Consider the function f(x) = ( 2x)tan^ 2 2x, x ≠ (π)/(4) . The value of f( (π)/(4) ) such that f is continuous at x = (π)/(4) is

Correct answer: B. 1 / e f((π)/(4)) = _ x → (π)/(4) ( 2x) ^ 2 2x = _ x → (π)/(4) e^ ^ 2 2x( 2x - 1) = _ x → (π)/(4) e 2x - 1 ^ 2 2x = e^ _ x → (π)/(4) 2 2x - 0 2 2x( - 2cosec^ 2 2x ) = _ x → (π)/(4) e - ^ 3 2x 2 = e(- 1)/(2)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

Consider the function f(x) = ( 2x)tan^ 2 2x, x ≠ (π)/(4) . The...

Consider the function

f\left(x\right) = \left(\sin 2x\right)tan^{2}2x, x \ne \dfrac{\pi}{4}

. The value of

f( \dfrac{\pi}{4} )

such that

f

is continuous at

x = \dfrac{\pi}{4}

Options

A $\sqrt{e}$
B $1 / \sqrt{e}$
C $2$
DNone of these

Correct answer

B. $1 / \sqrt{e}$

Step-by-step solution

f\left(\dfrac{\pi}{4}\right) = \lim_{x \to \dfrac{\pi}{4}} \left(\sin 2x\right)\tan^{2} 2x

= \lim_{x \to \dfrac{\pi}{4}} e^{\tan^{2} 2x\left(\sin 2x - 1\right)} = \lim_{x \to \dfrac{\pi}{4}} e\dfrac{\sin 2x - 1}{\cot^{2} 2x}

= e^{\lim_{x \to \dfrac{\pi}{4}}\dfrac{2\cos 2x - 0}{2\cot 2x( - 2cosec^{2}2x )}} = \lim_{x \to \dfrac{\pi}{4}}e\dfrac{- \sin^{3}2x}{2} = e\dfrac{- 1}{2}

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