc (cx + d)^{m – 1} …………….(ii)
Similarly, (ax + b)ⁿ = na(ax + b)^{n – 1} ………….(iii)
Therefore, from eqs. (i), (ii) and (iii), we obtain
f'(x) = (ax + b)ⁿ {mc (cx + d)^{m – 1}} + (cx + d)^m {na(ax + b)^{n – 1}}
= (ax + b)^{n – 1} (a + d)^{m – 1} [mc (ax + b) + na (cx + d)].

Question 14.
sin (x + a).
Answer.
Let f(x) = sin (x + a), f(x + h) = sin (x + h + a)
By first principle,

Question 15.
cosec x cot x
Answer.
Let f(x) = cosec x cot x
By Leibnitz product rule,
f’(x) = cosec x(cot x’ + cot x (cosec x)’ ……………(i)
Let f₁(x) = cot x,
Accordingly, f₁(x + h) = cot(x + h)
By first prinicple,
f₁‘(x) = \(\lim -{h \rightarrow 0} \frac{f-{1}(x+h)-f-{1}(x)}{h}\)

= \(\lim -{h \rightarrow 0} \frac{\cot (x+h)-\cot x}{h}\)

= \(\lim -{h \rightarrow 0} \frac{1}{h}\left(\frac{\cos (x+h)}{\sin (x+h)}-\frac{\cos x}{\sin x}\right)\)

Question 16.
\(\frac{\cos x}{1+\sin x}\)
Answer.

Question 17.
\(\frac{\sin x+\cos x}{\sin x-\cos x}\)
Answer.

Question 18.
\(\frac{\sec x-1}{\sec x+1}\)
Answer.
Let f(x) = \(\frac{\sec x-1}{\sec x+1}\)

Question 19.
sinⁿ x.
Ans.
Let y = sinⁿ x
Accordingly, for n = 1, y = sin x.
∴ \(\frac{d y}{d x}\) = cos x, i.e., \(\frac{d}{d x}\) sin x = cos x
For n = 2, y = sin² x
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (sin x sin x)
= (sin x)’ sin x + sin x (sin x)’ [by Leibnitz product rule]
= cos x sin x + sin x cos x
= 2 sin x cos x ………………..(i)
For n = 3, y = sin³ x
∴ \(\frac{d y}{d x}\) = \(\frac{d}{d x}\) (sin x sin² x)
= (sin x)’ sin² x + sin x (sin² x)’ [by Leibnitz product rule]
= cos x sin² x + sin x(2 sin x cos x) [using eq. (i)]
= cos x sin² x + 2 sin² x cos x
= 3 sin² x cos x
We assert that \(\frac{d}{d x}\) (sinⁿ x) = n sin^{(n – 1)} x cos x
Let our assertion be true for n = k.
ie., \(\frac{d}{d x}\) (sin^k x) = k sin^{(k – 1)} x cos x …………….(ii)
Consider, (sin^{k + 1} x) = \(\frac{d}{d x}\) (sin x sin^k x)
= (sin x)’ sin^k x + sin x (sin^k x)’ [by Leibnitz product rule]
= cos x sin^k x + sin x (k sin^{(k – 1)} x cos x) [using eq. (ii)]
= cos x sin^k x + k sin^k x cos x
= (k + 1) sin^k x cos x
Thus, our assertion is true for n = k + 1.
Hence, by mathematical induction, \(\frac{d}{d x}\) (sinⁿ x) = n sin^{(n – 1)} x cos x.

Question 20.
\(\frac{a+b \sin x}{c+d \cos x}\)
Answer.

Question 21.
\(\frac{\sin (x+a)}{\cos x}\)
Answer.

Question 22.
x⁴ (5 sin x – 3 cos x)
Answer.
Let f(x) = x⁴ (5 sin x – 3 cos x)
By product rule,
f’(x) = x⁴ \(\frac{d}{d x}\) (5 sin x – 3 cos x) + (5 sin x – 3 cos x) \(\frac{d}{d x}\) (x⁴)
= x⁴ [5 \(\frac{d}{d x}\) (sin x) – 3 (cos x)] + (5 sin x – 3 cos x) (4x³)
= x⁴[5 cos x – 3(- sin x)] + (5 sin x – 3 cos x) (4x³)
= x³ [ 5x cos x + 3x sin x +20 sin x – 12 cos x].

Question 23.
(x² + 1) cos x.
Answer.
Let f(x) = (x² + 1) cos x
By product rule,
f’(x) = (x² + 1) \(\frac{d}{d x}\) (cos x) + cos x \(\frac{d}{d x}\) (x² + 1)
= (x² + 1) (- sin x) + cos x (2x)
= – x² sin x – sin x + 2x cos x.

Question 24.
(ax² + sin x) (p + q cos x).
Answer.
Let f(x) = (ax² + sin x) (p + q cos x)
By product rule,
f’(x) = (ax² + sin x) (p + q cos x) + (p + q cos x) (ax² + sin x)
= (ax² + sin x)(- q sin x) + (p + q cos x)(2ax + cos x)
= – q sin x (ax² + sin x) + (p + q cos x) (2ax + cos x).

Question 25.
(x + cos x) (x – tan x).
Answer.
Let y = (x + cos x) (x – tan x)
On differentiating both sides w.r.t. x, we get
ciy d d
= (x + cos x) . \(\frac{d}{d x}\) (x – tan x) + (x – tan x) (x + cos x)
[∵ \(\frac{d}{d x}\) (u . v) = u \(\frac{d v}{d x}\) + v \(\frac{d u}{d x}\)]
= (x + cos x) (1 – sec² x) + (x – tan x) (1 – sin x)
= (x + cos x) (- tan 2x) + (x – tan x) (1 – sin x)

Question 26.
\(\frac{4 x+5 \sin x}{3 x+7 \cos x}\)
Answer

Question 27.
\(\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}\)
Answer.
Let f(x) = \(\frac{x^{2} \cos \left(\frac{\pi}{4}\right)}{\sin x}\)

By quotient rule, f'(x) = \(\cos \frac{\pi}{4} \cdot\left[\frac{\sin x \frac{d}{d x}\left(x^{2}\right)-x^{2} \frac{d}{d x}(\sin x)}{\sin ^{2} x}\right]\)

= \(\cos \frac{\pi}{4} \cdot\left[\frac{\sin x \cdot 2 x-x^{2} \cos x}{\sin ^{2} x}\right]\)

= \(\frac{x \cos \frac{\pi}{4}[2 \sin x-x \cos x]}{\sin ^{2} x}\)

Question 28.
\(\frac{x}{1+\tan x}\)
Answer.

