RBSE Solutions Class 12 Maths Chapter 7 Differentiation Exercise 7.6

Question 1. Verify the Rolle's theorem, for the following functions:
(i) \( f(x) = e^x (\sin x - \cos x), x \in \left[ \frac{\pi}{4}, \frac{5\pi}{4} \right] \)
(ii) \( f(x)=(x-a)^m (x - b)^n, x \in [a, b], m, n \in N \)
(iii) \( f(x) = |x|, x \in [-1,1] \)
(iv) \( f(x) = x^2 + 2x - 8, x \in [-4,2] \)
(v) \( f(x) = \begin{cases} x^2+1 & 0 \le x \le 1 \\ 3-x & 1 < x \le 2 \end{cases} \)
(vi) \( f(x) = [x], x \in [-2, 2] \)
Answer:
(i) Given function: \( f(x) = e^x (\sin x - \cos x) \).
Since \( e^x \), \( \sin x \), and \( \cos x \) are continuous functions, their product and difference are also continuous. So, \( f(x) \) is continuous in the interval \( \left[ \frac{\pi}{4}, \frac{5\pi}{4} \right] \).
Now, let's find the derivative: \( f'(x) = e^x (\cos x + \sin x) + (\sin x - \cos x) e^x \)
\( f'(x) = e^x (2 \sin x) \)
This derivative exists for all \( x \in \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \), so \( f(x) \) is differentiable in this open interval.
Next, let's check \( f(a) \) and \( f(b) \):
\( f\left( \frac{\pi}{4} \right) = e^{\pi/4} \left( \sin \frac{\pi}{4} - \cos \frac{\pi}{4} \right) = e^{\pi/4} \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right) = 0 \)
\( f\left( \frac{5\pi}{4} \right) = e^{5\pi/4} \left( \sin \frac{5\pi}{4} - \cos \frac{5\pi}{4} \right) = e^{5\pi/4} \left( -\frac{1}{\sqrt{2}} - (-\frac{1}{\sqrt{2}}) \right) = e^{5\pi/4} (0) = 0 \)
So, \( f\left( \frac{\pi}{4} \right) = f\left( \frac{5\pi}{4} \right) \).
All three conditions for Rolle's theorem are satisfied. Thus, there exists a point \( c \in \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \) such that \( f'(c) = 0 \).
Let's find c: \( f'(c) = e^c (2 \sin c) = 0 \).
Since \( e^c \) is never zero, \( 2 \sin c = 0 \implies \sin c = 0 \).
For \( c \in \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \), the value \( c = \pi \) satisfies \( \sin c = 0 \).
And \( \pi \in \left( \frac{\pi}{4}, \frac{5\pi}{4} \right) \). Hence, Rolle's theorem is verified.
(ii) Given function: \( f(x) = (x-a)^m (x - b)^n \), where \( x \in [a, b] \) and \( m, n \in N \).
This function is a product of two polynomial functions, \( (x-a)^m \) and \( (x-b)^n \). Polynomials are always continuous and differentiable everywhere. Therefore, \( f(x) \) is continuous in the closed interval \( [a, b] \) and differentiable in the open interval \( (a, b) \).
Now, let's check \( f(a) \) and \( f(b) \):
\( f(a) = (a-a)^m (a-b)^n = 0^m (a-b)^n = 0 \)
\( f(b) = (b-a)^m (b-b)^n = (b-a)^m 0^n = 0 \)
So, \( f(a) = f(b) \).
All three conditions for Rolle's theorem are satisfied. Thus, there exists a point \( c \in (a, b) \) such that \( f'(c) = 0 \).
