Class 12: Mean Value Theorems

$\displaystyle \textbf{Question 1: }~\text{Discuss the applicability of Rolle's theorem for the following functions} \\ \text{on the indicated intervals:}$

$\displaystyle (i)\;f(x)=3+(x-2)^{2/3}\text{ on }[1,3]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=3+(x-2)^{\tfrac{2}{3}}.$

$\displaystyle \text{Differentiating with respect to } x, \text{ we get}$

$\displaystyle f'(x)=\frac{2}{3}(x-2)^{-\tfrac{1}{3}}$

$\displaystyle \Rightarrow\; f'(x)=\frac{2}{3\,(x-2)^{\tfrac{1}{3}}}$

$\displaystyle \text{Clearly, we observe that at } x=2,\, f'(x) \text{ does not exist.}$

$\displaystyle \text{Since } 2\in[1,3], \text{ the function is not differentiable on } [1,3].$

$\displaystyle \text{Therefore, } f(x) \text{ is not differentiable on } [1,3].$

$\displaystyle \text{Hence, Rolle's Theorem is not applicable for the given function.}$

$\displaystyle (ii)\;f(x)=[x]\text{ for }-1\le x\le 1,\text{ where }[x]\text{ denotes the greatest integer not exceeding }x$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=[x].$

$\displaystyle \text{The domain of } f \text{ is } [-1,1].$

$\displaystyle \text{Let } c\in[-1,1] \text{ such that } c \text{ is not an integer.}$

$\displaystyle \text{Then, } \lim_{x\to c} f(x)=f(c).$

$\displaystyle \text{Thus, } f(x) \text{ is continuous at } x=c.$

$\displaystyle \text{Now, let } c=0.$

$\displaystyle \text{Then, } \lim_{x\to 0^-} f(x)=-1 \neq 0=f(0).$

$\displaystyle \text{Thus, } f \text{ is discontinuous at } x=0.$

$\displaystyle \text{Therefore, } f(x) \text{ is not continuous on } [-1,1].$

$\displaystyle \text{Hence, Rolle's Theorem is not applicable for the given function.}$

$\displaystyle (iii)\;f(x)=\sin\!\left(\frac{1}{x}\right)\text{ for }-1\le x\le 1$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\sin\!\left(\frac{1}{x}\right).$

$\displaystyle \text{The domain of } f \text{ is } [-1,1]\setminus\{0\}.$

$\displaystyle \text{It is known that } \lim_{x\to 0}\sin\!\left(\frac{1}{x}\right) \text{ does not exist.}$

$\displaystyle \text{Thus, } f(x) \text{ is discontinuous at } x=0 \text{ on } [-1,1].$

$\displaystyle \text{Hence, Rolle's Theorem is not applicable for the given function.}$

$\displaystyle (iv)\;f(x)=2x^2-5x+3\text{ on }[1,3]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=2x^2-5x+3 \text{ on } [1,3].$

$\displaystyle \text{The domain of } f \text{ is } [1,3].$

$\displaystyle \text{It is a polynomial function.}$

$\displaystyle \text{Thus, it is continuous on } [1,3] \text{ and differentiable on } (1,3).$

$\displaystyle \text{But } f(1)=0 \text{ and } f(3)=6.$

$\displaystyle \Rightarrow\; f(1)\neq f(3).$

$\displaystyle \text{Hence, Rolle's Theorem is not applicable for the given function.}$

$\displaystyle (v)\;f(x)=x^{2/3}\text{ on }[-1,1]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=x^{\tfrac{2}{3}} \text{ on } [-1,1].$

$\displaystyle \text{The domain of } f \text{ is } [-1,1].$

$\displaystyle \text{Differentiating } f(x) \text{ with respect to } x, \text{ we get}$

$\displaystyle f'(x)=\frac{2}{3}x^{-\tfrac{1}{3}}$

$\displaystyle \text{We observe that at } x=0,\, f'(x) \text{ is not defined.}$

$\displaystyle \text{Hence, Rolle's Theorem is not applicable for the given function.}$

$\displaystyle (vi)\;f(x)=\begin{cases}-4x+5,&0\le x\le 1\\[4pt]2x-3,&1<x\le 2\end{cases}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)= \begin{cases} -4x+5, & 0\le x\le 1,\\ 2x-3, & 1<x\le 2. \end{cases}$

$\displaystyle \text{At } x=1, \text{ we have}$

$\displaystyle \lim_{x\to 1^-} f(x) =\lim_{h\to 0}\,f(1-h) =\lim_{h\to 0}\,[ -4(1-h)+5 ] =1.$

$\displaystyle \text{And}$

$\displaystyle \lim_{x\to 1^+} f(x) =\lim_{h\to 0}\,f(1+h) =\lim_{h\to 0}\,[ 2(1+h)-3 ] =-1.$

$\displaystyle \therefore\; \lim_{x\to 1^-} f(x)\neq \lim_{x\to 1^+} f(x).$

$\displaystyle \text{Thus, } f(x) \text{ is discontinuous at } x=1.$

$\displaystyle \text{Hence, Rolle's Theorem is not applicable for the given function.}$

$\displaystyle \textbf{Question 2: }~\text{Verify Rolle's theorem for each of the following functions on the indicated intervals:}$

$\displaystyle (i)\;f(x)=x^2-8x+12\text{ on }[2,6]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } f(x)=x^2-8x+12.$

$\displaystyle \text{We know that a polynomial function is everywhere differentiable and hence continuous.}$

$\displaystyle \text{Therefore, being a polynomial function, } f(x) \text{ is continuous on } [2,6] \text{ and differentiable on } (2,6).$

$\displaystyle \text{Also,}$

$\displaystyle f(2)=(2)^2-8(2)+12=4-16+12=0.$

$\displaystyle f(6)=(6)^2-8(6)+12=36-48+12=0.$

$\displaystyle \therefore\; f(2)=f(6)=0.$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Now, we show that there exists } c\in(2,6) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=x^2-8x+12.$

$\displaystyle \Rightarrow\; f'(x)=2x-8.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 2x-8=0 \Rightarrow x=4.$

$\displaystyle \text{Thus, } c=4\in(2,6) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (ii)\;f(x)=x^2-4x+3\text{ on }[1,3] \hspace{7.0cm} \;[\text{CBSE 2007}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } f(x)=x^2-4x+3.$

