Class 12: Mean Value Theorems

$\displaystyle \textbf{Question 1: }~\text{Verify Lagrange's mean value theorem for the following functions on the } \\ \text{indicated intervals. In each case find a point }c\text{ in the indicated interval as stated} \\ \text{by the Lagrange's mean value theorem:}$
$\displaystyle \text{(i) }\;f(x)=x^2-1\text{ on }[2,3]$

$\displaystyle \text{Answer:}$

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

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

$\displaystyle \text{Thus, both conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{So, there exists at least one real number } c \in (2,3) \text{ such that } f'(c) = \frac{f(3) - f(2)}{3 - 2}.$

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

$\displaystyle f(3) = (3)^2 - 1 = 8, \quad f(2) = (2)^2 - 1 = 3.$

$\displaystyle \frac{f(3) - f(2)}{3 - 2} = \frac{8 - 3}{1} = 5.$

$\displaystyle \text{Hence, } f'(c) = 5 \Rightarrow 2c = 5.$

$\displaystyle \Rightarrow c = \frac{5}{2}.$

$\displaystyle \text{Since } \frac{5}{2} \in (2,3), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle \text{(ii) }\;f(x)=x^3-2x^2-x+3\text{ on }[0,1]$

$\displaystyle \text{Answer:}$

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

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

$\displaystyle \text{Thus, both conditions of Lagrange's Mean Value Theorem are satisfied.}$

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

$\displaystyle \text{Now, } f(x) = x^3 - 2x^2 - x + 3 \Rightarrow f'(x) = 3x^2 - 4x - 1.$

$\displaystyle f(1) = 1, \quad f(0) = 3.$

$\displaystyle \frac{f(1) - f(0)}{1 - 0} = \frac{1 - 3}{1} = -2.$

$\displaystyle \text{Hence, } f'(c) = -2 \Rightarrow 3c^2 - 4c - 1 = -2.$

$\displaystyle \Rightarrow 3c^2 - 4c + 1 = 0.$

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

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

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

$\displaystyle \text{Since } c = \frac{1}{3} \in (0,1), \text{ Lagrange's Mean Value Theorem is verified.}$

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

$\displaystyle \text{Answer:}$

$\displaystyle \text{We have } f(x) = x(x-1), \text{ which can be rewritten as } f(x) = x^2 - x.$

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

$\displaystyle \text{Thus, both conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{So, there exists at least one real number } c \in (1,2) \text{ such that } f'(c) = \frac{f(2) - f(1)}{2 - 1}.$

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

$\displaystyle f(2) = 2, \quad f(1) = 0.$

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

$\displaystyle \text{Hence, } f'(c) = 2 \Rightarrow 2c - 1 = 2.$

$\displaystyle \Rightarrow 2c = 3.$

$\displaystyle \Rightarrow c = \frac{3}{2}.$

$\displaystyle \text{Since } \frac{3}{2} \in (1,2), \text{ Lagrange's Mean Value Theorem is verified.}$

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

$\displaystyle \text{Answer:}$

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

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

$\displaystyle \text{Thus, both conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{So, there exists at least one real number } c \in (-1,2) \text{ such that } f'(c) = \frac{f(2) - f(-1)}{2 - (-1)}.$

$\displaystyle \text{Now, } f(x) = x^2 - 3x + 2 \Rightarrow f'(x) = 2x - 3.$

$\displaystyle f(2) = 0, \quad f(-1) = (-1)^2 - 3(-1) + 2 = 6.$

$\displaystyle \frac{f(2) - f(-1)}{2 - (-1)} = \frac{0 - 6}{3} = -2.$

$\displaystyle \text{Hence, } f'(c) = -2 \Rightarrow 2c - 3 = -2.$

$\displaystyle \Rightarrow 2c - 1 = 0.$

$\displaystyle \Rightarrow c = \frac{1}{2}.$

$\displaystyle \text{Since } \frac{1}{2} \in (-1,2), \text{ Lagrange's Mean Value Theorem is verified.}$

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

$\displaystyle \text{Answer:}$

$\displaystyle \text{We have } f(x) = 2x^2 - 3x + 1.$

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

$\displaystyle \text{Thus, both conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{So, there exists at least one real number } c \in (1,3) \text{ such that } f'(c) = \frac{f(3) - f(1)}{3 - 1}.$