Let's find the derivative \( f'(x) \):
\( f'(x) = m(x-a)^{m-1}(x-b)^n + (x-a)^m n(x-b)^{n-1} \)
\( f'(x) = (x-a)^{m-1}(x-b)^{n-1} [m(x-b) + n(x-a)] \)
Now set \( f'(c) = 0 \):
\( (c-a)^{m-1}(c-b)^{n-1} [m(c-b) + n(c-a)] = 0 \)
Since \( c \in (a, b) \), \( (c-a) \neq 0 \) and \( (c-b) \neq 0 \). So we must have:
\( m(c-b) + n(c-a) = 0 \)
\( mc - mb + nc - na = 0 \)
\( (m+n)c = mb + na \)
\( c = \frac{mb + na}{m+n} \)
Since \( m, n \) are positive integers, \( m+n \neq 0 \).
Also, \( a < \frac{mb+na}{m+n} < b \) because it's a weighted average of \( a \) and \( b \). So, \( c \in (a,b) \). Hence, Rolle's theorem is verified.
(iii) Given function: \( f(x) = |x|, x \in [-1,1] \)
The modulus function \( f(x) = |x| \) can be written as:
\( f(x) = \begin{cases} -x & -1 \le x < 0 \\ x & 0 \le x \le 1 \end{cases} \)
Continuity: The function \( |x| \) is continuous for all real numbers, so it is continuous in \( [-1,1] \).
Differentiability: Let's check differentiability at \( x=0 \).
Right Hand Derivative (RHD) at \( x=0 \):
\( R f'(0) = \lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{|h| - |0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \)
Left Hand Derivative (LHD) at \( x=0 \):
\( L f'(0) = \lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{|h| - |0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 \)
Since \( R f'(0) \neq L f'(0) \), the function \( f(x) = |x| \) is not differentiable at \( x=0 \).
For Rolle's theorem to apply, the function must be differentiable in the open interval \( (-1, 1) \). Since it is not differentiable at \( x=0 \), Rolle's theorem is not satisfied for this function.
(iv) Given function: \( f(x) = x^2 + 2x - 8, x \in [-4,2] \)
This is a polynomial function. Polynomials are continuous and differentiable everywhere. So, \( f(x) \) is continuous in \( [-4,2] \) and differentiable in \( (-4,2) \).
Now, let's check \( f(a) \) and \( f(b) \):
\( f(-4) = (-4)^2 + 2(-4) - 8 = 16 - 8 - 8 = 0 \)
\( f(2) = (2)^2 + 2(2) - 8 = 4 + 4 - 8 = 0 \)
So, \( f(-4) = f(2) \).
All three conditions for Rolle's theorem are satisfied. Thus, there exists a point \( c \in (-4, 2) \) such that \( f'(c) = 0 \).
Let's find the derivative \( f'(x) \):
\( f'(x) = 2x + 2 \)
Now set \( f'(c) = 0 \):
\( 2c + 2 = 0 \)
\( 2c = -2 \)
\( c = -1 \)
Since \( -1 \in (-4, 2) \), Rolle's theorem is verified.
(v) Given function: \( f(x) = \begin{cases} x^2+1 & 0 \le x \le 1 \\ 3-x & 1 < x \le 2 \end{cases} \)
The function is defined in \( [0,2] \).
Continuity: For \( 0 \le x < 1 \), \( f(x) = x^2+1 \) is continuous. For \( 1 < x \le 2 \), \( f(x) = 3-x \) is continuous. We need to check continuity at \( x=1 \).
\( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2+1) = (1)^2+1 = 2 \)
\( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3-x) = 3-1 = 2 \)
\( f(1) = (1)^2+1 = 2 \)
Since \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2 \), the function is continuous at \( x=1 \). Thus, \( f(x) \) is continuous in \( [0,2] \).
Differentiability: We need to check differentiability in \( (0,2) \). The individual parts are differentiable. We check at \( x=1 \).
Right Hand Derivative (RHD) at \( x=1 \):
\( R f'(1) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} \)
For \( h > 0 \), \( 1+h > 1 \), so \( f(1+h) = 3-(1+h) \). Also \( f(1) = 2 \).
