JEE Main 2014 11 April Online
Exam JEE Main
Year 2014
Date 11 April
Mode Online
Subject Mathematics
Chapter Limits JEE Main 2014 · 11 April · Online · Mathematics · Limits
If _ x arrow 2 (x-2\ x^2+k+2 x-2 k\ . x^2-4 x+4 =5 , then k is...
If
lim x → 2 tan ( x − 2 { x 2 + k + 2 x − 2 k } x 2 − 4 x + 4 = 5 \lim _{x \rightarrow 2} \frac{\tan \left(x-2\left\{x^2+k+2 x-2 k\right\}\right.}{x^2-4 x+4}=5 lim x → 2 x 2 − 4 x + 4 t a n ( x − 2 { x 2 + k + 2 x − 2 k } = 5 , then
k \mathrm{k} k is equal to:
Step-by-step solution
lim x → 2 tan ( x − 2 ) { x 2 + ( k − 2 ) x − 2 k } x 2 − 4 x + 4 = 5 ⇒ lim x → 2 tan ( x − 2 ) { x 2 + k x − 2 x − 2 k } ( x − 2 ) 2 = 5 ⇒ lim x → 2 tan ( x − 2 ) { x ( x − 2 ) + k ( x − 2 ) } ( x − 2 ) × ( x − 2 ) = 5 ⇒ lim x → 2 ( tan ( x − 2 ) ( x − 2 ) ) × lim x → 2 ( ( k + x ) ( x − 2 ) ( x − 2 ) ) = 5 ⇒ 1 × lim x → 2 ( k + x ) = 5 { ∵ lim h → 0 tan h h = 1 } or k + 2 = 5 ⇒ k = 3 \begin{aligned}& \lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^2+(k-2) x-2 k\right\}}{x^2-4 x+4}=5 \\
\Rightarrow & \lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^2+k x-2 x-2 k\right\}}{(x-2)^2}=5 \\
\Rightarrow & \lim _{x \rightarrow 2} \frac{\tan (x-2)\{x(x-2)+k(x-2)\}}{(x-2) \times(x-2)}=5 \\
\Rightarrow \lim _{x \rightarrow 2}\left(\frac{\tan (x-2)}{(x-2)}\right) \times \lim _{x \rightarrow 2}\left(\frac{(k+x)(x-2)}{(x-2)}\right)=5 \\
\Rightarrow & 1 \times \lim _{x \rightarrow 2}(k+x)=5 \\
&\left\{\because \lim _{h \rightarrow 0} \frac{\tan h}{h}=1\right\} \\
& \text{ or }k+2=5 \\
\Rightarrow k=3
\end{aligned} ⇒ ⇒ ⇒ x → 2 lim ( ( x − 2 ) tan ( x − 2 ) ) × x → 2 lim ( ( x − 2 ) ( k + x ) ( x − 2 ) ) = 5 ⇒ ⇒ k = 3 x → 2 lim x 2 − 4 x + 4 tan ( x − 2 ) { x 2 + ( k − 2 ) x − 2 k } = 5 x → 2 lim ( x − 2 ) 2 tan ( x − 2 ) { x 2 + k x − 2 x − 2 k } = 5 x → 2 lim ( x − 2 ) × ( x − 2 ) tan ( x − 2 ) { x ( x − 2 ) + k ( x − 2 )} = 5 1 × x → 2 lim ( k + x ) = 5 { ∵ h → 0 lim h tan h = 1 } or k + 2 = 5
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