$\displaystyle \text{We know that a polynomial function is everywhere differentiable and hence continuous.}$

$\displaystyle \text{Therefore, being a polynomial function, } f(x) \text{ is continuous on } [1,3] \text{ and differentiable on } (1,3).$

$\displaystyle \text{Also,}$

$\displaystyle f(1)=(1)^2-4(1)+3=1-4+3=0.$

$\displaystyle f(3)=(3)^2-4(3)+3=9-12+3=0.$

$\displaystyle \therefore\; f(1)=f(3)=0.$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Now, we show that there exists } c\in(1,3) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=x^2-4x+3.$

$\displaystyle \Rightarrow\; f'(x)=2x-4.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 2x-4=0 \Rightarrow x=2.$

$\displaystyle \text{Thus, } c=2\in(1,3) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (iii)\;f(x)=(x-1)(x-2)^2\text{ on }[1,2]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } f(x)=(x-1)(x-2)^2.$

$\displaystyle \text{Expanding,}$

$\displaystyle f(x)=(x-1)(x^2-4x+4)=x^3-5x^2+8x-4.$

$\displaystyle \text{We know that a polynomial function is everywhere differentiable and hence continuous.}$

$\displaystyle \text{Therefore, being a polynomial function, } f(x) \text{ is continuous on } [1,2] \text{ and differentiable on } (1,2).$

$\displaystyle \text{Also,}$

$\displaystyle f(1)=(1)^3-5(1)^2+8(1)-4=0.$

$\displaystyle f(2)=(2)^3-5(2)^2+8(2)-4=0.$

$\displaystyle \therefore\; f(1)=f(2)=0.$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Now, we show that there exists } c\in(1,2) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=x^3-5x^2+8x-4.$

$\displaystyle \Rightarrow\; f'(x)=3x^2-10x+8.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 3x^2-10x+8=0.$

$\displaystyle \Rightarrow\; 3x^2-6x-4x+8=0.$

$\displaystyle \Rightarrow\; 3x(x-2)-4(x-2)=0.$

$\displaystyle \Rightarrow\; (x-2)(3x-4)=0.$

$\displaystyle \Rightarrow\; x=2,\;\frac{4}{3}.$

$\displaystyle \text{Thus, } c=\frac{4}{3}\in(1,2) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (iv)\;f(x)=x(x-1)^2\text{ on }[0,1]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } f(x)=x(x-1)^2.$

$\displaystyle \Rightarrow\; f(x)=x(x^2-2x+1).$

$\displaystyle \therefore\; f(x)=x^3-2x^2+x.$

$\displaystyle \text{We know that a polynomial function is everywhere differentiable and hence continuous.}$

$\displaystyle \text{Therefore, being a polynomial function, } f(x) \text{ is continuous on } [0,1] \text{ and differentiable on } (0,1).$

$\displaystyle \text{Also,}$

$\displaystyle f(0)=0 \text{ and } f(1)=0.$

$\displaystyle \therefore\; f(0)=f(1).$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Now, we show that there exists } c\in(0,1) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=x^3-2x^2+x.$

$\displaystyle \Rightarrow\; f'(x)=3x^2-4x+1.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 3x^2-4x+1=0.$

$\displaystyle \Rightarrow\; 3x^2-3x-x+1=0.$

$\displaystyle \Rightarrow\; 3x(x-1)-1(x-1)=0.$

$\displaystyle \Rightarrow\; (x-1)(3x-1)=0.$

$\displaystyle \Rightarrow\; x=1,\;\frac{1}{3}.$

$\displaystyle \text{Thus, } c=\frac{1}{3}\in(0,1) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (v)\;f(x)=(x^2-1)(x-2)\text{ on }[-1,2]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } f(x)=(x^2-1)(x-2).$

$\displaystyle \text{Expanding,}$

$\displaystyle f(x)=x^3-2x^2-x+2.$

$\displaystyle \text{We know that a polynomial function is everywhere differentiable and hence continuous.}$

$\displaystyle \text{Therefore, being a polynomial function, } f(x) \text{ is continuous on } [-1,2] \text{ and differentiable on } (-1,2).$

$\displaystyle \text{Also,}$

$\displaystyle f(-1)=0 \text{ and } f(2)=0.$

$\displaystyle \therefore\; f(-1)=f(2).$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Now, we show that there exists } c\in(-1,2) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=x^3-2x^2-x+2.$

$\displaystyle \Rightarrow\; f'(x)=3x^2-4x-1.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 3x^2-4x-1=0.$

$\displaystyle \Rightarrow\; x=\frac{4\pm\sqrt{16+12}}{6}.$

$\displaystyle \Rightarrow\; x=\frac{4\pm\sqrt{28}}{6}.$

$\displaystyle \Rightarrow\; x=\frac{4\pm 2\sqrt{7}}{6}.$

$\displaystyle \Rightarrow\; x=\frac{2\pm \sqrt{7}}{3}.$

$\displaystyle \text{Thus, } c=\frac{1}{3}(2-\sqrt{7}),\;\frac{1}{3}(2+\sqrt{7})\in(-1,2) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (vi)\;f(x)=x(x-4)^2\text{ on }[0,4]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given function is } f(x)=x(x-4)^2, \text{ which can be rewritten as } f(x)=x^3-8x^2+16x.$

$\displaystyle \text{We know that a polynomial function is everywhere differentiable and hence continuous.}$

$\displaystyle \text{Therefore, being a polynomial function, } f(x) \text{ is continuous on } [0,4] \text{ and differentiable on } (0,4).$

$\displaystyle \text{Also,}$

$\displaystyle f(0)=0 \text{ and } f(4)=0.$

$\displaystyle \therefore\; f(0)=f(4).$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Now, we show that there exists } c\in(0,4) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=x^3-8x^2+16x.$

$\displaystyle \Rightarrow\; f'(x)=3x^2-16x+16.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 3x^2-16x+16=0.$

$\displaystyle \Rightarrow\; 3x^2-12x-4x+16=0.$

$\displaystyle \Rightarrow\; 3x(x-4)-4(x-4)=0.$

$\displaystyle \Rightarrow\; (x-4)(3x-4)=0.$

$\displaystyle \Rightarrow\; x=4,\;\frac{4}{3}.$

$\displaystyle \text{Thus, } c=\frac{4}{3}\in(0,4) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (vii)\;f(x)=x(x-2)^2\text{ on }[0,2]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=x(x-2)^2, \text{ which can be rewritten as } f(x)=x^3-4x^2+4x.$