Question 29.
(x + secx)(x-tanx).
Answer.
Let f(x) = (x + sec x) (x – tan x)
By product rule,
f’(x) = (x + sec x) \(\frac{d}{d x}\) (x – tan x) + (x – tan x) \(\frac{d}{d x}\) (x + sec x)
= (x + sec x) [\(\frac{d}{d x}\) (x) – \(\frac{d}{d x}\) tan x] + (x – tan x) [\(\frac{d}{d x}\) (x) + \(\frac{d}{d x}\) sec x]
= (x + sec x) [1 – \(\frac{d}{d x}\) tan x] + (x – tan x) [1 + \(\frac{d}{d x}\) sec x] ……………..(i)
Let f₁(x) = tan x, f₂(x) = sec x
Accordingly, f₁ (x + h) = tan (x + h) and f₂ (x + h) = sec (x + h)

From eqs. (i), (ii) and (iii), we obtain
f'(x) = (x + sec x) (1 – sec² x) + (x – tan x)(1 + sec x tan x).

Question 30.
\(\frac{x}{\sin ^{n} x}\)
Answer.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 13 Limits and Derivatives Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.2

Question 1.
Find the derivative of x² – 2 at x = 10.
Answer.
We have, f(x) = x² – 2
By using first principle of derivative,
f'(10) = \(\lim _{h \rightarrow 0} \frac{f(10+h)-f(10)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{\left[(10+h)^{2}-2\right]-\left[(10)^{2}-2\right]}{h}\)

= \(\lim _{h \rightarrow 0} \frac{100+20 h+h^{2}-2-100+2}{h}\)

= \(\lim _{h \rightarrow 0} \frac{h^{2}+20 h}{h}\)

= \(\lim _{h \rightarrow 0}\) (h + 20) = 20.

Question 2.
Find the derivative of 99x at x = 100.
Answer.
Let f(x) = 99x. Accordingly,
f'(100) = \(\lim _{h \rightarrow 0} \frac{f(100+h)-f(100)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{99(100+h)-99(100)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{99 \times 100+99 h-99 \times 100}{h}\)

= \(\lim _{h \rightarrow 0} \frac{99 h}{h}\)

= \(\lim _{h \rightarrow 0}\) (99) = 99
Thus, the derivative of 99x at x = 100 is 99.

Question 3.
Find the derivative of x at x = 1.
Answer.
Let f(x) = x. Accordingly,
f(1) = \(\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}\)

= \(\lim _{h \rightarrow 0} \frac{(1+h)-1}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=\lim _{h \rightarrow 0}(1)\) = 1.
Thus, the derivative of x at x = 1 is 1.

Question 4.
Find the derivative of the following functions from first principle.
(i) x³ – 27
(ii) (x – 1) (x – 2)
(iii) \(\frac{1}{x^{2}}\)
(iv) \(\frac{x+1}{x-1}\)
Answer.
(i) We have, f(x) = x³ – 27
By using first principle of derivative,

(ii) We have, f(x) = (x – 1) (x – 2) = x² – 3x + 2
By first principle of derivative, we have

(iii) Let f(x) = \(\frac{1}{x^{2}}\).
Accordingly, from the first, principle,

(iv) Let f(x) = \(\frac{x+1}{x-1}\)
Accordingly, from the first principle,

Question 5.
For the function f(x) = \(\). Prove that f'(1) = 100 f'(0).
Answer.

At x = 0, f'(0) = 1
At x = 1, f'(1) = 1⁹⁹ + 1⁹⁸ + ………….. + 1 + 1
= [1 + 1 + 1 + …………… + 1 + 1]_{100 terms}
= 1 × 100 = 100.

Question 6.
Find the derivative of xⁿ + ax^{n – 1} + a² x^{n – 2} + ………….. + a^{n – 1} x + aⁿ for some fixed real number a.
Answer.
Let f(x) = xⁿ + ax^{n – 1} + a² x^{n – 2} + ………….. + a^{n – 1} x + aⁿ
On differentiating both sides, we get
f'(x) = nx^{n – 1} + a(n – 1)x^{n – 2} + a² (n – 2)x^{n – 3} + …………. + a^{n – 1} . 1 + 0
On putting x = a both sides, we get
f'(a) = na^{n – 1} + a (n – 1) x^{n – 2} + a² (n – 2) a^{n – 3} + ……………. + a^{n – 1}
= n a^{n – 1} + (n – 1) a^{n – 1} + (n – 2) a^{n – 1} + ……………. + a^{n – 1}
= a^{n – 1} [n + (n – 1) + (n – 2) + ………….. + 1]
[∵ sum of n natural numbers = \(\frac{n(n+1)}{2}\)]
= \(a^{n-1} \frac{n(n+1)}{2}\).

Question 7.
For some constants a and b, find the derivative of
(i) (x – a) (x – b)
(ii)(ax +b)
(iii) \(\frac{x-a}{x-b}\)
Ans.
(i) Let f(x) = (x – a) (x – b)
f(x)= x² – (a + b) x + ab
∴ f'(x) = \(\frac{d}{d x}\) [x² – (a + b) x + ab]
= \(\frac{d}{d x}\) (x²) – (a + b) \(\frac{d}{d x}\)(x) + \(\frac{d}{d x}\) (ab)
On using theorem \(\frac{d}{d x}\) (xⁿ) = nx^{n – 1}, we obtain
f'(x) = 2x – (a + b) + 0 = 2x – a – b.

