JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 10)
Let $$f:\left( { - {\pi \over 4},{\pi \over 4}} \right) \to R$$ be defined as $$f(x) = \left\{ {\matrix{
{{{(1 + |\sin x|)}^{{{3a} \over {|\sin x|}}}}} & , & { - {\pi \over 4} < x < 0} \cr
b & , & {x = 0} \cr
{{e^{\cot 4x/\cot 2x}}} & , & {0 < x < {\pi \over 4}} \cr
} } \right.$$
If f is continuous at x = 0, then the value of 6a + b2 is equal to :
If f is continuous at x = 0, then the value of 6a + b2 is equal to :
1 $$-$$ e
e $$-$$ 1
1 + e
e
Explanation
$$\mathop {\lim }\limits_{x \to 0} f(x) = b$$
$$\mathop {\lim }\limits_{x \to {0^ + }} x{e^{{{\cot 4x} \over {\cot 2x}}}} = {e^{{1 \over 2}}} = b$$
$$\mathop {\lim }\limits_{x \to {0^ - }} {(1 + |\sin x|)^{{{3a} \over {|\sin x|}}}} = {e^{3a}} = {e^{{1 \over 2}}}$$
$$a = {1 \over 6} \Rightarrow 6a = 1$$
$$ \therefore $$ $$(6a + {b^2}) = (1 + e)$$
$$\mathop {\lim }\limits_{x \to {0^ + }} x{e^{{{\cot 4x} \over {\cot 2x}}}} = {e^{{1 \over 2}}} = b$$
$$\mathop {\lim }\limits_{x \to {0^ - }} {(1 + |\sin x|)^{{{3a} \over {|\sin x|}}}} = {e^{3a}} = {e^{{1 \over 2}}}$$
$$a = {1 \over 6} \Rightarrow 6a = 1$$
$$ \therefore $$ $$(6a + {b^2}) = (1 + e)$$
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