$\displaystyle \text{We know that a polynomial function is everywhere differentiable and hence continuous.}$

$\displaystyle \text{Therefore, being a polynomial function, } f(x) \text{ is continuous on } [0,2] \text{ and differentiable on } (0,2).$

$\displaystyle \text{Also,}$

$\displaystyle f(0)=0 \text{ and } f(2)=0.$

$\displaystyle \therefore\; f(0)=f(2).$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Now, we show that there exists } c\in(0,2) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=x^3-4x^2+4x.$

$\displaystyle \Rightarrow\; f'(x)=3x^2-8x+4.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 3x^2-8x+4=0.$

$\displaystyle \Rightarrow\; 3x^2-6x-2x+4=0.$

$\displaystyle \Rightarrow\; 3x(x-2)-2(x-2)=0.$

$\displaystyle \Rightarrow\; (x-2)(3x-2)=0.$

$\displaystyle \Rightarrow\; x=2,\;\frac{2}{3}.$

$\displaystyle \text{Thus, } c=\frac{2}{3}\in(0,2) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (viii)\;f(x)=x^2+5x+6\text{ on }[-3,-2]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=x^2+5x+6.$

$\displaystyle \text{We know that a polynomial function is everywhere differentiable and hence continuous.}$

$\displaystyle \text{Therefore, being a polynomial function, } f(x) \text{ is continuous on } [-3,-2] \\ \text{ and differentiable on } (-3,-2).$

$\displaystyle \text{Also,}$

$\displaystyle f(-3)=(-3)^2+5(-3)+6=9-15+6=0.$

$\displaystyle f(-2)=(-2)^2+5(-2)+6=4-10+6=0.$

$\displaystyle \therefore\; f(-3)=f(-2)=0.$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Now, we show that there exists } c\in(-3,-2) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=x^2+5x+6.$

$\displaystyle \Rightarrow\; f'(x)=2x+5.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 2x+5=0.$

$\displaystyle \Rightarrow\; x=-\frac{5}{2}.$

$\displaystyle \text{Thus, } c=-\frac{5}{2}\in(-3,-2) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle \textbf{Question 3: }~\text{Verify Rolle's theorem for each of the following functions on the indicated intervals:}$

$\displaystyle (i)\;f(x)=\cos 2\!\left(x-\frac{\pi}{4}\right)\text{ on }\left[0,\frac{\pi}{2}\right]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\cos 2\!\left(x-\frac{\pi}{4}\right)=\cos\!\left(2x-\frac{\pi}{2}\right)=\sin 2x.$

$\displaystyle \text{Since } \sin 2x \text{ is everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } \sin 2x \text{ is continuous on } \left[0,\frac{\pi}{2}\right] \text{ and differentiable on } \left(0,\frac{\pi}{2}\right).$

$\displaystyle \text{Also,}$

$\displaystyle f\!\left(\frac{\pi}{2}\right)=0 \text{ and } f(0)=0.$

$\displaystyle \therefore\; f\!\left(\frac{\pi}{2}\right)=f(0).$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Now, we show that there exists } c\in\left(0,\frac{\pi}{2}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\sin 2x.$

$\displaystyle \Rightarrow\; f'(x)=2\cos 2x.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 2\cos 2x=0.$

$\displaystyle \Rightarrow\; \cos 2x=0.$

$\displaystyle \Rightarrow\; 2x=\frac{\pi}{2}.$

$\displaystyle \Rightarrow\; x=\frac{\pi}{4}.$

$\displaystyle \text{Thus, } c=\frac{\pi}{4}\in\left(0,\frac{\pi}{2}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (ii)\;f(x)=\sin 2x\text{ on }\left[0,\frac{\pi}{2}\right]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\sin 2x.$

$\displaystyle \text{Since } \sin 2x \text{ is everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } \sin 2x \text{ is continuous on } \left[0,\frac{\pi}{2}\right] \text{ and differentiable on } \left(0,\frac{\pi}{2}\right).$

$\displaystyle \text{Also,}$

$\displaystyle f\!\left(\frac{\pi}{2}\right)=0 \text{ and } f(0)=0.$

$\displaystyle \therefore\; f\!\left(\frac{\pi}{2}\right)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in\left(0,\frac{\pi}{2}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\sin 2x.$

$\displaystyle \Rightarrow\; f'(x)=2\cos 2x.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 2\cos 2x=0.$

$\displaystyle \Rightarrow\; \cos 2x=0.$

$\displaystyle \Rightarrow\; 2x=\frac{\pi}{2}.$

$\displaystyle \Rightarrow\; x=\frac{\pi}{4}.$

$\displaystyle \text{Thus, } c=\frac{\pi}{4}\in\left(0,\frac{\pi}{2}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (iii)\;f(x)=\cos 2x\text{ on }\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\cos 2x.$

$\displaystyle \text{Since } \cos 2x \text{ is everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } \cos 2x \text{ is continuous on } \left[-\frac{\pi}{4},\frac{\pi}{4}\right] \text{ and differentiable on } \left(-\frac{\pi}{4},\frac{\pi}{4}\right).$

$\displaystyle \text{Also,}$

$\displaystyle f\!\left(\frac{\pi}{4}\right)=0 \text{ and } f\!\left(-\frac{\pi}{4}\right)=0.$

$\displaystyle \therefore\; f\!\left(\frac{\pi}{4}\right)=f\!\left(-\frac{\pi}{4}\right).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\cos 2x.$

$\displaystyle \Rightarrow\; f'(x)=-2\sin 2x.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow -2\sin 2x=0.$

$\displaystyle \Rightarrow\; \sin 2x=0.$

$\displaystyle \Rightarrow\; 2x=0.$

$\displaystyle \Rightarrow\; x=0.$

$\displaystyle \text{Since } c=0\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (iv)\;f(x)=e^x\sin x\text{ on }[0,\pi]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=e^x\sin x.$

$\displaystyle \text{Since } \sin x \text{ and } e^x \text{ are everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, being a product of these two functions, } f(x) \text{ is continuous on } [0,\pi] \text{ and differentiable on } (0,\pi).$

$\displaystyle \text{Also,}$

$\displaystyle f(\pi)=0 \text{ and } f(0)=0.$

$\displaystyle \therefore\; f(\pi)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=e^x\sin x.$

