Test Index

AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Section: Mathematics

© examsnet.com

Question : 63 of 160

Marks: +1, -0

If α =

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

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

then

α = β
2α = β
α = 2β
α = 3β

Solution:

Consider the equation

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

And,

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

Simplify the first equation as,

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

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

\;\; =\lim\limits_{x\rightarrow 0}\;\frac{2^{x}-1}{2\sin^{2}\frac{x}{2}}

\;\; =\;\frac{1}{2 x\rightarrow 0}\lim\limits_{x}\;\frac{\frac{2^{x}-1}{x}}{\frac{\sin^{2}\frac{x}{2}}{x^{2}}}

Solve further

\alpha =\;\frac{1}{2}\lim\limits_{x\rightarrow 0}\;\frac{\frac{2^{x}-1}{x}}{\left(\frac{\sin\frac{1}{2}x}{x}\right)^{2}}

=\;\frac{1}{2}\left[\lim\limits_{x\rightarrow 0}\frac{2^{x}-1}{x}\times\frac{1}{\lim\limits_{x\rightarrow 0}\left(\frac{\sin\frac{1}{2}x}{\frac{x}{2}}\right)^{2}}\;\;\times\;\;\frac{1}{4}\right]

=\;\frac{1}{2}\times\log2\times\frac{1}{1\times\frac{1}{4}}

=2\log2

Also

\beta \;\; =\lim\limits_{x\rightarrow 0}\;\frac{x\cdot2^{x}-x}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\times\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}

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

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

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

Solve further

\beta \;\; =\;\frac{1}{2}\times\log2\times2

\beta \;\; =\log2

2\beta \;\; =2\log2

2\beta \;\; =\alpha

© examsnet.com

Go to Question:

Prev Question

Next Question