$\displaystyle \text{Now, } f(x) = 2x^2 - 3x + 1 \Rightarrow f'(x) = 4x - 3.$

$\displaystyle f(3) = 2(3)^2 - 3(3) + 1 = 10, \quad f(1) = 2(1)^2 - 3(1) + 1 = 0.$

$\displaystyle \frac{f(3) - f(1)}{3 - 1} = \frac{10 - 0}{2} = 5.$

$\displaystyle \text{Hence, } f'(c) = 5 \Rightarrow 4c - 3 = 5.$

$\displaystyle \Rightarrow 4c - 8 = 0.$

$\displaystyle \Rightarrow c = 2.$

$\displaystyle \text{Since } 2 \in (1,3), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle \text{(vi) }\;f(x)=x^2-2x+4\text{ on }[1,5]$

$\displaystyle \text{Answer:}$

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

$\displaystyle \text{Since a polynomial function is everywhere continuous and differentiable, } \\ f(x) \text{ is continuous on } [1,5] \text{ and differentiable on } (1,5).$

$\displaystyle \text{Thus, both conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{So, there exists at least one real number } c \in (1,5) \text{ such that } f'(c) = \frac{f(5) - f(1)}{5 - 1}.$

$\displaystyle \text{Now, } f(x) = x^2 - 2x + 4 \Rightarrow f'(x) = 2x - 2.$

$\displaystyle f(5) = 25 - 10 + 4 = 19, \quad f(1) = 1 - 2 + 4 = 3.$

$\displaystyle \frac{f(5) - f(1)}{5 - 1} = \frac{19 - 3}{4} = 4.$

$\displaystyle \text{Hence, } f'(c) = 4 \Rightarrow 2c - 2 = 4.$

$\displaystyle \Rightarrow 2c - 6 = 0.$

$\displaystyle \Rightarrow c = 3.$

$\displaystyle \text{Since } 3 \in (1,5), \text{ Lagrange's Mean Value Theorem is verified.}$

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

$\displaystyle \text{Answer:}$

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

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

$\displaystyle \text{Thus, both conditions of Lagrange's Mean Value Theorem are satisfied.}$

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

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

$\displaystyle f(1) = 2 - 1 = 1, \quad f(0) = 0.$

$\displaystyle \frac{f(1) - f(0)}{1 - 0} = \frac{1 - 0}{1} = 1.$

$\displaystyle \text{Hence, } f'(c) = 1 \Rightarrow 2 - 2c = 1.$

$\displaystyle \Rightarrow -2c = -1.$

$\displaystyle \Rightarrow c = \frac{1}{2}.$

$\displaystyle \text{Since } \frac{1}{2} \in (0,1), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle \text{(viii) }\;f(x)=(x-1)(x-2)(x-3)\text{ on }[0,4]$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We have } f(x) = (x-1)(x-2)(x-3), \text{ which can be rewritten as } \\ f(x) = x^3 - 6x^2 + 11x - 6.$

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

$\displaystyle \text{Thus, both conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{So, there exists at least one real number } c \in (0,4) \text{ such that } f'(c) = \frac{f(4) - f(0)}{4 - 0}.$

$\displaystyle \text{Now, } f(x) = x^3 - 6x^2 + 11x - 6 \Rightarrow f'(x) = 3x^2 - 12x + 11.$

$\displaystyle f(0) = -6, \quad f(4) = 64 - 96 + 44 - 6 = 6.$

$\displaystyle \frac{f(4) - f(0)}{4 - 0} = \frac{6 - (-6)}{4} = 3.$

$\displaystyle \text{Hence, } f'(c) = 3 \Rightarrow 3c^2 - 12c + 11 = 3.$

$\displaystyle \Rightarrow 3c^2 - 12c + 8 = 0.$

$\displaystyle \Rightarrow c = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm 4\sqrt{3}}{6}.$

$\displaystyle \Rightarrow c = 2 \pm \frac{2}{\sqrt{3}}.$

$\displaystyle \text{Since } 2 - \frac{2}{\sqrt{3}} \in (0,4), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle \text{(ix) }\;f(x)=\sqrt{25-x^2}\text{ on }[-3,4]$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We have } f(x) = \sqrt{25 - x^2}.$

