Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.8

Note: Please note that Exercise 5.8 from Chapter 5, "Continuity and Differentiability" in the NCERT Solutions, has been removed from the revised syllabus. As a result, this exercise will no longer be a part of your study curriculum.

Exercise 5.8 focuses on the application of derivatives to find tangents and normals to curves. This exercise explores how the concepts of differentiation can be used to determine the equations of lines tangent or normal to a given curve at a specific point. It helps students understand the geometric interpretation of derivatives and their practical applications in analytic geometry.

Question 1. Verify Rolle’s theorem for the function f(x) = x²+ 2x – 8, x ∈ [– 4, 2].

Solution:

Now f(x) = x² + 2x – 8 is a polynomial

So, f(x) is continuous in the interval [-4,2] and differentiable in the interval (- 4,2)

f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 = 0

f(2) = 2² + 4 – 8 = 8 – 8 = 0

f(-4) = f(2)

As Conditions of Rolle’s theorem are satisfied.

Then there exists some c in (-4, 2) such that f′(c) = 0

f'(x) = 2x + 2

f’ (c) = 2c + 2 = 0

c = – 1, 

and -1 ∈ [-4,2]

Hence, f’ (c) = 0 at c = – 1.

Question 2. Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?

(i) f(x) = [x] for x ∈ [5, 9] 

Solution:

In the interval [5, 9], 

Now, f (x) = [x] which is neither continuous nor derivable at Integers. 

f (x) is neither continuous nor derivable at x = 6,7,8 

Hence, Rolle’s theorem is NOT applicable

(ii) f(x) = [x] for x ∈ [– 2, 2]

Solution:

In the interval [– 2, 2],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = -1,0,1

Hence, Rolle’s theorem is NOT applicable

(iii) f(x) = x²– 1 for x ∈ [1, 2]

Solution:

Now f(x) = x² - 1 is a polynomial

So, f(x) is continuous in the interval [1, 2] and differentiable in the interval (1,2)

f(1) = (1)²  – 1 = 0

f(2) = 2² – 1 = 3

f(-4) ≠ f(2)

As Conditions of Rolle’s theorem are NOT satisfied.

Hence, Rolle’s theorem is NOT applicable

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:

For Rolle’s theorem

f is continuous in [a, b] .........(1)

f is derivable in [a, b] .........(2)

f (a) = f (b)  .........(3)

then f’ (c)=0, c ∈ (a, b)

So as, f is continuous and derivable

but f '(c) ≠ 0 

It concludes, f(a) ≠ f(b) 

 f(-5) ≠ f(5)

Question 4. Verify Mean Value Theorem, if f(x) = x²– 4x – 3 in the interval [a, b], where a = 1 and b = 4.

Solution:

Now f(x) = x² - 4x -3 is a polynomial

So, f(x) is continuous in the interval [1,4] and differentiable in the interval (1,4)

f(1) = (1)² - 4(1) – 3 = -6

f(4) = 4² - 4(4) – 3 = -3

f′(c) = 2c - 4

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,4) such that 

f′(c) = \frac{f(b)-f(a)}{b-a}

= \frac{-3-(-6)}{4-1}

= 1

2c - 4 = 1

c = 5/2

and c = 5/2 ∈ (1,4)

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:

Now f(x) = x³– 5x²– 3x is a polynomial

So, f(x) is continuous in the interval [1,3] and differentiable in the interval (1,3)

f(1) = (1)³– 5(1)²– 3(1) = -7

f(3) = 3³– 5(3)²– 3(3) = -27

f′(c) = 3c² - 5(2c) - 3

f′(c) = 3c² - 10c - 3

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,3) such that

f′(c) = \frac{f(b)-f(a)}{b-a}

= \frac{f(3)-f(1)}{b-a}

= \frac{-27-(-7)}{3-1}

= \frac{-20}{2}

3c² - 10c - 3 = -10

3c² - 10c + 7 = 0

3c2 - 3c - 7c + 7 = 0

3c (c-1) - 7(c -1) = 0

(3c -7) (c-1) = 0

c = 7/3 or c = 1

As, 1 ∉ (1,3)

So, c = 7/3 ∈ (1,3)

According to the Rolle's Theorem

As, f(3) ≠ f(1), Then there does not exist some c ∈ (1,3) such that f′(c) = 0

Question 6. Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

(i) f(x) = [x] for x ∈ [5, 9]

Solution:

In the interval [5, 9],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = 6,7,8

Hence, Mean value theorem is NOT applicable

(ii) f(x) = [x] for x ∈ [– 2, 2]

Solution:

In the interval [– 2, 2],

Now, f (x) = [x] which is neither continuous nor derivable at Integers.

f (x) is neither continuous nor derivable at x = -1,0,1

Hence, Mean value theorem is NOT applicable

(iii) f(x) = x²– 1 for x ∈ [1, 2]

Solution:

Now f(x) = x² - 1 is a polynomial

So, f(x) is continuous in the interval [1,2] and differentiable in the interval (1,2)

f(1) = (1)² - 1 = 0

f(2) = 2² -1 = 3

f′(c) = 2c

As Conditions of Mean Value Theorem are satisfied.

Then there exists some c in (1,2) such that

f′(c) = \frac{f(b)-f(a)}{b-a}

= \frac{f(2)-f(1)}{2-1}

= \frac{3-0}{2-1}

= 3

2c = 3

c = 3/2

and c = 3/2 ∈ (1,4)

Summary

Exercise 5.8 applies the concept of derivatives to find equations of tangents and normals to various curves. It covers a wide range of curve types, including polynomial functions, trigonometric functions, and parametric equations. This exercise helps students understand the geometric significance of derivatives and how they can be used to analyze the behavior of curves at specific points.