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AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

Section: Mathematics

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Question : 62 of 160

Marks: +1, -0

If α =

\lim\limits_{x\to 0}\frac{x\cdot 2^x - x}{1 - \cos x}

\lim\limits_{x\to 0}\frac{x\cdot 2^x - x}{\sqrt{1+x^2} - \sqrt{1-x^2}}

, then

α = 5β
α = 2β
β = $2\alpha^{2}$
β = $\frac{1}{6}$ α

Solution:

It is given that,

\alpha = \lim\limits_{x\to 0} \frac{x\cdot 2^{x} - x}{1 - \cos x}

Apply L hospital rule,

\alpha = \lim\limits_{x\to 0} \frac{2^{x} + x\cdot 2^{x} \log 2 - 1}{\sin x}

Again, apply L hospital rule,

\alpha = \lim\limits_{x\to 0} \frac{2^{x} \log x + 2^{x} \log 2 + x\cdot 2^{x} (\log 2)^{2}}{\cos x}

\alpha = \log 2 + \log 2

\alpha = 2\log 2\;\cdots\;(I)

And

\beta = \lim\limits_{x\to 0} \frac{x\cdot 2^{x} - x}{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}

Apply L hospital rule,

\beta = \lim\limits_{x\to 0} \frac{2^{x} + x\cdot 2^{x} \log 2 - 1}{\left(\;\frac{2x}{2\sqrt{1+x^{2}}}-\;\frac{2x}{2\sqrt{1-x^{2}}}\right)}

Again, apply L hospital rule,

\beta = \lim\limits_{x\to 0} \frac{2^{x} \log 2 + 2^{x} \log 2 + x\cdot 2^{x} (\log 2)^{2}}{\left(\sqrt{1+x^{2}} - \;\frac{2x}{2\sqrt{1+x^{2}}}\right) + \left(\sqrt{1-x^{2}} + \;\frac{2x}{2\sqrt{1-x^{2}}}\right)}

\beta = \;\frac{2\log 2}{(1-0)+(1+0)} = \;\frac{2\log 2}{2}

\beta = \log 2 \;\cdots\;(II)

From equation

(I) and (II),

\alpha = 2\beta

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