$\displaystyle \text{Here, } f(x) \text{ exists if } 25 - x^2 \ge 0 \Rightarrow x^2 \le 25 \Rightarrow -5 \le x \le 5.$

$\displaystyle \text{Since for each } x \in [-3,4], \text{ the function } f(x) \text{ attains a unique definite value, } \\ f(x) \text{ is continuous on } [-3,4].$

$\displaystyle \text{Also, } f'(x) = \frac{1}{2\sqrt{25 - x^2}}(-2x) = -\frac{x}{\sqrt{25 - x^2}} \text{ exists for all } x \in (-3,4).$

$\displaystyle \text{Thus, } f(x) \text{ is differentiable on } (-3,4).$

$\displaystyle \text{Hence, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{Therefore, there exists at least one real number } c \in (-3,4) \text{ such that } f'(c) = \frac{f(4) - f(-3)}{4 - (-3)}.$

$\displaystyle \text{Now, } f(4) = \sqrt{25 - 16} = 3, \quad f(-3) = \sqrt{25 - 9} = 4.$

$\displaystyle \frac{f(4) - f(-3)}{4 - (-3)} = \frac{3 - 4}{7} = -\frac{1}{7}.$

$\displaystyle \text{Hence, } -\frac{c}{\sqrt{25 - c^2}} = -\frac{1}{7}.$

$\displaystyle \Rightarrow \frac{c}{\sqrt{25 - c^2}} = \frac{1}{7}.$

$\displaystyle \Rightarrow 49c^2 = 25 - c^2.$

$\displaystyle \Rightarrow 50c^2 = 25.$

$\displaystyle \Rightarrow c^2 = \frac{1}{2} \Rightarrow c = \pm \frac{1}{\sqrt{2}}.$

$\displaystyle \text{Since } \pm \frac{1}{\sqrt{2}} \in (-3,4), \text{ Lagrange's Mean Value Theorem is verified.}$

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

$\displaystyle \text{Answer:}$

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

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

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

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

$\displaystyle \text{Now, } f(x) = \tan^{-1} x \Rightarrow f'(x) = \frac{1}{1 + x^2}.$

$\displaystyle f(1) = \tan^{-1}(1) = \frac{\pi}{4}, \quad f(0) = \tan^{-1}(0) = 0.$

$\displaystyle \frac{f(1) - f(0)}{1 - 0} = \frac{\pi}{4}.$

$\displaystyle \text{Hence, } f'(c) = \frac{\pi}{4} \Rightarrow \frac{1}{1 + c^2} = \frac{\pi}{4}.$

$\displaystyle \Rightarrow 1 + c^2 = \frac{4}{\pi}.$

$\displaystyle \Rightarrow c^2 = \frac{4 - \pi}{\pi}.$

$\displaystyle \Rightarrow c = \sqrt{\frac{4 - \pi}{\pi}}.$

$\displaystyle \text{Since } \sqrt{\frac{4 - \pi}{\pi}} \in (0,1), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle \text{(xi) }\;f(x)=x+\frac{1}{x}\text{ on }[1,3] \hspace{7.0cm} \;[\text{CBSE 2000}]$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We have } f(x) = x + \frac{1}{x} = \frac{x^2 + 1}{x}.$

$\displaystyle \text{Clearly, } f(x) \text{ is continuous on } [1,3] \text{ and differentiable on } (1,3).$

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one real number } c \in (1,3) \text{ such that } f'(c) = \frac{f(3) - f(1)}{3 - 1}.$

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

$\displaystyle f(1) = 2, \quad f(3) = \frac{10}{3}.$

$\displaystyle \frac{f(3) - f(1)}{3 - 1} = \frac{\frac{10}{3} - 2}{2} = \frac{4}{6} = \frac{2}{3}.$

$\displaystyle \text{Hence, } f'(c) = \frac{2}{3} \Rightarrow \frac{c^2 - 1}{c^2} = \frac{2}{3}.$