(ii) Let f(x) = (ax² + b)²
f(x) = a²x⁴ + 2abx² + b²
∴ f'(x) = \(\frac{d}{d x}\) (a²x⁴ + 2abx² + b²)
= a² \(\frac{d}{d x}\) (x⁴) + 2ab \(\frac{d}{d x}\) (x²) + \(\frac{d}{d x}\) (b²)
On using theorem \(\frac{d}{d x}\) (xⁿ) = nx^{n – 1}, we obtain
f’(x) = a² (4x³) + 2ab (2x) + b2(O) = 4a2x3 + 4abx= 4ax (ax2 + b).

(iii) Let f(x) = \(\frac{x-a}{x-b}\)

Question 8.
Find the derivative of \(\frac{x^{n}-a^{n}}{x-a}\) for some constant a.
Answer.

Question 9.
Find the derivative of
(i) 2x – \(\frac{3}{4}\)
(ii) (5x³ + 3x – 1) (x – 1)
(iii) x^{– 3} (5 + 3x)
(iv) x⁵ (3 – 6x^{– 9})
(v) x^-4 (3 – 4x^{– 5})
(vi) \(\frac{2}{x+1}-\frac{x^{2}}{3 x-1}\)
Answer.
(i) Let f(x) = 2x – \(\frac{3}{4}\)
f'(x) = \(\frac{d}{d x}\left(2 x-\frac{3}{4}\right)\)

= \(2 \frac{d}{d x}(x)-\frac{d}{d x}\left(\frac{3}{4}\right)\)

= 2 – 0 = 2

(ii) Let f(x) = (5x³ + 3x – 1) (x – 1)
By Leibnitz product rule,
f(x) = (5x³ + 3x – 1) \(\frac{d}{d x}\) (x -1) + (x – 1) \(\frac{d}{d x}\) (5x³ + 3x – 1)
= (5x³ + 3x – 1) (1) + (x – 1) (53x² + 3 – 0)
= (5x³ +3x – 1) +(x – 1) (15x² + 3)
= 5x³ + 3x – 1 + 15x³3 + 3x – 15x² – 3
= 20x³ – 15x² + 6x – 4.

(iii) Let f(x) = x^{– 3} (5 + 3x) .
By Leibnitz product rule,
f(x) = x^{– 3} \(\frac{d}{d x}\) (5 + 3x) + (5 + 3x) \(\frac{d}{d x}\) (x^{– 3})
= x^{– 3} (0 + 3) + (5 + 3x) (- 3x^{– 3 – 1})
= x^{– 3} (3) + (5 + 3x) (- 3x^{– 4})
= – 3x^{– 3} (2 + \(\frac{5}{x}\))
= \(\frac{-3 x^{-3}}{x}\) (2x + 5)
= \(=\frac{-3}{x^{4}}\) (5 + 2x)

(iv) Let f(x) = x⁵ (3 – 6x^-9)
By Leibnitz product rule,
f'(x) = \(x^{5} \frac{d}{d x}\left(3-6 x^{-9}\right)+\left(3-6 x^{-9}\right) \frac{d}{d x}\left(x^{5}\right)\)
= x⁵ {0 – 6 (- 9) x^{– 9 – 1}} + (3 – 6x^{– 9}) (5x⁴)
= x⁵ (54x^{– 10}) + 15x⁴ – 30x^{– 5}
= 54x^{– 5} + 15x⁴ – 30x^{– 5}
= 24x^{– 5} + 15x⁴
= 15x⁴ + \(\frac{24}{x^{5}}\)

(v) Let f(x) = x^{– 4} (3 – 4x^{– 5})
By Leibnitz product rule,
f’(x) = x^{– 4} \(\frac{d}{d x}\) (3 – 4x^{– 5}) + (3 – 4x^{– 5}) \(\frac{d}{d x}\) (x^{– 4})
= x^{– 4} {0 – 4 (- 5) x^{– 5 – 1}} + (3 – 4x^{– 5}) (- 4)x^{– 4 – 1}
= x^{– 4} (20 x^{– 6}) + (3 – 4x^{– 5}) (- 4 x^{– 5})
= 20x^{– 10} – 12x^{– 5} + 16x^{– 10}
= 36x^{– 10} – 12x^{– 5}
= \(-\frac{12}{x^{5}}+\frac{36}{x^{10}}\).

(vi)

Question 10.
Find the derivative of cos x from first principle.
Answer.
Let f(x) = cos x.
Accordingly, from the first principle, /'(*) = Um /(* +10-/00 = ^ COSUM 111 <mx

= – cos x (0) – sin x(1)
[∵ \(\lim _{h \rightarrow 0} \frac{1-\cos h}{h}\) = 0 and \(\lim _{h \rightarrow 0} \frac{\sin h}{h}\) = 1]
= – sin x
∴ f'(x) = – sin x

Question 11.
Find the derivative of the following functions:
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3 cot x + 5 cosec x
(vi) 5 sin x – 6 cos x+7
(vii) 2 tan x – 7 sec x
Answer.
(i) Let f(x) = sin x cos x.
Accordingly, from the first principle,

(ii) Let (x) = sec x.
Accordingly, from the first principle.

= \(\frac{1}{\cos x} \lim _{\frac{h}{2} \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)} \lim _{h \rightarrow 0} \frac{\sin \left(\frac{2 x+h}{2}\right)}{\cos (x+h)}\)

= \(\frac{1}{\cos x} \cdot 1 \cdot \frac{\sin x}{\cos x}\)

= sec x tan x.