$\displaystyle \Rightarrow\; f'(x)=e^x(\sin x+\cos x).$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow e^x(\sin x+\cos x)=0.$

$\displaystyle \Rightarrow\; \sin x+\cos x=0.$

$\displaystyle \Rightarrow\; \tan x=-1.$

$\displaystyle \Rightarrow\; x=\frac{3\pi}{4}.$

$\displaystyle \text{Since } c=\frac{3\pi}{4}\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (v)\;f(x)=e^x\cos x\text{ on }\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=e^x\cos x.$

$\displaystyle \text{Since } \cos x \text{ and } e^x \text{ are everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, being a product of these two functions, } f(x) \text{ is continuous on } \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \text{ and differentiable on } \left(-\frac{\pi}{2},\frac{\pi}{2}\right).$

$\displaystyle \text{Also,}$

$\displaystyle f\!\left(-\frac{\pi}{2}\right)=0 \text{ and } f\!\left(\frac{\pi}{2}\right)=0.$

$\displaystyle \therefore\; f\!\left(-\frac{\pi}{2}\right)=f\!\left(\frac{\pi}{2}\right).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=e^x\cos x.$

$\displaystyle \Rightarrow\; f'(x)=e^x(\cos x-\sin x).$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow e^x(\cos x-\sin x)=0.$

$\displaystyle \Rightarrow\; \cos x-\sin x=0.$

$\displaystyle \Rightarrow\; \tan x=1.$

$\displaystyle \Rightarrow\; x=\frac{\pi}{4}.$

$\displaystyle \text{Since } c=\frac{\pi}{4}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (vi)\;f(x)=\cos 2x\text{ on }[0,\pi]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\cos 2x.$

$\displaystyle \text{Since } \cos 2x \text{ is everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } f(x) \text{ is continuous on } [0,\pi] \text{ and differentiable on } (0,\pi).$

$\displaystyle \text{Also,}$

$\displaystyle f(\pi)=\cos 2\pi=1 \text{ and } f(0)=\cos 0=1.$

$\displaystyle \therefore\; f(\pi)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\cos 2x.$

$\displaystyle \Rightarrow\; f'(x)=-2\sin 2x.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow -2\sin 2x=0.$

$\displaystyle \Rightarrow\; \sin 2x=0.$

$\displaystyle \Rightarrow\; 2x=\pi.$

$\displaystyle \Rightarrow\; x=\frac{\pi}{2}.$

$\displaystyle \text{Thus, } c=\frac{\pi}{2}\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (vii)\;f(x)=\frac{\sin x}{e^x}\text{ on }0\le x\le \pi$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\frac{\sin x}{e^x}.$

$\displaystyle \text{Since } \sin x \text{ and } e^x \text{ are everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, being the quotient of these two functions, } f(x) \text{ is continuous on } [0,\pi] \text{ and differentiable on } (0,\pi).$

$\displaystyle \text{Also,}$

$\displaystyle f(\pi)=\frac{\sin \pi}{e^\pi}=0 \text{ and } f(0)=\frac{\sin 0}{e^0}=0.$

$\displaystyle \therefore\; f(\pi)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\frac{\sin x}{e^x}.$

$\displaystyle \Rightarrow\; f'(x)=\frac{e^x\cos x-e^x\sin x}{e^{2x}}.$

$\displaystyle \Rightarrow\; f'(x)=\frac{\cos x-\sin x}{e^x}.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow \frac{\cos x-\sin x}{e^x}=0.$

$\displaystyle \Rightarrow\; \cos x-\sin x=0.$

$\displaystyle \Rightarrow\; \tan x=1.$

$\displaystyle \Rightarrow\; x=\frac{\pi}{4}.$

$\displaystyle \text{Thus, } c=\frac{\pi}{4}\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (viii)\;f(x)=\sin 3x\text{ on }[0,\pi]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\sin 3x.$

$\displaystyle \text{Since } \sin 3x \text{ is everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } f(x) \text{ is continuous on } [0,\pi] \text{ and differentiable on } (0,\pi).$

$\displaystyle \text{Also,}$

$\displaystyle f(\pi)=\sin 3\pi=0 \text{ and } f(0)=\sin 0=0.$

$\displaystyle \therefore\; f(\pi)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\sin 3x.$

$\displaystyle \Rightarrow\; f'(x)=3\cos 3x.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 3\cos 3x=0.$

$\displaystyle \Rightarrow\; \cos 3x=0.$

$\displaystyle \Rightarrow\; 3x=\frac{\pi}{2},\;\frac{3\pi}{2}.$

$\displaystyle \Rightarrow\; x=\frac{\pi}{6},\;\frac{\pi}{2}.$

$\displaystyle \text{Thus, } c=\frac{\pi}{6},\;\frac{\pi}{2}\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (ix)\;f(x)=e^{\,1-x^2}\text{ on }[-1,1]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=e^{\,1-x^2}.$

$\displaystyle \text{Since an exponential function is everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } f(x)=e^{\,1-x^2} \text{ is continuous on } [-1,1] \text{ and differentiable on } (-1,1).$

$\displaystyle \text{Also,}$

$\displaystyle f(1)=e^{\,1-1}=1 \text{ and } f(-1)=e^{\,1-1}=1.$

$\displaystyle \therefore\; f(1)=f(-1).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in(-1,1) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=e^{\,1-x^2}.$

$\displaystyle \Rightarrow\; f'(x)=e^{\,1-x^2}\cdot(-2x)=-2x\,e^{\,1-x^2}.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow -2x\,e^{\,1-x^2}=0.$

$\displaystyle \Rightarrow\; x=0.$

$\displaystyle \text{Thus, } c=0\in(-1,1) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (x)\;f(x)=\log(x^2+2)-\log 3\text{ on }[-1,1]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\log(x^2+2)-\log 3, \text{ which can be rewritten as } f(x)=\log\!\left(\frac{x^2+2}{3}\right).$

$\displaystyle \text{Since a logarithmic function is differentiable and hence continuous in its domain,}$

$\displaystyle \text{therefore, } f(x)=\log\!\left(\frac{x^2+2}{3}\right) \text{ is continuous on } [-1,1] \text{ and differentiable on } (-1,1).$