$\displaystyle \Rightarrow 3(c^2 - 1) = 2c^2.$

$\displaystyle \Rightarrow c^2 = 3.$

$\displaystyle \Rightarrow c = \sqrt{3}.$

$\displaystyle \text{Since } \sqrt{3} \in (1,3), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle \text{(xii) }\;f(x)=x(x+4)^2\text{ on }[0,4]$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We have } f(x) = x(x+4)^2 = x(x^2 + 8x + 16) = x^3 + 8x^2 + 16x.$

$\displaystyle \text{Since } f(x) \text{ is a polynomial function, it is everywhere continuous and differentiable.}$

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

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one real number } c \in (0,4) \text{ such that } f'(c) = \frac{f(4) - f(0)}{4 - 0}.$

$\displaystyle \text{Now, } f(x) = x^3 + 8x^2 + 16x \Rightarrow f'(x) = 3x^2 + 16x + 16.$

$\displaystyle f(4) = 64 + 128 + 64 = 256, \quad f(0) = 0.$

$\displaystyle \frac{f(4) - f(0)}{4 - 0} = \frac{256}{4} = 64.$

$\displaystyle \text{Hence, } f'(c) = 64 \Rightarrow 3c^2 + 16c + 16 = 64.$

$\displaystyle \Rightarrow 3c^2 + 16c - 48 = 0.$

$\displaystyle \Rightarrow c = \frac{-16 \pm \sqrt{256 + 576}}{6} = \frac{-16 \pm 4\sqrt{13}}{6}.$

$\displaystyle \Rightarrow c = \frac{-8 \pm 2\sqrt{13}}{3}.$

$\displaystyle \text{Since } \frac{-8 + 2\sqrt{13}}{3} \in (0,4), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle \text{(xiii) }\;f(x)=\sqrt{x^2-4}\text{ on }[2,4] \hspace{7.0cm} \;[\text{CBSE 2002}]$

$\displaystyle \text{Answer:}$

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

$\displaystyle \text{Here, } f(x) \text{ exists if } x^2 - 4 \ge 0 \Rightarrow x \le -2 \text{ or } x \ge 2.$

$\displaystyle \text{Since for each } x \in [2,4], \text{ the function } f(x) \text{ attains a unique definite value, } \\ f(x) \text{ is continuous on } [2,4].$

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

$\displaystyle \text{Thus, } f(x) \text{ is differentiable on } (2,4).$

$\displaystyle \text{Hence, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one real number } c \in (2,4) \text{ such that } f'(c) = \frac{f(4) - f(2)}{4 - 2}.$

$\displaystyle \text{Now, } f(4) = \sqrt{16 - 4} = 2\sqrt{3}, \quad f(2) = 0.$

$\displaystyle \frac{f(4) - f(2)}{4 - 2} = \frac{2\sqrt{3}}{2} = \sqrt{3}.$

$\displaystyle \text{Hence, } \frac{c}{\sqrt{c^2 - 4}} = \sqrt{3}.$

$\displaystyle \Rightarrow \frac{c^2}{c^2 - 4} = 3.$

$\displaystyle \Rightarrow c^2 = 3c^2 - 12.$

$\displaystyle \Rightarrow c^2 = 6 \Rightarrow c = \pm \sqrt{6}.$

$\displaystyle \text{Since } \sqrt{6} \in (2,4), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle \text{(xiv) }\;f(x)=x^2+x-1\text{ on }[0,4] \hspace{7.0cm} \;[\text{CBSE 2002}]$

$\displaystyle \text{Answer:}$

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

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

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one real number } c \in (0,4) \text{ such that } f'(c) = \frac{f(4) - f(0)}{4 - 0}.$

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

$\displaystyle f(4) = 16 + 4 - 1 = 19, \quad f(0) = -1.$

$\displaystyle \frac{f(4) - f(0)}{4 - 0} = \frac{19 - (-1)}{4} = \frac{20}{4} = 5.$

$\displaystyle \text{Hence, } f'(c) = 5 \Rightarrow 2c + 1 = 5.$

$\displaystyle \Rightarrow 2c = 4.$

$\displaystyle \Rightarrow c = 2.$

$\displaystyle \text{Since } 2 \in (0,4), \text{ Lagrange's Mean Value Theorem is verified.}$

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

$\displaystyle \text{Answer:}$

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

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

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

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

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

$\displaystyle \text{Now, } f(x) = \sin x - \sin 2x - x \Rightarrow f'(x) = \cos x - 2\cos 2x - 1.$