(iii) Let f(x) = 5 sec x + 4 cos x.
Accordingly, from the first principle,

= \(\frac{5}{\cos x}\left[\lim _{h \rightarrow 0} \frac{\sin \left(\frac{2 x+h}{2}\right)}{\cos (x+h)} \lim _{\frac{h}{2} \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}\right]\) -4 sin x

= \(\frac{5}{\cos x} \cdot \frac{\sin x}{\cos x} \cdot 1\) – 4 sin x

= 5 sec x tan x – 4 sin x

(iv) Let f(x) = cosec x
Accordingly, from the first principle,

(v) Let f(x) = 3cot x – 5 cosec x
Accordingly, from the first principle,

(vi) Let f(x) = 5sin x – 6cos x + 7.
Accordingly, from the first principle,
f’(x) = \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
= \(\lim _{h \rightarrow 0} \frac{1}{h}\) [5 sin (x + h) – 6 cos (x + h) + 7 – 5 sin x + 6 cos x – 7]

= \(\lim _{h \rightarrow 0} \frac{1}{h}\) [5 {sin (x + h) – sin x) – 6 {cos (x + h) – cos x}]

= 5 \(\lim _{h \rightarrow 0} \frac{1}{h}\) [sin(x + h) – sin x] – 6 \(\lim _{h \rightarrow 0} \frac{1}{h}\) [cos(x + h) – cos x]

= \(5 \lim _{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{x+h+x}{2}\right) \sin \left(\frac{x+h-x}{2}\right)\right]\) – 6 \(\lim _{h \rightarrow 0} \frac{\cos x \cos h-\sin x \sin h-\cos x}{h}\)

= 5 \(\lim _{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{(2 x+h)}{2}\right) \sin \frac{h}{2}\right]\) – 6 \(\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-\cos x(1-\cos h)-\sin x \sin h}{h}\right]\)

= 5 cos x . 1 – 6 [(- cos x) . (0) – sin x . 1]
= 5 cos x + 6 sin x

(vi) Let f(x) = 2 tan x – 7 sec x
Accordingly, from the first principle,

= \(\) \(\)

= \(\)

= 2 sec² x – 7 sec x tan x.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 13 Limits and Derivatives Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.1

Question 1.
Evaluate the given limit: \(\lim _{x \rightarrow 3}\) x + 3.
Answer.
\(\lim _{x \rightarrow 3}\) x + 3 = 3 + 3 = 6.

Question 2.
Evaluate the given limit: \(\lim _{x \rightarrow \pi}\) (x – \(\frac{22}{7}\)).
Answer.
\(\lim _{x \rightarrow \pi}\) (x – \(\frac{22}{7}\)) = (π – \(\frac{22}{7}\)).

Question 3.
Evaluate the given limit : \(\lim _{x \rightarrow 1}\) πr².
Answer.
\(\lim _{x \rightarrow 1}\) πr² = π (1)² = π.

Question 4.
Evaluate the given limit : \(\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}\).
Answer.
\(\lim _{x \rightarrow 4} \frac{4 x+3}{x-2}=\frac{4(4)+3}{4-2}=\frac{16+3}{2}=\frac{19}{2}\)

Question 5.
Evaluate the given limit: \(\lim _{x \rightarrow-1} \frac{x^{10}+x^{5}+1}{x-1}\)
Answer.
\(\lim _{x \rightarrow-1} \frac{x^{10}+x^{5}+1}{x-1}=\frac{(-1)^{10}+(-1)^{5}+1}{-1-1}\)

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

Question 6.
Evaluate the given limit : \(\lim _{x \rightarrow 0} \frac{(x+1)^{5}-1}{x}\).
Answer.

Question 7.
Evaluate the given limit : \(\lim _{x \rightarrow 2} \frac{3 x^{2}-x-10}{x^{2}-4}\).
Answer.
At x = 2, the value of the given rational function takes the form \(\frac{0}{0}\).
∴ \(\lim _{x \rightarrow 2} \frac{3 x^{2}-x-10}{x^{2}-4}=\lim _{x \rightarrow 2} \frac{(x-2)(3 x+5)}{(x-2)(x+2)}\)

\(\lim _{x \rightarrow 2} \frac{3 x+5}{x+2}=\frac{3(2)+5}{2+2}=\frac{11}{4}\)

Question 8.
Evaluate the given limit: \(\lim _{x \rightarrow 3} \frac{x^{4}-81}{2 x^{2}-5 x-3}\).
Answer.
At x = 3, the value of the given rational function takes the form \(\frac{0}{0}\).
∴ \(\lim _{x \rightarrow 3} \frac{x^{4}-81}{2 x^{2}-5 x-3}\) = \(\lim _{x \rightarrow 3} \frac{(x-3)(x+3)\left(x^{2}+9\right)}{(x-3)(2 x+1)}\)

\(\lim _{x \rightarrow 3} \frac{(x+3)\left(x^{2}+9\right)}{2 x+1}\) = \(\frac{(3+3)\left(3^{2}+9\right)}{2(3)+1}\)

= \(\frac{6 \times 18}{7}=\frac{108}{7}\)

Question 9.
Evaluate the given limit; \(\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}\).
Answer.
\(\lim _{x \rightarrow 0} \frac{a x+b}{c x+1}=\frac{a(0)+b}{c(0)+1}\) = .

Question 10.
Evaluate the given limit: \(\lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\).
Answer.

Question 11.
Evaluate the given limit: \(\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}\), a + b + c ≠ 0.
Answer.
\(\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}=\frac{a(1)^{2}+b(1)+c}{c(1)^{2}+b(1)+a}\)

= \(\frac{a+b+c}{a+b+c}\) = 1. [a + b + c ≠ 0]

Question 12.
Evaluate the given limit: \(\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}\)
Answer.

Question 13.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\sin a x}{b x}\).
Answer.

Question 14.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\), a, b ≠ 0.
Answer.

Question 15.
Evaluate the given limit: \(\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}\)
Answer.
\(\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}\)
put π – x = θ, As x → π, θ → 0 (zero)
\(\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\pi \theta}=\lim _{\theta \rightarrow 0} \frac{1}{\pi} \frac{(\sin \theta)}{\theta}=\frac{1}{\pi}\)

Question 16.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}\).
Answer.
\(\lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}=\frac{\cos 0}{\pi-0}=\frac{1}{\pi}\)

Question 17.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{\cos 2 x-1}{\cos x-1}\).
Answer.

Question 18.
Evaluate the given limit: \(\lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}\)
Answer.

Question 19.
Evaluate the given limit: \(\lim _{x \rightarrow 0}\) x sec x.
Answer.
\(\lim _{x \rightarrow 0}\) x sec x = \(\lim _{x \rightarrow 0} \frac{x}{\cos x}=\frac{0}{\cos 0}=\frac{0}{1}\) = 0.