$\displaystyle \text{Also,}$

$\displaystyle f(1)=\log\!\left(\frac{1+2}{3}\right)=\log 1=0 \text{ and } f(-1)=\log\!\left(\frac{1+2}{3}\right)=\log 1=0.$

$\displaystyle \therefore\; f(1)=f(-1).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in(-1,1) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\log\!\left(\frac{x^2+2}{3}\right).$

$\displaystyle \Rightarrow\; f'(x)=\frac{1}{\frac{x^2+2}{3}}\cdot\frac{2x}{3}=\frac{2x}{x^2+2}.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow \frac{2x}{x^2+2}=0.$

$\displaystyle \Rightarrow\; x=0.$

$\displaystyle \text{Thus, } c=0\in(-1,1) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (xi)\;f(x)=\sin x+\cos x\text{ on }\left[0,\frac{\pi}{2}\right]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\sin x+\cos x.$

$\displaystyle \text{Since } \sin x \text{ and } \cos x \text{ are everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } f(x)=\sin x+\cos x \text{ is continuous on } \left[0,\frac{\pi}{2}\right] \text{ and differentiable on } \left(0,\frac{\pi}{2}\right).$

$\displaystyle \text{Also,}$

$\displaystyle f\!\left(\frac{\pi}{2}\right)=\sin\!\left(\frac{\pi}{2}\right)+\cos\!\left(\frac{\pi}{2}\right)=1 \text{ and } f(0)=\sin 0+\cos 0=1.$

$\displaystyle \therefore\; f\!\left(\frac{\pi}{2}\right)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in\left(0,\frac{\pi}{2}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\sin x+\cos x.$

$\displaystyle \Rightarrow\; f'(x)=\cos x-\sin x.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow \cos x-\sin x=0.$

$\displaystyle \Rightarrow\; \tan x=1.$

$\displaystyle \Rightarrow\; x=\frac{\pi}{4}.$

$\displaystyle \text{Thus, } c=\frac{\pi}{4}\in\left(0,\frac{\pi}{2}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (xii)\;f(x)=2\sin x+\sin 2x\text{ on }[0,\pi]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=2\sin x+\sin 2x.$

$\displaystyle \text{Since } \sin x \text{ and } \sin 2x \text{ are everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } f(x) \text{ is continuous on } [0,\pi] \text{ and differentiable on } (0,\pi).$

$\displaystyle \text{Also,}$

$\displaystyle f(\pi)=2\sin \pi+\sin 2\pi=0 \text{ and } f(0)=2\sin 0+\sin 0=0.$

$\displaystyle \therefore\; f(\pi)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=2\sin x+\sin 2x.$

$\displaystyle \Rightarrow\; f'(x)=2\cos x+2\cos 2x.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 2\cos x+2\cos 2x=0.$

$\displaystyle \Rightarrow\; \cos x+\cos 2x=0.$

$\displaystyle \Rightarrow\; \cos x+(2\cos^2 x-1)=0.$

$\displaystyle \Rightarrow\; 2\cos^2 x+\cos x-1=0.$

$\displaystyle \Rightarrow\; (\cos x+1)(2\cos x-1)=0.$

$\displaystyle \Rightarrow\; \cos x=-1 \text{ or } \cos x=\frac{1}{2}.$

$\displaystyle \Rightarrow\; x=\pi \text{ or } x=\frac{\pi}{3}.$

$\displaystyle \text{Thus, } c=\frac{\pi}{3}\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (xiii)\;f(x)=\frac{x}{2}-\sin\!\left(\frac{\pi x}{6}\right)\text{ on }[-1,0]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\frac{x}{2}-\sin\!\left(\frac{\pi x}{6}\right).$

$\displaystyle \text{Since } \frac{x}{2} \text{ and } \sin\!\left(\frac{\pi x}{6}\right) \text{ are everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } f(x) \text{ is continuous on } [-1,0] \text{ and differentiable on } (-1,0).$

$\displaystyle \text{Also,}$

$\displaystyle f(-1)=-\frac{1}{2}-\sin\!\left(-\frac{\pi}{6}\right)=-\frac{1}{2}+\frac{1}{2}=0 \text{ and } f(0)=0-\sin 0=0.$

$\displaystyle \therefore\; f(-1)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in(-1,0) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\frac{x}{2}-\sin\!\left(\frac{\pi x}{6}\right).$

$\displaystyle \Rightarrow\; f'(x)=\frac{1}{2}-\frac{\pi}{6}\cos\!\left(\frac{\pi x}{6}\right).$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow \frac{1}{2}-\frac{\pi}{6}\cos\!\left(\frac{\pi x}{6}\right)=0.$

$\displaystyle \Rightarrow\; \cos\!\left(\frac{\pi x}{6}\right)=\frac{3}{\pi}.$

$\displaystyle \Rightarrow\; \frac{\pi x}{6}=\cos^{-1}\!\left(\frac{3}{\pi}\right).$

$\displaystyle \Rightarrow\; x=\frac{6}{\pi}\cos^{-1}\!\left(\frac{3}{\pi}\right).$

$\displaystyle \text{Thus, one value of } c \text{ in } (-1,0) \text{ is } c=-\frac{6}{\pi}\cos^{-1}\!\left(\frac{3}{\pi}\right), \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (xiv)\;f(x)=\frac{6x}{\pi}-4\sin^2 x\text{ on }\left[0,\frac{\pi}{6}\right]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\frac{6x}{\pi}-4\sin^2 x.$

$\displaystyle \text{Since } \frac{6x}{\pi} \text{ and } \sin^2 x \text{ are everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } f(x) \text{ is continuous on } \left[0,\frac{\pi}{6}\right] \text{ and differentiable on } \left(0,\frac{\pi}{6}\right).$

$\displaystyle \text{Also,}$

$\displaystyle f\!\left(\frac{\pi}{6}\right)=\frac{6}{\pi}\cdot\frac{\pi}{6}-4\sin^2\!\left(\frac{\pi}{6}\right)=1-4\cdot\frac{1}{4}=0 \text{ and } f(0)=0.$

$\displaystyle \therefore\; f\!\left(\frac{\pi}{6}\right)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in\left(0,\frac{\pi}{6}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\frac{6x}{\pi}-4\sin^2 x.$

$\displaystyle \Rightarrow\; f'(x)=\frac{6}{\pi}-8\sin x\cos x.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow \frac{6}{\pi}-8\sin x\cos x=0.$

