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  •  If  <img alt="\lim _{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}{1-\cos (2 x)}=17" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarr

 If  \lim _{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{c x e^{-c x}}{2}}{1-\cos (2 x)}=17, then 5 \mathrm{a}^2+\mathrm{b}^2  is equal to

Option: 1

76


Option: 2

72


Option: 3

64


Option: 4

68


Answers (1)

best_answer

\lim _{x \rightarrow 0} \frac{e^{a x}-\cos b x-\frac{c x e^{-c x}}{2}}{1-\cos 2 x}=17

On expansion

\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(1+a x+\frac{(a x)^2}{2 !}+\ldots\right)-\left(1-\frac{(b x)^2}{2 !}+\ldots\right)-\frac{c x}{2}\left(1-c x+\frac{(c x)^2}{2 !}\right)}{\left(\frac{1-\cos 2 x}{(2 x)^2}\right) \times(2 x)^2}=17 \\ & \lim _{x \rightarrow 0} \frac{x\left(a-\frac{c}{2}\right)+x^2\left(\frac{a^2}{2}+\frac{b^2}{2}+\frac{c^2}{2}\right)}{\frac{1}{2}\left(4 x^2\right)}=17 \end{aligned}

For limit to be exist

\mathrm{a}-\frac{\mathrm{c}}{2}=0 \Rightarrow \mathrm{c}=2 \mathrm{a}

\begin{aligned} & \Rightarrow \frac{\frac{\mathrm{a}^2}{2}+\frac{\mathrm{b}^2}{2}+\frac{\mathrm{c}^2}{2}}{2}=17 \\ & \Rightarrow \frac{\mathrm{a}^2}{2}+\frac{\mathrm{b}^2}{2}+\frac{4 \mathrm{a}^2}{2}=34 \\ & \Rightarrow 5 \mathrm{a}^2+\mathrm{b}^2=68 \end{aligned}

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shivangi.shekhar

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