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

$\displaystyle \frac{f(\pi) - f(0)}{\pi - 0} = \frac{-\pi}{\pi} = -1.$

$\displaystyle \text{Hence, } f'(c) = -1 \Rightarrow \cos c - 2\cos 2c - 1 = -1.$

$\displaystyle \Rightarrow \cos c - 2\cos 2c = 0.$

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

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

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

$\displaystyle \Rightarrow \cos c = \frac{1 \pm \sqrt{33}}{8}.$

$\displaystyle \Rightarrow c = \cos^{-1}\!\left(\frac{1 \pm \sqrt{33}}{8}\right).$

$\displaystyle \text{Since } \cos^{-1}\!\left(\frac{1 \pm \sqrt{33}}{8}\right) \in (0,\pi), \text{ Lagrange's Mean Value Theorem is verified.}$

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

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

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

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one real number } c \in (1,3) \text{ such that } f'(c) = \frac{f(3) - f(1)}{3 - 1}.$

$\displaystyle \text{Now, } f(x) = x^3 - 5x^2 - 3x \Rightarrow f'(x) = 3x^2 - 10x - 3.$

$\displaystyle f(3) = 27 - 45 - 9 = -27, \quad f(1) = 1 - 5 - 3 = -7.$

$\displaystyle \frac{f(3) - f(1)}{3 - 1} = \frac{-27 - (-7)}{2} = \frac{-20}{2} = -10.$

$\displaystyle \text{Hence, } f'(c) = -10 \Rightarrow 3c^2 - 10c - 3 = -10.$

$\displaystyle \Rightarrow 3c^2 - 10c + 7 = 0.$

$\displaystyle \Rightarrow c = \frac{10 \pm \sqrt{100 - 84}}{6} = \frac{10 \pm 4}{6}.$

$\displaystyle \Rightarrow c = 1, \ \frac{7}{3}.$

$\displaystyle \text{Since } \frac{7}{3} \in (1,3), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle \textbf{Question 2: }~\text{Discuss the applicability of Lagrange's mean value theorem } \\ \text{for the function }f(x)=|x|\text{ on }[-1,1].$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given } f(x) = |x|.$

$\displaystyle \text{If Lagrange's Mean Value Theorem were applicable to the given function, then } \\ f(x) \text{ would be continuous on } [-1,1] \text{ and differentiable on } (-1,1).$

$\displaystyle \text{But it is known that } f(x) = |x| \text{ is not differentiable at } x = 0 \in (-1,1).$

$\displaystyle \text{Thus, our supposition is wrong.}$

$\displaystyle \text{Therefore, Lagrange's Mean Value Theorem is not applicable for the given function.}$

$\displaystyle \textbf{Question 3: }~\text{Show that the Lagrange's mean value theorem is not applicable } \\ \text{to the function }f(x)=\frac{1}{x}\text{ on }[-1,1].$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given } f(x) = \frac{1}{x}.$

$\displaystyle \text{Clearly, } f(x) \text{ does not exist at } x = 0.$

$\displaystyle \text{Hence, the given function is not continuous on } [-1,1].$

$\displaystyle \text{Therefore, Lagrange's Mean Value Theorem is not applicable for the given function on } [-1,1].$

$\displaystyle \textbf{Question 4: }~\text{Verify the hypothesis and conclusion of Lagrange's mean value } \\ \text{theorem for the function }f(x)=\frac{1}{4x-1},\;1\le x\le 4.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given } f(x) = \frac{1}{4x - 1}.$

$\displaystyle \text{Since for each } x \in [1,4], \text{ the function } f(x) \text{ attains a unique definite value, } \\ f(x) \text{ is continuous on } [1,4].$

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

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one real number } c \in (1,4) \text{ such that } f'(c) = \frac{f(4) - f(1)}{4 - 1}.$

$\displaystyle \text{Now, } f(4) = \frac{1}{15}, \quad f(1) = \frac{1}{3}.$

$\displaystyle \frac{f(4) - f(1)}{4 - 1} = \frac{\frac{1}{15} - \frac{1}{3}}{3} = \frac{-4}{45}.$