\( R f'(1) = \lim_{h \to 0^+} \frac{(3-(1+h)) - 2}{h} = \lim_{h \to 0^+} \frac{3-1-h-2}{h} = \lim_{h \to 0^+} \frac{-h}{h} = -1 \)
Left Hand Derivative (LHD) at \( x=1 \):
\( L f'(1) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} \)
For \( h < 0 \), \( 1+h < 1 \), so \( f(1+h) = (1+h)^2+1 \). Also \( f(1) = 2 \).
\( L f'(1) = \lim_{h \to 0^-} \frac{((1+h)^2+1) - 2}{h} = \lim_{h \to 0^-} \frac{(1+2h+h^2+1) - 2}{h} \)
\( L f'(1) = \lim_{h \to 0^-} \frac{2h+h^2}{h} = \lim_{h \to 0^-} (2+h) = 2 \)
Since \( R f'(1) \neq L f'(1) \), the function \( f(x) \) is not differentiable at \( x=1 \).
Therefore, Rolle's theorem is not satisfied for this function.
(vi) Given function: \( f(x) = [x], x \in [-2, 2] \)
The greatest integer function \( f(x) = [x] \) is not continuous at integer points. In the interval \( [-2, 2] \), it is not continuous at \( x = -1, 0, 1, 2 \).
For Rolle's theorem to apply, the function must be continuous in the closed interval \( [-2, 2] \). Since it is not continuous, the first condition of Rolle's theorem is not met.
Therefore, Rolle's theorem is not satisfied for this function.
In simple words: For Rolle's theorem, a function must be smooth (continuous and differentiable) and start and end at the same height. If any of these conditions are not met, the theorem does not apply. In parts (iii), (v), and (vi), the functions were not smooth enough, so the theorem could not be verified. In parts (i), (ii), and (iv), the functions followed all rules, and we found the special point 'c' predicted by the theorem.

🎯 Exam Tip: When verifying Rolle's theorem, always check all three conditions in order: continuity, differentiability, and then \( f(a) = f(b) \). If any condition fails, Rolle's theorem cannot be applied.

Question 2. Prove Rolle's theorem for following functions :
(i) \( f(x) = x^2 + 5x + 6, x \in [-3, -2] \)
(ii) \( f(x) = e^x \sin x, x \in [0, \pi] \)
(iii) \( f(x) = \sqrt{x(1 - x)}, x \in [0, 1] \)
(iv) \( f(x) = \cos 2x, x \in [0, \pi] \)
Answer:
(i) Given function: \( f(x) = x^2 + 5x + 6, x \in [-3, -2] \)
Since \( f(x) \) is a polynomial function, it is continuous in the closed interval \( [-3, -2] \) and differentiable in the open interval \( (-3, -2) \).
Now, let's check \( f(a) \) and \( f(b) \):
\( f(-3) = (-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0 \)
\( f(-2) = (-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0 \)
So, \( f(-3) = f(-2) \).
All three conditions for Rolle's theorem are satisfied. Thus, there exists a point \( c \in (-3, -2) \) such that \( f'(c) = 0 \).
Let's find the derivative \( f'(x) \):
\( f'(x) = 2x + 5 \)
Now set \( f'(c) = 0 \):
\( 2c + 5 = 0 \)
\( 2c = -5 \)
\( c = -\frac{5}{2} \)
Since \( -\frac{5}{2} = -2.5 \), which is in \( (-3, -2) \), Rolle's theorem is proved.
(ii) Given function: \( f(x) = e^x \sin x, x \in [0, \pi] \)
The functions \( e^x \) and \( \sin x \) are continuous for all real numbers. Therefore, their product \( f(x) = e^x \sin x \) is continuous in the closed interval \( [0, \pi] \) and differentiable in the open interval \( (0, \pi) \).