$\displaystyle \Rightarrow\; 8\sin x\cos x=\frac{6}{\pi}.$

$\displaystyle \Rightarrow\; \sin 2x=\frac{3}{2\pi}.$

$\displaystyle \Rightarrow\; 2x=\sin^{-1}\!\left(\frac{3}{2\pi}\right).$

$\displaystyle \Rightarrow\; x=\frac{1}{2}\sin^{-1}\!\left(\frac{3}{2\pi}\right).$

$\displaystyle \text{Thus, } c=\frac{1}{2}\sin^{-1}\!\left(\frac{3}{2\pi}\right)\in\left(0,\frac{\pi}{6}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (xv)\;f(x)=4^{\sin x}\text{ on }[0,\pi]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=4^{\sin x}.$

$\displaystyle \text{Since the sine function and the exponential function are everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } f(x)=4^{\sin x} \text{ is continuous on } [0,\pi] \text{ and differentiable on } (0,\pi).$

$\displaystyle \text{Also,}$

$\displaystyle f(\pi)=4^{\sin \pi}=1 \text{ and } f(0)=4^{\sin 0}=1.$

$\displaystyle \therefore\; f(\pi)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=4^{\sin x}.$

$\displaystyle \Rightarrow\; f'(x)=4^{\sin x}(\cos x)\log 4.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 4^{\sin x}(\cos x)\log 4=0.$

$\displaystyle \Rightarrow\; \cos x=0.$

$\displaystyle \Rightarrow\; x=\frac{\pi}{2}.$

$\displaystyle \text{Thus, } c=\frac{\pi}{2}\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (xvi)\;f(x)=x^2-5x+4\text{ on }[1,4] \hspace{7.0cm} \;[\text{CBSE 2007}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{According to Rolle's Theorem, if } f(x) \text{ is a real-valued function defined on } [a,b] \text{ such that it is continuous on } [a,b],$

$\displaystyle \text{differentiable on } (a,b) \text{ and } f(a)=f(b), \text{ then there exists a real number } c\in(a,b) \text{ such that } f'(c)=0.$

$\displaystyle \text{Now, } f(x) \text{ is defined for all } x\in[1,4].$

$\displaystyle \text{At each point of } [1,4], \text{ the limit of } f(x) \text{ is equal to the value of the function. Therefore, } f(x) \text{ is continuous on } [1,4].$

$\displaystyle \text{Also, } f'(x)=2 \text{ exists for all } x\in(1,4).$

$\displaystyle \text{So, } f(x) \text{ is differentiable on } (1,4).$

$\displaystyle \text{Also,}$

$\displaystyle f(1)=2(1)-5=-3 \text{ and } f(4)=2(4)-5=3.$

$\displaystyle \text{Since } f(1)\neq f(4), \text{ the condition } f(a)=f(b) \text{ is not satisfied.}$

$\displaystyle \text{Hence, Rolle's Theorem is not applicable for the given function.}$

$\displaystyle (xvii)\;f(x)=\sin^4 x+\cos^4 x\text{ on }\left[0,\frac{\pi}{2}\right]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The given function is } f(x)=\sin^4 x+\cos^4 x.$

$\displaystyle \text{Since } \sin x \text{ and } \cos x \text{ are everywhere continuous and differentiable,}$

$\displaystyle \text{therefore, } f(x)=\sin^4 x+\cos^4 x \text{ is continuous on } \left[0,\frac{\pi}{2}\right] \text{ and differentiable on } \left(0,\frac{\pi}{2}\right).$

$\displaystyle \text{Also,}$

$\displaystyle f\!\left(\frac{\pi}{2}\right)=\sin^4\!\left(\frac{\pi}{2}\right)+\cos^4\!\left(\frac{\pi}{2}\right)=1 \text{ and } f(0)=\sin^4 0+\cos^4 0=1.$

$\displaystyle \therefore\; f\!\left(\frac{\pi}{2}\right)=f(0).$

$\displaystyle \text{Thus, } f(x) \text{ satisfies all the conditions of Rolle's Theorem.}$

$\displaystyle \text{Now, we show that there exists } c\in\left(0,\frac{\pi}{2}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{We have } f(x)=\sin^4 x+\cos^4 x.$

$\displaystyle \Rightarrow\; f'(x)=4\sin^3 x\cos x-4\cos^3 x\sin x.$

$\displaystyle \therefore\; f'(x)=0 \Rightarrow 4\sin^3 x\cos x-4\cos^3 x\sin x=0.$

$\displaystyle \Rightarrow\; \sin^3 x\cos x-\cos^3 x\sin x=0.$

$\displaystyle \Rightarrow\; \tan^3 x-\tan x=0.$

$\displaystyle \Rightarrow\; \tan x(\tan^2 x-1)=0.$

$\displaystyle \Rightarrow\; \tan x=0 \text{ or } \tan x=\pm 1.$

$\displaystyle \Rightarrow\; x=0,\;\frac{\pi}{4},\;\frac{3\pi}{4}.$

$\displaystyle \text{Thus, } c=\frac{\pi}{4}\in\left(0,\frac{\pi}{2}\right) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle (xviii)\;f(x)=\sin x-\sin 2x\text{ on }[0,\pi]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{First, we state the conditions for the applicability of Rolle's Theorem:}$

$\displaystyle \text{For a real-valued function } f:$

$\displaystyle \text{(a) } f \text{ is continuous on the closed interval } [a,b].$

$\displaystyle \text{(b) } f \text{ is differentiable on the open interval } (a,b).$

$\displaystyle \text{(c) } f(a)=f(b).$

$\displaystyle \text{Then, there exists at least one } c\in(a,b) \text{ such that } f'(c)=0.$

$\displaystyle \text{The given function is } f(x)=\sin x-\sin 2x \text{ on } [0,\pi].$

$\displaystyle \text{We know that the sine function is continuous and differentiable on } \mathbb{R}.$

$\displaystyle \text{Hence, } f(x) \text{ is continuous on } [0,\pi] \text{ and differentiable on } (0,\pi).$

$\displaystyle \text{Now, we check the values of } f(x) \text{ at the endpoints.}$

$\displaystyle f(0)=\sin 0-\sin 0=0.$

$\displaystyle f(\pi)=\sin \pi-\sin 2\pi=0.$

$\displaystyle \therefore\; f(0)=f(\pi).$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Now, we find } f'(x).$