Question 20.
Evaluate the given limit: \(\) a, b, a + b ≠ 0.
Answer.

Question 21.
Evaluate the given limit: \(\lim _{x \rightarrow 0}\) (cosec x – cot x).
Answer.
\(\lim _{x \rightarrow 0}\) (cosec x – cot x)
At x = 0, the value of the given function takes the form ∞ – ∞.

Question 22.
Evaluate the given limit \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}\).
Answer.
\(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}\)

At x = \(\frac{\pi}{2}\), the value of the given function takes the form \(\frac{0}{0}\).

Question 23.
Find \(\lim _{x \rightarrow 0}\) f(x) and \(\lim _{x \rightarrow 1}\) f(x), where f(x) = .
Answer.
Given, f(x) =
(i) Now, LHL = \(\lim _{x \rightarrow 0^{-}}\) f(x) = \(\lim _{x \rightarrow 0^{-}}\) (2x + 3)
= \(\lim _{h \rightarrow 0}\) [2 (0 – h) + 3] = 3
[putting x = 0 – h as x → 0, then h → 0]
RHL = \(\lim _{x \rightarrow 0^{+}}\) f(x) = \(\lim _{x \rightarrow 0^{+}}\) 3(x + 1)
= \(\lim _{h \rightarrow 0}\) [3(0 + h) + 1] = 3
[putting x = 0 + h as x → 0,then h → 0]J
Here, LHL = RHL
∴ \(\lim _{x \rightarrow 0}\) f(x) = 3

(ii) We have to find \(\lim _{x \rightarrow 1}\) f(x)
\(\lim _{x \rightarrow 1}\) f(x) = \(\lim _{x \rightarrow 1}\) 3 (x + 1)
= 3 (1 + 1) = 6

Question 24.
Find \(\lim _{x \rightarrow 1}\) f(x), where f(x) =
Answer.
Given, f(x) =
At x = 1,
RHL = \(\lim _{x \rightarrow 1^{+}}\) f(x) = \(\lim _{h \rightarrow 0}\) f(1 + h)
= \(\) (1 + h)² – 1 [put x = 1 + h]
= – (1 + 0)² – 1
= – 1 – 1 = – 2
LHL = \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{h \rightarrow 0}\) f(1 – h)
= \(\lim _{h \rightarrow 0}\) (1 – h)² – 1 [put x = 1 – h]
= (1 – 0)² – 1 = 1 – 1 = 0
RHL ≠ LHL
Hence, at x = 1 , limit does not exist.

Question 25.
Evaluate \(\lim _{x \rightarrow 0}\) f(x), where f(x) = .
Answer.

Question 26.
Find \(\lim _{x \rightarrow 0}\) f(x), where f(x) = .
Answer.

Question 27.
Find \(\lim _{h \rightarrow 5}\) f(x), where f(x) = |x| – 5.
Answer.
The given function is f(x) = |x| – 5
when x > 5, put x = 5 + h, where h is small
|x| = |5 + h| = 5 + h
∴ \(\lim _{x \rightarrow 5^{+}}\) f(x) = \(\lim _{h \rightarrow 0}\) [(5 + h) – 5] = \(\lim _{h \rightarrow 0}\) h = 0
when x < 5, put x = 5 – h, where h is small
∴ |5 – h| = 5 – h
∴ \(\lim _{x \rightarrow 5^{-}}\) f(x) = \(\lim _{h \rightarrow 0}\) (5 – h – 5) = \(\lim _{h \rightarrow 0}\) (- h) = 0
∴ \(\lim _{x \rightarrow 5^{-}}\) f(x) = \(\lim _{x \rightarrow 5^{+}}\) f(x) = 0
∴ \(\lim _{h \rightarrow 5}\) f(x) = 0

Question 28.
Suppose f(x) = and if \(\lim _{h \rightarrow 1}\) f(x) = f(1) what are possible values of a and b ?
Answer.
The given function is f(x) =
\(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1}\) (a + bx) = a + b
\(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1}\) (b – ax) = b – a
f(1) = 4
It is given that \(\lim _{x \rightarrow 1}\) f(x) = f(1).
∴ \(\lim _{x \rightarrow 1^{-}}\) f(x) = \(\lim _{x \rightarrow 1^{+}}\) f(x)
= \(\lim _{x \rightarrow 1}\) f(x) = f(1)
a + b = 4 and b – a = 4.
On solving these two equations, we obtain a = 0 and b = 4.
Thus, the respective possible values of a and b are 0 and 4.

Question 29.
Let a₁, a₂, ………. a, be fixed real numbers and define a function f(x) = (x – a₁) (x – a₂) ………….. (x – a_n). What is \(\lim _{x \rightarrow a_{1}}\) f(x)? For some a ≠ a₁, a₂, ………………. a_n compute \(\lim _{x \rightarrow a}\) f(x).
Answer.

Question 30.
If f(x) = For what value(s) of a does f(x) exists?
Answer.

Question 31.
If the function f(x) satisfies \(\lim _{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}\) = π, evaluate \(\lim _{x \rightarrow 1}\) f(x).
Answer.

Question 32.
If f(x) = For what integers m and n does \(\lim _{x \rightarrow 0}\) f(x) and \(\lim _{x \rightarrow 1}\) f(x) exist?
Answer.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Miscellaneous Exercise

Question 1.
Three vertices of a parallelogram ABCD are A (3,- 1, 2), B(1, 2, – 4) and C (- 1, 1, 2). Find the coordinates of the fourth vertex.
Answer.
The three vertices of a parallelogramABCD are given as A (3,- 1, 2), B (1, 2, – 4) and C (- 1, 1, 2).
Let the coordinates of the fourth vertex be D(x, y, z).

We know that the diagonals of a parallelogram bisect each other.
Therefore, in parallelogram ABCD, AC and BD bisect each other.
∴ Mid-point of AC = Mid-point of BD
\(\left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)=\left(\frac{x+1}{2}, \frac{y+2}{2}, \frac{z-4}{2}\right)\)
(1, 0, 2) = \(\left(\frac{x+1}{2}, \frac{y+2}{2}, \frac{z-4}{2}\right)\)
\(\frac{x+1}{2}\) = 1, \(\frac{y+2}{2}\) = 0, and \(\frac{z-4}{2}\) = 2
x = 1 ,y = – 2, and z = 8
Thus, the coordinates of the fourth vertex are (1, – 2, 8).