Now, let's check \( f(a) \) and \( f(b) \):
\( f(0) = e^0 \sin(0) = 1 \times 0 = 0 \)
\( f(\pi) = e^\pi \sin(\pi) = e^\pi \times 0 = 0 \)
So, \( f(0) = f(\pi) \).
All three conditions for Rolle's theorem are satisfied. Thus, there exists a point \( c \in (0, \pi) \) such that \( f'(c) = 0 \).
Let's find the derivative \( f'(x) \):
\( f'(x) = e^x \cos x + e^x \sin x = e^x (\cos x + \sin x) \)
Now set \( f'(c) = 0 \):
\( e^c (\cos c + \sin c) = 0 \)
Since \( e^c \) is never zero, we must have \( \cos c + \sin c = 0 \).
\( \sin c = -\cos c \)
\( \tan c = -1 \)
For \( c \in (0, \pi) \), the value \( c = \frac{3\pi}{4} \) satisfies \( \tan c = -1 \).
Since \( \frac{3\pi}{4} \in (0, \pi) \), Rolle's theorem is proved.
(iii) Given function: \( f(x) = \sqrt{x(1 - x)}, x \in [0, 1] \)
For \( f(x) \) to be defined, \( x(1-x) \ge 0 \). This is true for \( x \in [0, 1] \).
The function \( g(x) = x(1-x) \) is a polynomial, so it's continuous. The square root function \( \sqrt{y} \) is continuous for \( y \ge 0 \). Therefore, \( f(x) = \sqrt{x(1-x)} \) is continuous in the closed interval \( [0, 1] \).
Now, let's find the derivative \( f'(x) \):
\( f'(x) = \frac{1}{2\sqrt{x(1-x)}} (1-2x) = \frac{1-2x}{2\sqrt{x(1-x)}} \)
For \( f(x) \) to be differentiable in \( (0, 1) \), \( f'(x) \) must exist. \( f'(x) \) exists for all \( x \in (0, 1) \) because the denominator \( 2\sqrt{x(1-x)} \) is not zero in this open interval.
Now, let's check \( f(a) \) and \( f(b) \):
\( f(0) = \sqrt{0(1-0)} = \sqrt{0} = 0 \)
\( f(1) = \sqrt{1(1-1)} = \sqrt{0} = 0 \)
So, \( f(0) = f(1) \).
All three conditions for Rolle's theorem are satisfied. Thus, there exists a point \( c \in (0, 1) \) such that \( f'(c) = 0 \).
Now set \( f'(c) = 0 \):
\( \frac{1-2c}{2\sqrt{c(1-c)}} = 0 \)
This means the numerator must be zero: \( 1-2c = 0 \)
\( 2c = 1 \)
\( c = \frac{1}{2} \)
Since \( \frac{1}{2} \in (0, 1) \), Rolle's theorem is proved.
(iv) Given function: \( f(x) = \cos 2x, x \in [0, \pi] \)
The cosine function is continuous for all real numbers. Therefore, \( f(x) = \cos 2x \) is continuous in the closed interval \( [0, \pi] \) and differentiable in the open interval \( (0, \pi) \).
Now, let's check \( f(a) \) and \( f(b) \):
\( f(0) = \cos(2 \times 0) = \cos(0) = 1 \)
\( f(\pi) = \cos(2 \times \pi) = \cos(2\pi) = 1 \)
So, \( f(0) = f(\pi) \).
All three conditions for Rolle's theorem are satisfied. Thus, there exists a point \( c \in (0, \pi) \) such that \( f'(c) = 0 \).
Let's find the derivative \( f'(x) \):
\( f'(x) = -2 \sin 2x \)
Now set \( f'(c) = 0 \):
\( -2 \sin 2c = 0 \)
\( \sin 2c = 0 \)
For \( 2c \) to be \( 0, \pi, 2\pi, ... \), and since \( c \in (0, \pi) \), it means \( 2c \in (0, 2\pi) \).
In this interval, \( \sin 2c = 0 \) when \( 2c = \pi \).