$\displaystyle \text{Hence, } -\frac{4}{(4c - 1)^2} = -\frac{4}{45}.$

$\displaystyle \Rightarrow \frac{1}{(4c - 1)^2} = \frac{1}{45}.$

$\displaystyle \Rightarrow (4c - 1)^2 = 45.$

$\displaystyle \Rightarrow 4c - 1 = \pm 3\sqrt{5}.$

$\displaystyle \Rightarrow c = \frac{1 \pm 3\sqrt{5}}{4}.$

$\displaystyle \text{Since } \frac{1 + 3\sqrt{5}}{4} \in (1,4), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle \textbf{Question 5: }~\text{Find a point on the parabola }y=(x-4)^2,\text{ where the tangent is parallel } \\ \text{to the chord joining }(4,0)\text{ and }(5,1).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } f(x) = (x - 4)^2 = x^2 - 8x + 16.$

$\displaystyle \text{The tangent to the curve is parallel to the chord joining the points } (4,0) \text{ and } (5,1).$

$\displaystyle \text{Assume that the chord joins the points } (a, f(a)) \text{ and } (b, f(b)).$

$\displaystyle \text{Here, } a = 4 \text{ and } b = 5.$

$\displaystyle \text{Since } f(x) \text{ is a polynomial function, it is everywhere continuous and differentiable.}$

$\displaystyle \text{Therefore, } f(x) \text{ is continuous on } [4,5] \text{ and differentiable on } (4,5).$

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one real number } c \in (4,5) \text{ such that } f'(c) = \frac{f(5) - f(4)}{5 - 4}.$

$\displaystyle \text{Now, } f(x) = x^2 - 8x + 16 \Rightarrow f'(x) = 2x - 8.$

$\displaystyle f(5) = 1, \quad f(4) = 0.$

$\displaystyle \frac{f(5) - f(4)}{5 - 4} = 1.$

$\displaystyle \text{Hence, } f'(c) = 1 \Rightarrow 2c - 8 = 1.$

$\displaystyle \Rightarrow 2c = 9 \Rightarrow c = \frac{9}{2}.$

$\displaystyle \text{Since } \frac{9}{2} \in (4,5), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle f(c) = \left(\frac{9}{2} - 4\right)^2 = \frac{1}{4}.$

$\displaystyle \text{Thus, the point } \left(\frac{9}{2}, \frac{1}{4}\right) \text{ lies on the curve where the tangent is parallel} \\ \text{to the chord joining } (4,0) \text{ and } (5,1).$

$\displaystyle \textbf{Question 6: }~\text{Find a point on the curve }y=x^2+x,\text{ where the tangent is parallel } \\ \text{to the chord joining }(0,0)\text{ and }(1,2).$
$\displaystyle \text{Answer:}$

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

$\displaystyle \text{The tangent to the curve is parallel to the chord joining the points } (0,0) \text{ and } (1,2).$

$\displaystyle \text{Assume that the chord joins the points } (a, f(a)) \text{ and } (b, f(b)).$

$\displaystyle \text{Here, } a = 0 \text{ and } b = 1.$

$\displaystyle \text{Since } f(x) \text{ is a polynomial function, it is everywhere continuous and differentiable.}$

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

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

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

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

$\displaystyle f(1) = 2, \quad f(0) = 0.$

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

$\displaystyle \text{Hence, } f'(c) = 2 \Rightarrow 2c + 1 = 2.$

$\displaystyle \Rightarrow 2c = 1 \Rightarrow c = \frac{1}{2}.$

$\displaystyle \text{Since } \frac{1}{2} \in (0,1), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle f(c) = \left(\frac{1}{2}\right)^2 + \frac{1}{2} = \frac{3}{4}.$

$\displaystyle \text{Thus, the point } \left(\frac{1}{2}, \frac{3}{4}\right) \text{ lies on the given curve where the tangent is parallel} \\ \text{to the chord joining } (0,0) \text{ and } (1,2).$

$\displaystyle \textbf{Question 7: }~\text{Find a point on the parabola }y=(x-3)^2,\text{ where the tangent is parallel } \\ \text{to the chord joining }(3,0)\text{ and }(4,1).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } f(x) = (x-3)^2 = x^2 - 6x + 9.$

