If f(x)= 1- ^2x 3 ^2x ,&if x 2 Then f(x) is continuous at x= 2 , if: 1. a=1 3 , b=2 2. a=1 3 , b=8 3 3. a=2 3 , b=8 3 4. none of these

2. a=1 3 , b=8 3 Solution: Given, f(x)= 1- ^2x 3 ^2x ,&if x 2 We have (LHL at x= 2 )= _ x 2 ^- f(x)= _ h 0 f ( 2 -h ) = _ h 0 ( 1- ^2 ( 2 -h ) 3 ^2 ( 2 -h ) ) = _ h 0 ( 1- ^2h 3 ^2h ) =1 3 _ h 0 ( ^2h ^2h ) =1 3 (RHL at x= 2 )= _ x 2 ^+ f(x)= _ h 0 f ( 2 +h ) = _ h 0 ( b [1- ( 2 +h) ] [x-2 ( 2 +h ) ]^2 ) = _ h 0 ( b(1- ) [-2h]^2 ) = _ h 0 ( 2b ^ 2 h 2 4h^2 ) = _ h 0 ( 2b ^2 h 2 16 h^2 4 ) = b 8 _ h 0 ( h 2 h 2 )^2 = b 8 1 = b 8 Also, f ( 2 )=a If f(x) is continuous at x= 2 , then _ x 2 ^- f(x)= _ x 2 ^+ f(x)=f ( 2 ) 1 3 = b 8 =a =1 3 and b=8 3

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

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Question

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

$\text{a}=\frac{1}{3},\text{ b}=2$
$\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$
$\text{a}=\frac{2}{3},\text{ b}=\frac{8}{3}$
none of these

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Answer

$\text{a}=\frac{1}{3},\text{ b}=\frac{8}{3}$

Solution:
Given, $\text{f(x)}=\begin{cases}\frac{1-\sin^2\text{x}}{3\cos^2\text{x}},&\text{if}\text{ x}<\frac{\pi}{2}\\\text{a},&\text{if}\text{ x}=\frac{\pi}{2}\\\frac{\text{b}(1-\sin\text{x})}{(\pi-2\text{x})^2},&\text{if}\text{ x }>\frac{\pi}{2}\end{cases}$
We have
$\Big(\text{LHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}-\text{h}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{1-\sin^2\big(\frac{\pi}{2}-\text{h}\big)}{3\cos^2\big(\frac{\pi}{2}-\text{h}\big)}\Bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{1-\cos^2\text{h}}{3\sin^2\text{h}}\Big)$
$=\frac{1}{3}\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\sin^2\text{h}}{\sin^2\text{h}}\Big)$
$=\frac{1}{3}$
$\Big(\text{RHL at x}=\frac{\pi}{2}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\text{b}\big[1-\sin\big(\frac{\pi}{2}+\text{h})\big]}{\big[\text{x}-2\big(\frac{\pi}{2}+\text{h}\big)\big]^2}\Bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\text{b}(1-\cos\text{h})}{[-2\text{h}]^2}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^{2}\frac{\text{h}}{2}}{4\text{h}^2}\bigg)$
$=\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{2\text{b}\sin^2\frac{\text{h}}{2}}{16\frac{\text{h}^2}{4}}\bigg)$
$=\frac{\text{b}}{8}\lim\limits_{\text{h}\rightarrow0}\bigg(\frac{\sin\frac{\text{h}}{2}}{\frac{\text{h}}{2}}\bigg)^2$
$=\frac{\text{b}}{8}\times1$
$=\frac{\text{b}}{8}$
Also, $\text{f}\big(\frac{\pi}{2}\big)=\text{a}$
If f(x) is continuous at $\text{x}=\frac{\pi}{2},$ then
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^+}\text{f(x)}=\text{f}\big(\frac{\pi}{2}\big)$
$\Rightarrow\frac{1}{3}=\frac{\text{b}}{8}=\text{a}$
$\Rightarrow\text{a}=\frac{1}{3}\text{ and }\text{b}=\frac{8}{3}$

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