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Find \mathrm{a, b \: and\: c} such that

\mathrm{ \lim _{x \rightarrow 0} \frac{a x e^x-b \log (1+x)+c x e^{-x}}{x^2 \sin x}=2 . }

Option: 1

\mathrm{a=3,b=12,c=9}


Option: 2

\mathrm{a=12,b=3,c=9}


Option: 3

\mathrm{a=9,b=3,c=12}


Option: 4

\mathrm{a=3,b=9,c=12}


Answers (1)

best_answer

\mathrm{\text { Limit } =\lim _{x \rightarrow 0} \frac{a\left(e^x+x e^x\right)-b \cdot \frac{1}{1+x}+c\left(e^{-x}-x e^{-x}\right)}{2 x \sin x+x^2 \cos x} }

                                                                               {using L'Hospital's rule}

\mathrm{=\frac{a-b+c}{0}=\infty \text { unless } a-b+c=0 . }

But, the limit is given to be 2 . So, the indeterminate form \mathrm{\frac{0}{0}} should continue.

So, \mathrm{a-b+c=0.}                           ...........(1)

Then, limit

\mathrm{=\lim_{x\rightarrow 0}\frac{\left \{ a(1+x)e^{x}+ae^{x}+\frac{b}{\left ( 1+x \right )^{2}}+c\left ( -1 \right )e^{x}+c\left ( 1-x \right )e^{x}\left ( -1 \right ) \right \}}{2\sin x+2x\cos x+2x\cos x-x^{2}\sin x}}[using L' Hospital's rule]

\mathrm{=\lim_{x\rightarrow 0}\frac{a\left ( 2+x \right )e^{x}+\frac{b}{\left ( 1+x \right )^{2}}-c\left ( 2-x \right )e^{-x}}{2\sin x+4x\cos x-x^{2}\sin x}}

\mathrm{=\frac{2a+b-2c}{0}=\infty} unless \mathrm{2a+b-2c=0}

But the limit is given to be 2

so, \mathrm{2a+b-2c=0}         ...........(2)

 Then, \mathrm{limit=\lim_{x\rightarrow 0}\frac{\left \{ a\left ( 2+x \right )e^{x}+ae^{x}-\frac{2b}{\left ( 1+x \right )^{3}}+c\left ( 2-x \right )e^{-x}+ce^{-x} \right \}}{6\cos x-4x\sin x-2x\sin x-x^{2}\cos x}}

\mathrm{=\frac{2 a+a-2 b+2 c+c}{6}=2 \text { (given) } }

\mathrm{\therefore \quad 3 a-2 b+3 c=12 }              ..............(3)

\mathrm{ \text { (1) } \Rightarrow b=a+c }

\mathrm{ \text { (2) } \Rightarrow 2 a+a+c-2 c=0 \text {, i.e., } c=3 a }

\mathrm{\therefore \quad b=a+c }

\mathrm{=a+3 a=4 a }

\mathrm{ \therefore \text { (3) } \Rightarrow 3 a-2 \times 4 a+3 \times 3 a=12 ; \quad \therefore \quad a=3 }

\mathrm{ \therefore b=4 a=4 \times 3=12 }

\mathrm{c=3 a=3 \times 3=9.}

Thus,\mathrm{ a=3, b=12, c=9.}

Hence option 1 is correct.











 

 


 

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Gunjita

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