$\displaystyle \text{The tangent to the curve is parallel to the chord joining the points } (3,0) \text{ and } (4,1).$

$\displaystyle \text{Assume that the chord joins the points } (a, f(a)) \text{ and } (b, f(b)).$

$\displaystyle \text{Here, } a = 3 \text{ and } b = 4.$

$\displaystyle \text{Since } f(x) \text{ is a polynomial function, it is everywhere continuous and differentiable.}$

$\displaystyle \text{Therefore, } f(x) \text{ is continuous on } [3,4] \text{ and differentiable on } (3,4).$

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one real number } c \in (3,4) \text{ such that } f'(c) = \frac{f(4) - f(3)}{4 - 3}.$

$\displaystyle \text{Now, } f(x) = x^2 - 6x + 9 \Rightarrow f'(x) = 2x - 6.$

$\displaystyle f(3) = 0, \quad f(4) = 1.$

$\displaystyle \frac{f(4) - f(3)}{4 - 3} = 1.$

$\displaystyle \text{Hence, } f'(c) = 1 \Rightarrow 2c - 6 = 1.$

$\displaystyle \Rightarrow 2c = 7 \Rightarrow c = \frac{7}{2}.$

$\displaystyle \text{Since } \frac{7}{2} \in (3,4), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle f(c) = \left(\frac{7}{2} - 3\right)^2 = \frac{1}{4}.$

$\displaystyle \text{Thus, the point } \left(\frac{7}{2}, \frac{1}{4}\right) \text{ lies on the given curve where the tangent is} \\ \text{parallel to the chord joining } (3,0) \text{ and } (4,1).$

$\displaystyle \textbf{Question 8: }~\text{Find the points on the curve }y=x^3-3x,\text{ where the tangent to the curve is parallel } \\ \text{to the chord joining }(1,-2)\text{ and }(2,2).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } f(x) = x^3 - 3x.$

$\displaystyle \text{The tangent to the curve is parallel to the chord joining the points } (1,-2) \text{ and } (2,2).$

$\displaystyle \text{Assume that the chord joins the points } (a,f(a)) \text{ and } (b,f(b)).$

$\displaystyle \text{Here, } a = 1 \text{ and } b = 2.$

$\displaystyle \text{Since } f(x) \text{ is a polynomial function, it is everywhere continuous and differentiable.}$

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

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

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

$\displaystyle \text{Now, } f(x) = x^3 - 3x \Rightarrow f'(x) = 3x^2 - 3.$

$\displaystyle f(1) = -2, \quad f(2) = 2.$

$\displaystyle \frac{f(2) - f(1)}{2 - 1} = \frac{2 - (-2)}{1} = 4.$

$\displaystyle \text{Hence, } f'(c) = 4 \Rightarrow 3c^2 - 3 = 4.$

$\displaystyle \Rightarrow 3c^2 = 7 \Rightarrow c^2 = \frac{7}{3}.$

$\displaystyle \Rightarrow c = \sqrt{\frac{7}{3}}.$

$\displaystyle \text{Since } \sqrt{\frac{7}{3}} \in (1,2), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle f(c) = \left(\sqrt{\frac{7}{3}}\right)^3 - 3\sqrt{\frac{7}{3}} = \sqrt{\frac{7}{3}}\!\left(\frac{7}{3} - 3\right) = -\frac{2}{3}\sqrt{\frac{7}{3}}.$

$\displaystyle \text{Thus, the point } \left(\sqrt{\frac{7}{3}}, -\frac{2}{3}\sqrt{\frac{7}{3}}\right) \text{ lies on the given curve where the tangent} \\ \text{is parallel to the chord joining } (1,-2) \text{ and } (2,2).$

$\displaystyle \textbf{Question 9: }~\text{Find a point on the curve }y=x^3+1\text{ where the tangent is parallel } \\ \text{to the chord joining }(1,2)\text{ and }(3,28).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } f(x) = x^3 + 1.$

$\displaystyle \text{The tangent to the curve is parallel to the chord joining the points } (1,2) \text{ and } (3,28).$

$\displaystyle \text{Assume that the chord joins the points } (a,f(a)) \text{ and } (b,f(b)).$

