JEE Main 2018 15 April Evening Shift Online
Exam JEE Main
Year 2018
Date 15 April
Shift Evening Shift
Mode Online
Subject Mathematics
Chapter Limits JEE Main 2018 · 15 April · Evening Shift · Online · Mathematics · Limits
_ x arrow 0 (x 2 x-2 x x)/((1- 2 x)^2) equals.
lim x → 0 x tan 2 x − 2 x tan x ( 1 − cos 2 x ) 2 \lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2} lim x → 0 ( 1 − c o s 2 x ) 2 x t a n 2 x − 2 x t a n x equals.
Options A 1 B − 1 2 -\frac{1}{2} − 2 1 C 1 4 \frac{1}{4} 4 1 D 1 2 \frac{1}{2} 2 1 Correct answer D. 1 2 \frac{1}{2} 2 1
Step-by-step solution
Let,
L = lim x → 0 ( x tan 2 x − 2 x tan x ) ( 1 − cos 2 x ) 2 = lim x → 0 K (say) ⇒ K = x [ 2 tan x 1 − ( tan x ) 2 ] − 2 x tan x ( 1 − ( 1 − 2 sin 2 x ) ) 2 = 2 x tan x − [ 2 x tan x − 2 x tan 3 x ] 4 sin 4 x × ( 1 − tan 2 x ) = 2 x tan 3 x 4 sin 4 x × ( 1 − tan 2 x ) = 2 x tan 3 x 4 sin 4 x × ( cos 2 x − sin 2 x cos 2 x ) \begin{aligned}&L=\lim _{x \rightarrow 0} \frac{(x \tan 2 x-2 x \tan x)}{(1-\cos 2 x)^2}=\lim _{x \rightarrow 0}K \text{ (say) }\\
&\Rightarrow K=\frac{x\left[\frac{2 \tan x}{1-(\tan x)^2}\right]-2 x \tan x}{\left(1-\left(1-2 \sin ^2 x\right)\right)^2}\\
&=\frac{2 x \tan x-\left[2 x \tan x-2 x \tan ^3 x\right]}{4 \sin ^4 x \times\left(1-\tan ^2 x\right)}\\
&=\frac{2 x \tan ^3 x}{4 \sin ^4 x \times\left(1-\tan ^2 x\right)}\\
&=\frac{2 x \tan ^3 x}{4 \sin ^4 x \times\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}\right)}
\end{aligned} L = x → 0 lim ( 1 − cos 2 x ) 2 ( x tan 2 x − 2 x tan x ) = x → 0 lim K (say) ⇒ K = ( 1 − ( 1 − 2 sin 2 x ) ) 2 x [ 1 − ( t a n x ) 2 2 t a n x ] − 2 x tan x = 4 sin 4 x × ( 1 − tan 2 x ) 2 x tan x − [ 2 x tan x − 2 x tan 3 x ] = 4 sin 4 x × ( 1 − tan 2 x ) 2 x tan 3 x = 4 sin 4 x × ( c o s 2 x c o s 2 x − s i n 2 x ) 2 x tan 3 x = 2 x sin 3 x cos 3 x 4 sin 4 x × ( cos 2 x − sin 2 x cos 2 x ) ⇒ K = x 2 sin x × ( cos 2 x − sin 2 x ) cos x ∴ L = lim x → 0 x 2 sin x × lim x → 0 1 cos x ( cos 2 x − sin 2 x ) = lim x → 0 x 2 sin x × lim x → 0 1 cos 0 ( cos 2 0 − sin 2 0 ) = 1 2 \begin{aligned}&=\frac{2 x \frac{\sin ^3 x}{\cos ^3 x}}{4 \sin ^4 x \times\left(\frac{\cos ^2 x-\sin ^2 x}{\cos ^2 x}\right)}\\
& \Rightarrow K=\frac{x}{2 \sin x \times\left(\cos ^2 x-\sin ^2 x\right) \cos x}\\
\therefore L &=\lim _{x \rightarrow 0} \frac{x}{2 \sin x} \times \lim _{x \rightarrow 0} \frac{1}{\cos x\left(\cos ^2 x-\sin ^2 x\right)}\\
&=\lim _{x \rightarrow 0} \frac{x}{2 \sin x} \times \lim _{x \rightarrow 0} \frac{1}{\cos 0\left(\cos ^2 0-\sin ^2 0\right)}\\
&=\frac{1}{2}
\end{aligned} ∴ L = 4 sin 4 x × ( c o s 2 x c o s 2 x − s i n 2 x ) 2 x c o s 3 x s i n 3 x ⇒ K = 2 sin x × ( cos 2 x − sin 2 x ) cos x x = x → 0 lim 2 sin x x × x → 0 lim cos x ( cos 2 x − sin 2 x ) 1 = x → 0 lim 2 sin x x × x → 0 lim cos 0 ( cos 2 0 − sin 2 0 ) 1 = 2 1 $
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