\( c = \frac{\pi}{2} \)
Since \( \frac{\pi}{2} \in (0, \pi) \), Rolle's theorem is proved.
In simple words: To prove Rolle's theorem, you need to show three things: the function is smooth (no breaks or sharp points), and its value at the start of the interval is the same as its value at the end. Then, you find a point 'c' inside the interval where the slope of the function is zero, just as the theorem says. We did this for all four parts by checking each condition and finding 'c'.

🎯 Exam Tip: Remember that "proving" Rolle's theorem means showing that all its conditions are met and then finding a valid 'c' that satisfies \( f'(c) = 0 \) within the given interval. Clearly state each step: continuity, differentiability, and \( f(a)=f(b) \).

Question 3. Verify the Lagrange's mean value theorem for the following functions :
(i) \( f(x) = x + \frac{1}{x}, x \in [1, 3] \)
(ii) \( f(x) = \frac{x^2-4}{x-1}, x \in [0, 2] \)
(iii) \( f(x) = x^2 - 3x + 2, x \in [2, 3] \)
(iv) \( f(x) = \frac{1}{4x-1}, x \in [1, 4] \)
Answer:
(i) Given function: \( f(x) = x + \frac{1}{x}, x \in [1, 3] \)
The function \( f(x) = x + \frac{1}{x} \) is a rational function. It is continuous everywhere except where the denominator is zero, i.e., \( x=0 \). Since \( x=0 \) is not in the interval \( [1, 3] \), \( f(x) \) is continuous in \( [1, 3] \).
Now, let's find the derivative \( f'(x) \):
\( f'(x) = 1 - \frac{1}{x^2} \)
This derivative exists for all \( x \in (1, 3) \) because \( x^2 \) is not zero in this open interval. So, \( f(x) \) is differentiable in \( (1, 3) \).
Lagrange's Mean Value Theorem (LMVT) states that there exists a point \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
Here, \( a=1, b=3 \).
\( f(a) = f(1) = 1 + \frac{1}{1} = 2 \)
\( f(b) = f(3) = 3 + \frac{1}{3} = \frac{10}{3} \)
So, \( \frac{f(b) - f(a)}{b - a} = \frac{\frac{10}{3} - 2}{3 - 1} = \frac{\frac{10-6}{3}}{2} = \frac{\frac{4}{3}}{2} = \frac{4}{6} = \frac{2}{3} \)
Now, set \( f'(c) = \frac{2}{3} \):
\( 1 - \frac{1}{c^2} = \frac{2}{3} \)
\( 1 - \frac{2}{3} = \frac{1}{c^2} \)
\( \frac{3-2}{3} = \frac{1}{c^2} \)
\( \frac{1}{3} = \frac{1}{c^2} \)
\( c^2 = 3 \)
\( c = \pm \sqrt{3} \)
We need \( c \in (1, 3) \). \( \sqrt{3} \approx 1.732 \), which lies in \( (1, 3) \). \( -\sqrt{3} \) is not in the interval.
Thus, for \( c = \sqrt{3} \), LMVT is verified.
(ii) Given function: \( f(x) = \frac{x^2-4}{x-1}, x \in [0, 2] \)
The function is a rational function. It is continuous everywhere except where the denominator is zero, i.e., \( x=1 \). Since \( x=1 \) is in the interval \( [0, 2] \), the function is discontinuous at \( x=1 \).
For LMVT to apply, the function must be continuous in the closed interval \( [a, b] \). Since it is not continuous at \( x=1 \), LMVT is not satisfied for this function.
(iii) Given function: \( f(x) = x^2 - 3x + 2, x \in [2, 3] \)
Since \( f(x) \) is a polynomial function, it is continuous in the closed interval \( [2, 3] \) and differentiable in the open interval \( (2, 3) \).
LMVT states that there exists a point \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
Here, \( a=2, b=3 \).