$\displaystyle f'(x)=\cos x-2\cos 2x.$

$\displaystyle \text{Setting } f'(x)=0,$

$\displaystyle \cos x-2\cos 2x=0.$

$\displaystyle \Rightarrow\; \cos x-2(2\cos^2 x-1)=0.$

$\displaystyle \Rightarrow\; \cos x-4\cos^2 x+2=0.$

$\displaystyle \Rightarrow\; 4\cos^2 x-\cos x-2=0.$

$\displaystyle \Rightarrow\; \cos x=\frac{1\pm\sqrt{1+32}}{8}=\frac{1\pm\sqrt{33}}{8}.$

$\displaystyle \text{Thus, there exists } c\in(0,\pi) \text{ such that } f'(c)=0.$

$\displaystyle \text{Hence, Rolle's Theorem is verified.}$

$\displaystyle \textbf{Question 7: }~\text{Using Rolle's theorem, find points on the curve }y=16-x^2,\;x\in[-1,1],\text{ where tangent is parallel to }x\text{-axis.} \hspace{2.0cm} \;[\text{ CBSE 2000}]$

$\displaystyle \text{Answer:}$

$\displaystyle \text{The equation of the curve is } y=16-x^2. \text{1}$

$\displaystyle \text{Let } P(x_1,y_1) \text{ be a point on the curve where the tangent is parallel to the } x\text{-axis.}$

$\displaystyle \text{Then, } \left(\frac{dy}{dx}\right)_P=0. \text{2}$

$\displaystyle \text{Differentiating (1) with respect to } x, \text{ we get}$

$\displaystyle \frac{dy}{dx}=-2x.$

$\displaystyle \Rightarrow\; \left(\frac{dy}{dx}\right)_P=-2x_1.$

$\displaystyle \Rightarrow\; -2x_1=0 \; \text{(from (2))}.$

$\displaystyle \Rightarrow\; x_1=0.$

$\displaystyle \text{Since } P(x_1,y_1) \text{ lies on the curve } y=16-x^2,$

$\displaystyle \therefore\; y_1=16-x_1^2.$

$\displaystyle \text{When } x_1=0,$

$\displaystyle y_1=16.$

$\displaystyle \text{Hence, } (0,16) \text{ is the required point.}$

$\displaystyle \textbf{Question 8: }~\text{At what points on the following curves, is the tangent parallel to }x\text{-axis?}$

$\displaystyle (i)\;y=x^2\text{ on }[-2,2]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } f(x)=x^2.$

$\displaystyle \text{Since } f(x) \text{ is a polynomial function, it is continuous on } [-2,2] \text{ and differentiable on } (-2,2).$

$\displaystyle \text{Also,}$

$\displaystyle f(2)=2^2=4 \text{ and } f(-2)=(-2)^2=4.$

$\displaystyle \therefore\; f(2)=f(-2).$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one point } c\in(-2,2) \text{ such that } f'(c)=0.$

$\displaystyle \text{Now, } f'(x)=2x.$

$\displaystyle \therefore\; f'(c)=0 \Rightarrow 2c=0 \Rightarrow c=0.$

$\displaystyle \text{Hence, } f(c)=f(0)=0.$

$\displaystyle \text{By the geometrical interpretation of Rolle's Theorem, } (0,0) \text{ is the point on } y=x^2 \text{ where the tangent is parallel to the } x\text{-axis.}$

$\displaystyle (ii)\;y=e^{\,1-x^2}\text{ on }[-1,1]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } f(x)=e^{\,1-x^2}.$

$\displaystyle \text{Since } f(x) \text{ is an exponential function, it is continuous on } [-1,1] \text{ and differentiable on } (-1,1).$

$\displaystyle \text{Also,}$

$\displaystyle f(1)=e^{\,1-1}=1 \text{ and } f(-1)=e^{\,1-1}=1.$

$\displaystyle \therefore\; f(1)=f(-1).$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one point } c\in(-1,1) \text{ such that } f'(c)=0.$

$\displaystyle \text{Now, } f'(x)=-2x\,e^{\,1-x^2}.$

$\displaystyle \therefore\; f'(c)=0 \Rightarrow -2c\,e^{\,1-c^2}=0.$

$\displaystyle \Rightarrow\; c=0 \quad (\text{since } e^{\,1-c^2}\neq 0).$

$\displaystyle \therefore\; f(c)=f(0)=e.$

$\displaystyle \text{By the geometrical interpretation of Rolle's Theorem, } (0,e) \text{ is the point on } y=e^{\,1-x^2} \text{ where the tangent is parallel to the } x\text{-axis.}$

$\displaystyle (iii)\;y=12(x+1)(x-2)\text{ on }[-1,2]$

$\displaystyle \text{Let } f(x)=12(x+1)(x-2). \text{1}$

$\displaystyle \Rightarrow\; f(x)=12(x^2-x-2).$

$\displaystyle \Rightarrow\; f(x)=12x^2-12x-24.$

$\displaystyle \text{Since } f(x) \text{ is a polynomial function, it is continuous on } [-1,2] \text{ and differentiable on } (-1,2).$

$\displaystyle \text{Also,}$

$\displaystyle f(2)=12(3)(0)=0 \text{ and } f(-1)=12(0)(-3)=0.$

$\displaystyle \therefore\; f(2)=f(-1).$

$\displaystyle \text{Thus, all the conditions of Rolle's Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one point } c\in(-1,2) \text{ such that } f'(c)=0.$

$\displaystyle \text{Now, } f'(x)=24x-12.$

$\displaystyle \therefore\; f'(c)=0 \Rightarrow 24c-12=0 \Rightarrow c=\frac{1}{2}.$

$\displaystyle \text{Using (1),}$

$\displaystyle f\!\left(\frac{1}{2}\right)=12\!\left(\frac{3}{2}\right)\!\left(-\frac{3}{2}\right)=-27.$

$\displaystyle \text{By the geometrical interpretation of Rolle's Theorem, } \left(\frac{1}{2},-27\right) \text{ is the point on } y=12(x+1)(x-2)$

$\displaystyle \text{where the tangent is parallel to the } x\text{-axis.}$

$\displaystyle \textbf{Question 9: }~\text{If }f:[-5,5]\to\mathbb{R}\text{ is differentiable and if }f'(x) \text{ doesn't vanish anywhere,} \\ \text{ then prove that }f(-5)\ne f(5).$

