JEE Main 2018 15 April Evening Shift Online
Exam JEE Main
Year 2018
Date 15 April
Shift Evening Shift
Mode Online
Subject Mathematics
Chapter Continuity and Differentiability JEE Main 2018 · 15 April · Evening Shift · Online · Mathematics · Continuity and Differentiability
Let f(x)=\ array cc (x-1)^ (1)/(2-x) , & x>1, x ≠ 2 k, & x=2 array...
Let
f(x)=\left\{ ( x − 1 ) 1 2 − x , x > 1 , x ≠ 2 k , x = 2 \begin{array}{cc}(x-1)^{\frac{1}{2-x}}, & x>1, x \neq 2 \\ k, & x=2\end{array} ( x − 1 ) 2 − x 1 , k , x > 1 , x = 2 x = 2 \right.
The value of
k k k for which
f f f is continuous at
x = 2 x=2 x = 2 is
Options A e − 2 e^{-2} e − 2 B e e e C e − 1 e^{-1} e − 1 D 1 Correct answer C. e − 1 e^{-1} e − 1
Step-by-step solution
Since
f ( x ) f(x) f ( x ) is continuous at
x = 2 x=2 x = 2 .
∴ lim x → 2 f ( x ) = f ( 2 ) ⇒ lim x → 2 ( x − 1 ) 1 2 − x = k ( 1 ∞ form ) ∴ e l = k \begin{aligned}&\therefore \lim _{x \rightarrow 2}f(x)=f(2) \\
&\Rightarrow \lim _{x \rightarrow 2}(x-1)^{\frac{1}{2-x}}=k \quad \left(1^{\infty} \text{ form }\right)\\
&\therefore e^{l=k}
\end{aligned} ∴ x → 2 lim f ( x ) = f ( 2 ) ⇒ x → 2 lim ( x − 1 ) 2 − x 1 = k ( 1 ∞ form ) ∴ e l = k where
l = lim x → 2 ( x − 1 − 1 ) × 1 2 − x = lim x → 2 x − 2 2 − x = lim x → 2 ( x − 2 x − 2 ) ⇒ k = e − 1 \begin{aligned}&l=\lim _{x \rightarrow 2}(x-1-1) \times \frac{1}{2-x}=\lim _{x \rightarrow 2} \frac{x-2}{2-x}\\
&=\lim _{x \rightarrow 2} \left(\frac{x-2}{x-2}\right)\\
&\Rightarrow k=e^{-1}
\end{aligned} l = x → 2 lim ( x − 1 − 1 ) × 2 − x 1 = x → 2 lim 2 − x x − 2 = x → 2 lim ( x − 2 x − 2 ) ⇒ k = e − 1
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