Question

JEE ADVANCED MATHS QUESTION SEPTEMBER 2020

Answer 1

Answer:

1

Step-by-step explanation:

Solution is refer to attachment .

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Answer 2

ANSWER.

The value of the real number a for which the

right hand limit is equal to nonzero = 1.

EXPLANATION.

$\sf : \implies \: \lim_{x \to \: {0}^{ + } } \: \: \dfrac{(1 - x) {}^{ \dfrac{1}{x} } - e {}^{ - 1} }{ {x}^{a} }$

$\sf : \implies \: \lim_{x \to \: {0}^{ + } } \: \dfrac{ {e}^{ \dfrac{1}{x} ln(1 - x)} - {e}^{ - 1} }{ {x}^{a} } \\ \\ \sf : \implies \: \lim_{x \to \: {0}^{ + } } \: \frac{e {}^{ \dfrac{1}{x}( - x - \dfrac{x {}^{2} }{2} - \dfrac{ {x}^{3} }{3} .....} - {e}^{ - 1} }{ {x}^{ a } } \\ \\ \sf : \implies \: \lim_{x \to \: {0}^{ + } } \: \frac{e {}^{( - 1 - \dfrac{x}{2} - \dfrac{ {x}^{2} }{3} .....} - {e}^{ - 1} }{ {x}^{a} }$

$\sf : \implies \: \lim_{x \to \: {0}^{ + } } \: \dfrac{(e {}^{ - 1})(e {}^{ \dfrac{ - x}{2} - \dfrac{ {x}^{2} }{3}.... } - 1)}{( \dfrac{ - {x}^{} }{2} - \dfrac{ {x}^{2} }{3} .....)} \times \: \: \dfrac{\dfrac{ - {x}^{} }{2} - \dfrac{ {x}^{2} }{3} .....}{ {x}^{a} } \\ \\ \\ \sf : \implies \: \lim_{x \to \: {0}^{ + } } \: \: \frac{1}{e} ( \frac{ - 1}{2}x {}^{(1 - a) } - \frac{ 1}{3}x {}^{(2 - a)} .....) \\ \\ \sf : \implies \: \lim_{x \to \: {0}^{ + } } \: \: 1 - a \: = 0 \\ \\ \sf : \implies \: a \: = 1$