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  • Consider <img alt="f(x)=\left\{\begin{array}{cl}\frac{a+3 \cos x}{x^2}, & x<0 \\ b \tan \left\{\frac{\pi}{[x+3]}\right\}, & x \geq 0\end{array}\right." src="https://entrancecorner.oncode

Consider f(x)=\left\{\begin{array}{cl}\frac{a+3 \cos x}{x^2}, & x<0 \\ b \tan \left\{\frac{\pi}{[x+3]}\right\}, & x \geq 0\end{array}\right. where [.] represents the greatest integer function. If f(x) is continuous at x=0, then 4 b^2+a is equal to

Option: 1

1


Option: 2

2


Option: 3

0


Option: 4

-1


Answers (1)

best_answer

For x=0 and x=0+h ;[x+3]=3
\mathrm{\begin{aligned} & \therefore \quad f(0)=b \tan \frac{\pi}{3}=b \sqrt{3} \\ & \text { R.H.L. }=\lim _{h \rightarrow 0} b \tan \frac{\pi}{[h+3]}=b \tan \frac{\pi}{3}=b \sqrt{3} \\ & \text { L.H.L. }=\lim _{h \rightarrow 0} \frac{a+3 \cos (0-h)}{(-h)^2}=\lim _{h \rightarrow 0} \frac{a+3 \cos h}{h^2} \\ & =\lim _{h \rightarrow 0} \frac{a+3\left(1-\frac{h^2}{2 !}+\frac{h^4}{4 !}-\ldots\right)}{h^2} \\ & =\lim _{h \rightarrow 0}\left[\frac{(a+3)}{h^2}-\frac{3}{2}+\text { powers of } h\right] \end{aligned} }
For left hand limit to exist, we must have a+3=0
\mathrm{\therefore a=-3 \, and \, \, in\, \, that\, \, case\, \, the\, \, L.H.L. =-\frac{3}{2} }
For continuity, we have
\mathrm{\begin{aligned} & b \sqrt{3}=-\frac{3}{2}=b \sqrt{3} \\ & \Rightarrow a=-3 \text { and } b=-\frac{1}{2} \sqrt{3} \Rightarrow 4 b^2+a=0 \end{aligned} }

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seema garhwal

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