$\displaystyle \text{Here, } a = 1 \text{ and } b = 3.$

$\displaystyle \text{Since } f(x) \text{ is a polynomial function, it is everywhere continuous and differentiable.}$

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

$\displaystyle \text{Thus, both the conditions of Lagrange's Mean Value Theorem are satisfied.}$

$\displaystyle \text{Consequently, there exists at least one real number } c \in (1,3) \text{ such that } f'(c) = \frac{f(3) - f(1)}{3 - 1}.$

$\displaystyle \text{Now, } f(x) = x^3 + 1 \Rightarrow f'(x) = 3x^2.$

$\displaystyle f(1) = 2, \quad f(3) = 28.$

$\displaystyle \frac{f(3) - f(1)}{3 - 1} = \frac{28 - 2}{2} = 13.$

$\displaystyle \text{Hence, } f'(c) = 13 \Rightarrow 3c^2 = 13.$

$\displaystyle \Rightarrow c^2 = \frac{13}{3} \Rightarrow c = \sqrt{\frac{13}{3}}.$

$\displaystyle \text{Since } \sqrt{\frac{13}{3}} \in (1,3), \text{ Lagrange's Mean Value Theorem is verified.}$

$\displaystyle f(c) = \left(\sqrt{\frac{13}{3}}\right)^3 + 1 = \left(\frac{13}{3}\right)^{\frac{3}{2}} + 1.$

$\displaystyle \text{Thus, the point } \left(\sqrt{\frac{13}{3}},\, 1 + \left(\frac{13}{3}\right)^{\frac{3}{2}}\right) \text{ lies on the given curve where the tangent} \\ \text{is parallel to the chord joining } (1,2) \text{ and } (3,28).$

$\displaystyle \textbf{Question 10: } \text{Let } C\text{ be a curve defined parametrically as} x=a\cos^3\theta,\ y=a\sin^3\theta, \ \le\theta\le\frac{\pi}{2}. \text{ Determine a point } P \text{ on } C, \text{ where the tangent to } C \text{ is parallel } \\ \text{to the chord joining the points }(a,0) \text{ and } (0,a). \hspace{4.0cm} \;[\text{CBSE 2014}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the point be } P(x,y).$

$\displaystyle \text{Given the parametric equations } x = a\cos^3\theta, \quad y = a\sin^3\theta.$

$\displaystyle \text{Slope of the tangent } = \frac{dy}{dx}.$

$\displaystyle \frac{dx}{d\theta} = -3a\cos^2\theta \sin\theta, \quad \frac{dy}{d\theta} = 3a\sin^2\theta \cos\theta.$

$\displaystyle \therefore \frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \frac{3a\sin^2\theta \cos\theta}{-3a\cos^2\theta \sin\theta} = -\tan\theta.$

$\displaystyle \text{Thus, slope of the tangent } = -\tan\theta.$

$\displaystyle \text{Given two points } (a,0) \text{ and } (0,a).$

$\displaystyle \text{Slope of the chord } = \frac{y_2 - y_1}{x_2 - x_1} = \frac{a - 0}{0 - a} = -1.$

$\displaystyle \text{It is given that the tangent is parallel to the chord.}$

$\displaystyle \text{Hence, slope of tangent } = \text{ slope of chord}.$

$\displaystyle -\tan\theta = -1 \Rightarrow \tan\theta = 1.$

$\displaystyle \Rightarrow \theta = \frac{\pi}{4}.$

$\displaystyle x = a\cos^3\!\left(\frac{\pi}{4}\right) = a\left(\frac{1}{\sqrt{2}}\right)^3 = \frac{a}{2\sqrt{2}}.$

$\displaystyle y = a\sin^3\!\left(\frac{\pi}{4}\right) = a\left(\frac{1}{\sqrt{2}}\right)^3 = \frac{a}{2\sqrt{2}}.$

$\displaystyle \text{Hence, the point } P \text{ is } \left(\frac{a}{2\sqrt{2}}, \frac{a}{2\sqrt{2}}\right).$

$\displaystyle \textbf{Question 11: }~\text{Using Lagrange's mean value theorem, prove that } \\ (b-a)\sec^2 a<\tan b-\tan a<(b-a)\sec^2 b,\text{ where }0<a<b<\frac{\pi}{2}.$
$\displaystyle \text{Answer:}$