$\displaystyle \text{Answer:}$

$\displaystyle \text{It is given that } f:[-5,5]\to\mathbb{R} \text{ is a differentiable function.}$

$\displaystyle \text{Every differentiable function is continuous. Hence,}$

$\displaystyle \text{(a) } f \text{ is continuous on } [-5,5].$

$\displaystyle \text{(b) } f \text{ is differentiable on } (-5,5).$

$\displaystyle \text{Therefore, by the Mean Value Theorem, there exists } c\in(-5,5) \text{ such that}$

$\displaystyle f'(c)=\frac{f(5)-f(-5)}{5-(-5)}=\frac{f(5)-f(-5)}{10}.$

$\displaystyle \Rightarrow\; 10f'(c)=f(5)-f(-5).$

$\displaystyle \text{It is also given that } f'(x) \text{ does not vanish anywhere.}$

$\displaystyle \Rightarrow\; f'(c)\neq 0.$

$\displaystyle \Rightarrow\; 10f'(c)\neq 0.$

$\displaystyle \Rightarrow\; f(5)-f(-5)\neq 0.$

$\displaystyle \Rightarrow\; f(5)\neq f(-5).$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 10: }~\text{Examine if Rolle's theorem is applicable to any one of the following functions:}$

$\displaystyle (i)\;f(x)=[x]\text{ for }x\in[5,9]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{By Rolle's Theorem, for a function } f:[a,b]\to\mathbb{R}, \text{ if}$

$\displaystyle \text{(a) } f \text{ is continuous on } [a,b],$

$\displaystyle \text{(b) } f \text{ is differentiable on } (a,b), \text{ and}$

$\displaystyle \text{(c) } f(a)=f(b),$

$\displaystyle \text{then there exists some } c\in(a,b) \text{ such that } f'(c)=0.$

$\displaystyle \text{Therefore, Rolle's Theorem is not applicable to functions that do not satisfy any of the above conditions.}$

$\displaystyle \text{(i) Let } f(x)=[x] \text{ for } x\in[5,9].$

$\displaystyle \text{It is evident that the given function } f(x) \text{ is not continuous at every integer point.}$

$\displaystyle \text{In particular, } f(x) \text{ is not continuous at } x=5 \text{ and } x=9.$

$\displaystyle \text{Thus, } f(x) \text{ is not continuous on } [5,9].$

$\displaystyle \text{Also, } f(5)=[5]=5 \text{ and } f(9)=[9]=9.$

$\displaystyle \therefore\; f(5)\neq f(9).$

$\displaystyle \text{Now, we check the differentiability of } f \text{ on } (5,9).$

$\displaystyle \text{Let } n \text{ be an integer such that } n\in(5,9).$

$\displaystyle \text{The left-hand derivative of } f \text{ at } x=n \text{ is}$

$\displaystyle \lim_{h\to0^-}\frac{f(n+h)-f(n)}{h} =\lim_{h\to0^-}\frac{[n+h]-[n]}{h} =\lim_{h\to0^-}\frac{(n-1)-n}{h} =\lim_{h\to0^-}\frac{-1}{h} =\infty.$

$\displaystyle \text{The right-hand derivative of } f \text{ at } x=n \text{ is}$

$\displaystyle \lim_{h\to0^+}\frac{f(n+h)-f(n)}{h} =\lim_{h\to0^+}\frac{[n+h]-[n]}{h} =\lim_{h\to0^+}\frac{n-n}{h} =\lim_{h\to0^+}0 =0.$

$\displaystyle \text{Since the left-hand and right-hand derivatives at } x=n \text{ are not equal, }$

$\displaystyle f \text{ is not differentiable at } x=n.$

$\displaystyle \text{Thus, } f \text{ is not differentiable on } (5,9).$

$\displaystyle \text{Hence, } f(x)=[x] \text{ does not satisfy the hypotheses of Rolle's Theorem on } [5,9].$

$\displaystyle \text{Therefore, Rolle's Theorem is not applicable for } f(x)=[x] \text{ on } [5,9].$

$\displaystyle \text{(ii) Let } f(x)=[x] \text{ for } x\in[-2,2].$

$\displaystyle \text{It is evident that } f(x) \text{ is not continuous at every integer point.}$

$\displaystyle \text{In particular, } f(x) \text{ is not continuous at } x=-2 \text{ and } x=2.$

$\displaystyle \text{Thus, } f(x) \text{ is not continuous on } [-2,2].$

$\displaystyle \text{Also, } f(-2)=[-2]=-2 \text{ and } f(2)=[2]=2.$

$\displaystyle \therefore\; f(-2)\neq f(2).$

$\displaystyle \text{Now, we check the differentiability of } f \text{ on } (-2,2).$

$\displaystyle \text{Let } n \text{ be an integer such that } n\in(-2,2).$

$\displaystyle \text{The left-hand derivative of } f \text{ at } x=n \text{ is}$

$\displaystyle \lim_{h\to0^-}\frac{f(n+h)-f(n)}{h} =\lim_{h\to0^-}\frac{(n-1)-n}{h} =\lim_{h\to0^-}\frac{-1}{h} =\infty.$

$\displaystyle \text{The right-hand derivative of } f \text{ at } x=n \text{ is}$

$\displaystyle \lim_{h\to0^+}\frac{f(n+h)-f(n)}{h} =\lim_{h\to0^+}\frac{n-n}{h} =0.$

$\displaystyle \text{Since the left-hand and right-hand derivatives at } x=n \text{ are not equal,}$

$\displaystyle f \text{ is not differentiable at } x=n.$

$\displaystyle \text{Thus, } f \text{ is not differentiable on } (-2,2).$

$\displaystyle \text{Hence, } f(x)=[x] \text{ does not satisfy the hypotheses of Rolle's Theorem on } [-2,2].$

$\displaystyle \text{Therefore, Rolle's Theorem is not applicable for } f(x)=[x] \text{ on } [-2,2].$

$\displaystyle \text{Can you say something about the converse of Rolle's Theorem for these functions?}$

$\displaystyle \textbf{Question 11: }~\text{It is given that the Rolle's theorem holds for the function }f(x)=x^3+bx^2+cx,\;x\in[1,2]\text{ at the point }x=\frac{4}{3}.\text{ Find the values of }b\text{ and }c.$