Let f(x)= 1- ^3x 3 ^2x ,&if x 2 if f(x) is continuous at x= 2 , find a and b.

It is given that the function is continuous at x= 2 LHL=RHL=f ( 2 ) ...(i) Now, f ( 2 )=a LHL= _ x ^- 2 f(x)= _ h 0 f ( 2 -h ) = _ h 0 1- ^3 ( 2 -h ) 3 ^2 ( 2 -h ) = _ h 0 1- ^3h 3 ^2h = _ h 0 (1- )(1+ ^2h+ ) 3 ^2h = _ h 0 2 ^2 h 2 (1+ ^2h+ ) 3 ^2h = _ h 0 2 ( h 2 h 2 )^2 h^2 4 (1+ ^2h+ ) 3 ( h )^2 h^2 = _ h 0 2 1 4 (1+ ^2h+ ) 3 =1 2 RHL= _ x ^- 2 f(x)= _ h 0 f ( 2 +h ) = _ h 0 b (1- ( 2 +h ) ) ( -2 ( 2 +h ) )^2 = _ h 0 b(1- ) ( - -2h)^2 = _ h 0 b 2 ^2 h 2 (2h)^2 = _ h 0 b 2 ( h 2 h 2 )^2 1 4 = _ h 0 b 8 = b 8 Thus, using (i) we get, a=1 2 And b 8 =1 2 =4 Thus, a=1 2 and b=4

Let f(x)= 1- ^3x 3 ^2x ,&if x< 2 ,&if x= 2 b(1- ) ( -2x) ^2,&x> 2…

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Question

Let $\text{f(x)}=\begin{cases}\frac{1-\sin^3\text{x}}{3\cos^2\text{x}},&\text{if }\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if }\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})}^2,&\text{x}>\frac{\pi}{2}\end{cases}$ if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ find a and b.

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Answer

It is given that the function is continuous at $\text{x}=\frac{\pi}{2}$
$\therefore\ \text{LHL}=\text{RHL}=\text{f}\Big(\frac{\pi}{2}\Big)\ ...(\text{i})$
Now, $\text{f}\Big(\frac{\pi}{2}\Big)=\text{a}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow\frac{\pi^-}{2}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow 0}\text{f}\Big(\frac{\pi}{2}-\text{h}\Big)$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{1-\sin^3\Big(\frac{\pi}{2}-\text{h}\Big)}{3\cos^2\Big(\frac{\pi}{2}-\text{h}\Big)}=\lim_\limits{\text{h}\rightarrow 0}\frac{1-\cos^3\text{h}}{3\sin^2\text{h}}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{(1-\cos\text{h})(1+\cos^2\text{h}+\cos\text{h})}{3\sin^2\text{h}}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{2\sin^2\frac{\text{h}}{2}(1+\cos^2\text{h}+\cos\text{h})}{3\sin^2\text{h}}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{2\Bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\Bigg)^2\times\frac{\text{h}^2}{4}\times(1+\cos^2\text{h}+\cos\text{h})}{3\Big(\frac{\sin\text{h}}{\text{h}}\Big)^2\times\text{ h}^2}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{2\times\frac{1}{4}(1+\cos^2\text{h}+\cos\text{h})}{3}=\frac{1}{2}$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow\frac{\pi^-}{2}}\text{f(x)}=\lim_\limits{\text{h}\rightarrow 0}\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}\Big(1-\sin\Big(\frac{\pi}{2}+\text{h}\Big)\Big)}{\Big(\pi-2\Big(\frac{\pi}{2}+\text{h}\Big)\Big)^2}=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}(1-\cos\text{h})}{(\pi-\pi-2\text{h})^2}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b }\times\ 2\sin^2\frac{\text{h}}{2}}{(2\text{h})^2}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}}{2}\Bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\Bigg)^2\times\frac{1}{4}$
$=\lim_\limits{\text{h}\rightarrow 0}\frac{\text{b}}{8}=\frac{\text{b}}{8}$
Thus, using (i) we get,
$\text{a}=\frac{1}{2}$
And $\frac{\text{b}}{8}=\frac{1}{2}$
$\Rightarrow\text{b}=4$
Thus, $\text{a}=\frac{1}{2}$ and $\text{b}=4$

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