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

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.