Question 2.
Find the lengths of the medians of the trisìngle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).
Answer.
ABC is a thangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

[∵ Coordinates of mid-points = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)\)]

Let points D, E and F are the mid-points of BC, AC and AB, respectively.
So, AD, BE and CF will be the medians of the triangle.
Coordinates of point D = \(\left(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}\right)\) = (3, 2, 0)

Coordinates of point E = \(\left(\frac{0+6}{2}, \frac{0+0}{2}, \frac{6+0}{2}\right)\) = (3, 0, 3)

Coordinates of point F = \(\left(\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}\right)\) = (0, 2, 3)

Thus, the coordinates of the fourth vertex are (1, – 2, 8).

Now, length of median
AD = Distance between points A and D
AD = \(\sqrt{(0-3)^{2}+(0-2)^{2}+(6-0)^{2}}\)
[∵ distance = \(\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(z_{1}-z_{2}\right)^{2}}\)]
= \(\sqrt{9+4+36}\)
= √49 = 7

similarly, BE = \(\sqrt{(0-3)^{2}+(4-0)^{2}+(0+3)^{2}}\)
= \(\sqrt{9+16+9}\) = √34

and CF = \(\sqrt{(6-0)^{2}+(0-2)^{2}+(0-3)^{2}}\)
= \(\sqrt{36+4+9}\) = √49 = 7
Hence, length of the median are 7, √34 and 7.

Question 3.
If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (- 4, 3b, -10) and R (8, 14, 2c), then find the values of a, b and c.
Answer.

It is known that the coordinates of the centroid of the triangle, whose vertices are (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃),are \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right)\)

Therefore, coordinates of the centroid of ∆PQR = \(\left(\frac{2 a-4+8}{3}, \frac{2+3 b+14}{3}, \frac{6-10+2 c}{3}\right)\)

= \(\left(\frac{2 a+4}{3}, \frac{3 b+16}{3}, \frac{2 c-4}{3}\right)\)

It is given that the origin is the centroid of ∆PQR.
∴ (0, 0, 0) = \(\left(\frac{2 a+4}{3}, \frac{3 b+16}{3}, \frac{2 c-4}{3}\right)\)

\(\frac{2 a+4}{3}\) = 0, \(\frac{3 b+16}{3}\) = 0 and \(\frac{2 c-4}{3}\) = 0

a = – 2, b = – \(\frac{16}{3}\) and c = 2.

Thus, the respective values of a, b and c are – 2, – \(\frac{16}{3}\) and 2.

Question 4.
Find the coorlinsites of a point on y-axis which are at a distance of 5√2 from point P(3, – 2, 5).
Answer.
If a point is on the y-axis, then x-coordinate and the z-coordinate of the point are zero.
Let A (0, b, 0) be the point on the y-axis at a distance of 5√2 from point P (3, – 2, 5).
Accordingly, AP = 5√2
∴ AP² = 50
(3 – 0)² + (- 2 – b)² + (5 – 0)² = 50
⇒ 9 + 4 + b² + 4b + 25 = 50
⇒ b² + 4b – 12 = 0
⇒ b² + 6b – 2b – 12 = 0
⇒ (b + 6) (b – 2) = 0
⇒ b = – 6 o r2.
Thus, the coordinates of the required points are (0, 2, 0) and (0, – 6, 0).

Question 5.
A point R with x-coordinnte 4 lies on the line segment joining the points P (2, – 3, 4) and Q(8, 0, 10). Find the coordinates of the point R.
[Hint : Suppose R divides PQ in the ratio k : 1. The co-ordinates of the point R are given by, \(\left(\frac{8 k+2}{k+1}, \frac{-3}{k+1}, \frac{10 k+4}{k+1}\right)\)]
Answer.
Let the coordinates of R be (4, y, z) and R divides PQ in ratio k : 1
∴ Coordinate of R is \(\left(\frac{8 k+2}{k+1}, \frac{-3}{k+1}, \frac{10 k+4}{k+1}\right)\).
[∵ using internal ratio formula]

But x – coordinate of R is 4.
So, \(\frac{8 k+2}{k+1}\) = 4
⇒ 8k + 2 = 4k + 4
⇒ 4k = 2
⇒ k = \(\frac{1}{2}\)

img 5

Hence, coordinates of R are (4, – 2, 6).

Question 6.
If A and B be the points (3, 4, 5) and (- 1, 3, – 7), respectively, find the equation of the set of points P such that PA² + PB² = k², where k is a constant.
Answer.
The coordinates of points A and B are given as (3, 4, 5) and Q (- 1, 3, – 7) respectively.
Let the coordinates of point P be (x, y, z).
On using distance formula, we obtain .
PA² = (x – 3)² + (y – 4)² + (z – 5)²
= x² + 9 – 6x + y² +16 – 8y + z² + 25 – 10z
= x² – 6x + y² – 8y + z² – 10z + 50
PB² = (x + 1)² + (y – 3)² + (z + 7)²
= x² + 2x +1 + y² – 6y + 9 + z² + 14z + 49
= x² + 2x + y² – 6y + z² + 14z + 59
Now, if PA² + PB² = k², then
(x² – 6x + y² – 8y + z² – 10z + 50) + (x² + 2x + y² – 6y + z² + 14z + 59) = k²
⇒ 2x² + 2y² + 2z² – 4x – 14y + 4z +109 = k²
⇒ 2(x² + y² + z² – 2x – 7y + 2z) = k² – 109
⇒ x² + y² + z² – 2x – 7y + 2z = \(\frac{k^{2}-109}{2}\)
Thus, the required equation is x² + y² + z² – 2x + 7y + 2z = \(\frac{k^{2}-109}{2}\).

