JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 11)
If the function ƒ defined on , $$\left( {{\pi \over 6},{\pi \over 3}} \right)$$ by
$$$f(x) = \left\{ {\matrix{
{{{\sqrt 2 {\mathop{\rm cosx}\nolimits} - 1} \over {\cot x - 1}},} & {x \ne {\pi \over 4}} \cr
{k,} & {x = {\pi \over 4}} \cr
} } \right.$$$
is continuous, then
k is equal to
1
1 / $$\sqrt 2$$
$${1 \over 2}$$
2
Explanation
$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$$ = f($${{\pi \over 4}}$$) = k
$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$$ ($${0 \over 0}$$ form) = k
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - \sqrt 2 \sin x} \over {-\cos e{c^2}x}}$$ (Using L Hospital Rule)
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to {\pi \over 4}} \sqrt 2 {\sin ^3}x$$ = k
$$ \Rightarrow $$ k = $$\sqrt 2 {\left( {{1 \over {\sqrt 2 }}} \right)^3}$$ = $${1 \over 2}$$
$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{\sqrt 2 \cos x - 1} \over {\cot x - 1}}$$ ($${0 \over 0}$$ form) = k
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - \sqrt 2 \sin x} \over {-\cos e{c^2}x}}$$ (Using L Hospital Rule)
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to {\pi \over 4}} \sqrt 2 {\sin ^3}x$$ = k
$$ \Rightarrow $$ k = $$\sqrt 2 {\left( {{1 \over {\sqrt 2 }}} \right)^3}$$ = $${1 \over 2}$$
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