Test Index

BITSAT 2018 Slot 2 Solved Paper

Section: Mathematics

Question : 132 of 150

Marks: +1, -0

The value of

\alpha

\lim\limits_{x \rightarrow 0} \frac{1}{x^{2}} (e^{\alpha x} - e^{x} - x) = \frac{3}{2}

, is

Solution:

Since, the numerator tends to

\infty

x \rightarrow 0

\lim\limits_{x \rightarrow 0} \frac{1}{x^{2}} (e^{\alpha x} - e^{x} - x)

= \frac{1}{2} \lim\limits_{x \rightarrow 0} \frac{(\alpha e^{\alpha x} - e^{x} - 1)}{x}

For last limit to exist we must have,

\lim\limits_{x \rightarrow 0} (\alpha e^{\alpha x} - e^{x} - 1) = 0

\therefore \alpha - 1 - 1 = 0 \Rightarrow \alpha = 2

For

\alpha = 2

the last limit and equal to

= \frac{1}{2} \lim\limits_{x \rightarrow 0} \frac{(2 e^{2 x} - e^{x} - 1)}{x}

= \frac{1}{2} \lim\limits_{x \rightarrow 0} (4 e^{2 x} - e^{x}) = \frac{3}{2}

Hence, option (d) is correct.

Go to Question:

Prev Question

Next Question