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.3

Question 1.
Find the coordinates of the point which divides the line segment joining the points (- 2, 3, 5) and (1, – 4, 6) in the ratio (i) 2 : 3 internally,
(ii) 2 : 3 externally.
Answer.
Let P(x, y, z) be any point which divides the line segment joining points A(- 2, 3, 5) and B(1, – 4, 6) in the ratio 2 : 3 internally.

Here, the ratio is 2 : 3
m = 2, n = 3
The coordinates of point P = \(\left[\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}, \frac{m_{1} z_{2}+m_{2} z_{1}}{m_{1}+m_{2}}\right)\right]\)

= \(\left[\frac{2 \times 1+3 \times(-2)}{2+3}, \frac{2(-4)+3(3)}{2+3}, \frac{2(6)+3(5)}{2+3}\right]\)

= \(\left(\frac{2-6}{5}, \frac{-8+9}{5}, \frac{12+15}{5}\right)=\left(\frac{-4}{5}, \frac{1}{5}, \frac{27}{5}\right)\)

(ii) Let P(x, y, z) be any point which divides the line segment joining points A(- 2, 3, 5) and 8(1, – 4, 6) in the ratio 2 : 3 externally.

Here, the ratio is 2 : 3
∴ m = 2, n = 3
The coordinates of point P = \(\left[\left(\frac{m_{1} x_{2}-m_{2} x_{1}}{m_{1}-m_{2}}, \frac{m_{1} y_{2}-m_{2} y_{1}}{m_{1}-m_{2}}, \frac{m_{1} z_{2}-m_{2} z_{1}}{m_{1}-m_{2}}\right)\right]\)

= \(\left[\frac{2(1)+(-3)(-2)}{2+(-3)}, \frac{2(-4)+(-3) \times 3}{2+(-3)}, \frac{2(6)+(-3)(5)}{2+(-3)}\right]\)

= \(\left(\frac{2+6}{2-3}, \frac{-8-9}{2-3}, \frac{12-15}{2-3}\right)=\left(\frac{8}{-1}, \frac{-17}{-1}, \frac{-3}{-1}\right)\)

= (- 8, 17, 3).

Qiestion 2.
Given that P (3, 2, -4), Q (5, 4, – 6) and R (9, 8, – 10) are coimear. Find the ratio in which Q divides PR.
Answer.
Let point Q(5, 4, -6) divide the line segment joining point P (3, 2, – 4) and R (9, 8, – 10) in the ratio k : 1.
Therefore, by section formula:
(5, 4, – 6) = \(\left(\frac{k(9)+3}{k+1}, \frac{k(8)+2}{k+1}, \frac{k(-10)-4}{k+1}\right)\)

⇒ \(\frac{9 k+3}{k+1}\) = 5
⇒ 9k + 3 = 5k + 5
⇒ 4k = 2
⇒ k = \(\frac{2}{4}=\frac{1}{2}\).
Thus, point Q divides PR in the ratio 1 : 2.

Qiestion 3.
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (- 2, 4, 7) and (3, – 5, 8).
Answer.
Let the YZ plane divide the line segment joining points (- 2, 4, 7) and (3, – 5, 8) in the ratio k: 1.
Hence, by section formula, the coordinates of point of intersection are given by
\(\left(\frac{k(3)-2}{k+1}, \frac{k(-5)+4}{k+1}, \frac{k(8)+7}{k+1}\right)\)

On the YZ plane, the x-coordinate of any point is zero.
⇒ \(\frac{3 k-2}{k+1}\) = 0
⇒ 3k – 2 = 0
⇒ k = \(\frac{2}{3}\)
Thus, the YZ plane divides the line segment formed by joining the given points in the ratio 2 : 3.

Qiestion 4.
Using section formula, show that the points A (2, – 3, 4), B(- 1, 2, 1) and C(0, \(\frac{1}{3}\), 2) are collinear.
Answer.
Let C(0, \(\frac{1}{3}\), 2) divides the join of A(2, – 3, 4) and B(- 1, 2,1) in the ratio k : 1.

Then, coordinates of C are \(\left(\frac{-k+2}{k+1}, \frac{2 k-3}{k+1}, \frac{k+4}{k+1}\right)\) ………………. (i)
[using internal ratio formula]

But coordinates of C are (0, \(\frac{1}{3}\), 2) ………………(ii) [given]

\(\frac{-k+2}{k+1}\) = 0
⇒ – k + 2 = 0
⇒ k = 2

\(\frac{2 k-3}{k+1}=\frac{1}{3}\)
⇒ 6k – 9 = k + 1
⇒ k = 2

\(\frac{k+4}{k+1}\) = 2
⇒ k + 4 = 2k + 2
⇒ k = 2

From each of these equations, we get k = 2
Since, each of these equations give same value of k.
Therefore, the given points are collinear and C divides AB internally in the ratio 2 : 1.

Question 5.
Find the coordinates of the points which trisect the line segment joining the points P(4,2, -6) and Q (10, -16, 6).
Answer.
Let R and S be two points which trisect the joining of P and Q.

Point R divides the join of PQ in the ratio 1 : 2.
∴ Coordinates of R are \(\left[\frac{1(10)+2(4)}{1+2}, \frac{1(-16)+2(2)}{1+2}, \frac{1(6)+2(-6)}{1+2}\right]\)

= \(\left(\frac{10+8}{3}, \frac{-16+4}{3}, \frac{6-12}{3}\right)\)

= \(\left(\frac{18}{3}, \frac{-12}{3}, \frac{-6}{3}\right)\)

= (6, – 4, – 2).

Also, point S divides the join of PQ in the rano 2 : 1
∴ Coordinates of S are \(\left[\frac{2(10)+1(4)}{1+2}, \frac{2(-16)+1(2)}{1+2}, \frac{2(6)+1(-6)}{1+2}\right]\)

= \(\left(\frac{20+4}{3}, \frac{-32+2}{3}, \frac{12-6}{3}\right)\)

= \(\left(\frac{24}{3}, \frac{-30}{3}, \frac{6}{3}\right)\)

= (8, – 10, 2).