\( f(a) = f(2) = (2)^2 - 3(2) + 2 = 4 - 6 + 2 = 0 \)
\( f(b) = f(3) = (3)^2 - 3(3) + 2 = 9 - 9 + 2 = 2 \)
So, \( \frac{f(b) - f(a)}{b - a} = \frac{2 - 0}{3 - 2} = \frac{2}{1} = 2 \)
Now, let's find the derivative \( f'(x) \):
\( f'(x) = 2x - 3 \)
Set \( f'(c) = 2 \):
\( 2c - 3 = 2 \)
\( 2c = 5 \)
\( c = \frac{5}{2} \)
Since \( \frac{5}{2} = 2.5 \), which lies in \( (2, 3) \), LMVT is verified.
(iv) Given function: \( f(x) = \frac{1}{4x-1}, x \in [1, 4] \)
This is a rational function. It is continuous everywhere except where the denominator is zero, i.e., \( 4x-1=0 \implies x=\frac{1}{4} \). Since \( x=\frac{1}{4} \) is not in the interval \( [1, 4] \), \( f(x) \) is continuous in \( [1, 4] \).
Now, let's find the derivative \( f'(x) \):
\( f'(x) = -\frac{1}{(4x-1)^2} \times 4 = -\frac{4}{(4x-1)^2} \)
This derivative exists for all \( x \in (1, 4) \) because the denominator \( (4x-1)^2 \) is not zero in this open interval. So, \( f(x) \) is differentiable in \( (1, 4) \).
LMVT states that there exists a point \( c \in (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
Here, \( a=1, b=4 \).
\( f(a) = f(1) = \frac{1}{4(1)-1} = \frac{1}{3} \)
\( f(b) = f(4) = \frac{1}{4(4)-1} = \frac{1}{15} \)
So, \( \frac{f(b) - f(a)}{b - a} = \frac{\frac{1}{15} - \frac{1}{3}}{4 - 1} = \frac{\frac{1-5}{15}}{3} = \frac{-\frac{4}{15}}{3} = -\frac{4}{45} \)
Now, set \( f'(c) = -\frac{4}{45} \):
\( -\frac{4}{(4c-1)^2} = -\frac{4}{45} \)
\( \frac{1}{(4c-1)^2} = \frac{1}{45} \)
\( (4c-1)^2 = 45 \)
\( 4c-1 = \pm \sqrt{45} = \pm 3\sqrt{5} \)
Case 1: \( 4c-1 = 3\sqrt{5} \)
\( 4c = 1 + 3\sqrt{5} \)
\( c = \frac{1 + 3\sqrt{5}}{4} \approx \frac{1 + 3 \times 2.236}{4} = \frac{1 + 6.708}{4} = \frac{7.708}{4} \approx 1.927 \)
This value \( 1.927 \) is in \( (1, 4) \).
Case 2: \( 4c-1 = -3\sqrt{5} \)
\( 4c = 1 - 3\sqrt{5} \)
\( c = \frac{1 - 3\sqrt{5}}{4} \approx \frac{1 - 6.708}{4} = \frac{-5.708}{4} \approx -1.427 \)
This value \( -1.427 \) is not in \( (1, 4) \).
Thus, for \( c = \frac{1 + 3\sqrt{5}}{4} \), LMVT is verified.
In simple words: Lagrange's Mean Value Theorem (LMVT) says that if a function is continuous and smooth, you can always find a point inside an interval where the slope of the curve is the same as the average slope of the line connecting the start and end points of the interval. We checked each function to make sure it was smooth enough and then found that special point 'c' for which the slope matched the average slope. Part (ii) did not satisfy the theorem because it had a break in the function.

🎯 Exam Tip: For LMVT, carefully check continuity and differentiability first. If the function is not continuous or differentiable in the specified intervals, the theorem doesn't apply. Remember the formula for the average slope: \( \frac{f(b) - f(a)}{b - a} \).