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.2

Question 1.
Find the distance between the following pairs of points :
(i) (2, 3, 5) and (4, 3, 1)
(ii) (- 3, 7, 2) and (2, 4, – 1)
(iii) (- 1, 3, – 4) and (1, – 3, 4)
(iv) (2, – 1, 3) and (- 2, 1, 3).
Answer.
The distance between point P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) is given by
PQ = \(\)
(i) Distance between points (2, 3, 5) and (4, 3, 1)
= \(\sqrt{(4-2)^{2}+(3-3)^{2}+(1-5)^{2}}=\sqrt{(2)^{2}+(0)^{2}+(-4)^{2}}\)
= \(\sqrt{4+16}=\sqrt{20}=2 \sqrt{5}\)

(ii) Distance between points (- 3, 7, 2) and (2, 4, – 1)
= \(\sqrt{(2+3)^{2}+(4-7)^{2}+(-1-2)^{2}}=\sqrt{(5)^{2}+(-3)^{2}+(-3)^{2}}\)
= \(\sqrt{25+9+9}=\sqrt{43}\)

(iii) Distance between points ( – 1, 3, – 4) and (1, – 3, 4)
= \(\sqrt{(1+1)^{2}+(-3-3)^{2}+(4+4)^{2}}\)
= \(\sqrt{(2)^{2}+(-6)^{2}+(8)^{2}}\)
= \(\sqrt{4+36+64}=\sqrt{104}=2 \sqrt{26}\)

(iv) Distance between points (2, -1, 3) and (-2, 1, 3)
= \(\sqrt{(-2-2)^{2}+(1+1)^{2}+(3-3)^{2}}\)
= \(\sqrt{(-4)^{2}+(2)^{2}+(0)^{2}}\)
= \(\sqrt{16+4}=\sqrt{20}=2 \sqrt{5}\)

Question 2.
Show that the point (- 2, 3, 5), (1, 2, 3) and (7,0, – 1) are collinear.
Answer.
Let points (- 2, 3, 5), (1, 2, 3) and (7, 0, – 1) be denoted by P, Q, and R respectively.
Points P, Q, and R are collinear if they lie on a line.

Here, PQ + QP = \(\sqrt{14}+2 \sqrt{14}=3 \sqrt{14}\) = PR
Hence, points P (- 2, 3, 5), Q(1, 2, 3), and P (7, 0, – 1) are collinear.

Question 3.
Verify the following:
(i) (0, 7, – 10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) are the vertices of a right angled triangle.
(iii) (- 1, 2, 1), (1, – 2, 5), (4, – 7, 8) and (2, – 3, 4) are the vertices of a parallelogram.
Answer.
(i) Let points (0, 7, – 10), (1, 6, – 6) and (4, 9, – 6) be denoted by A, B and C respectively.

Here, AB = BC ≠ CA.
Thus, the given points are the vertices of an isosceles triangle.

(ii) Let (0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) be denoted by A, B, and C respectively.

Now, AB² + BC² = (3√2)² + (3√2)²
= 18 + 18 = 36 = AC²
Hence, the given points are the vertices of a right-angled triangle.

(iii) Let points (- 1, 2, 1), (1, – 2, 5), (4, – 7, 8) and (2, – 3, 4) be denoted by A, B, C and D respectively.

Now, AB = CD = 6, BC = AD = √43.
Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.
Therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.

Question 4.
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, – 1).
Answer.
Let P(x, y, z) be the point that is equidistant from points A(l, 2, 3) and B(3, 2, -1).
Accordingly, PA = PB
⇒ PA² = PB²
⇒ (x – 1)² + (y – 2)² + (z – 3)² = (x – 3)² + (y – 2)² + (z + 1)²
x² – 2x + 1 + y² – 4y + 4 + z² – 6z + 9 = x² – 6x + 9 + y² – 4y + 4 + z² + 2z +1
⇒ – 2x – 4y – 6z + 14 = – 6x – 4y + 2z + 14
⇒ – 2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0.

Question 5.
Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (- 4, 0, 0) is equal to 10.
Answer.
We have, a point P(x, y, z) such that, PA + PB = 0

On squaring both sides, we get
x² + y² + z² – 8x + 16 = 100 + x² + y² + z² + 8x + 16 – 20 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\)
⇒ – 16x – 100 = – 20 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\)
⇒ 4x + 25 = 5 \(\sqrt{x^{2}+y^{2}+z^{2}+8 x+16}\) [dividing both sides by -4]
Again squaring on both sides, we get
16x² + 200x + 625 = 25 (x² + y² + z² + 8x +16)
⇒ 16x² + 200x + 625 = 25x² + 25y² + 25z² +200x + 400
⇒ 9x² + 25y² + 25z² – 225 = 0.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Ex 12.1

Question 1.
A point is on the x-axis. What are its y – coordinates and z – coordinates?
Answer.
Coordinates of any point on the X-axis is (x, 0, 0).
Because at X-axis, both y and z – coordinates are zero.
So, its y and z – coordinates are zero.

Question 2.
A point is in the XZ – plane. What can you say about its y-coordinate?
Answer.
Any point on the XZ – plane will have the coordinate (x, 0, z), so its y – coordinate is 0.

Question 3.
Name the octants in which the following points lie : (1, 2, 3), (4, – 2, 3), (4, – 2, – 5), (4, 2, – 5), (- 4, 2, – 5), (- 4, 2, 5), (- 3, – 1, 6), (2, – 4, – 7).
Answer.
The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.
The x-coordinate, y-coordinate, and z-coordinate of point (4, – 2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.
The x-coordinate, y-coordinate, and z-coordinate of point (4, – 2, – 5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.
The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, – 5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.
The x-coordinate, y-coordinate, and z-coordinate of point (- 4, 2, – 5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.
The x-coordinate, y-coordinate, and z-coordinate of point (- 4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.
The x-coordinate, y-coordinate, and z-coordinate of point ( – 3, – 1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.
The x-coordinate, y-coordinate, and z-coordinate of point (2, – 4, – 7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

Question 4.
Fill in the blanks :
(i) The x-axis and y-axis taken together determine a plane known as
(ii) The coordinates of points in the XY-plane are of the form
(iii) Coordinate planes divide the space into octants.
Answer.
(i) XY plane
(ii) (x, y, 0)
(iii) eight