If f(x)= 1- ( -2x)^2 , when x 2 = then f(x) will be continuous function at x= 2 , where = 1. 1 8 2. 1 4 3. 1 2 4. none of these

1. 1 8 Solution: If f(x) is continuous at x= 2 , then _ x x 2 f(x)=f ( 2 ) 1- x 2 ( -2x)^2 =f ( 2 ) ...(i) suppose ( 2 -x )=t, then _ t 0 1- ( 2 -t ) (2t)^2 =f ( 2 ) [From eq.(i)] _ t 0 1- (2t)^2 =f ( 2 ) 1 4 _ t 0 2 ^2 ( t 2 ) t^2 =f ( 2 ) 1 4 _ t 0 2 4 ^ 2 ( t 2 ) t^ 2 4 = f ( 2 ) 1 8 _ t 0 2 4 ^ 2 ( t 2 ) t^ 2 4 = f ( 2 ) 1 8 _ t 0 ( t 2 ) t 2 ^2=f ( 2 ) ( 2 )= =1 8

If f(x)= 1- ( -2x)^2 , when x 2 = then f(x) will be continuous fu…

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Question

If $\text{f(x)}=\frac{1-\sin\text{x}}{(\pi-2\text{x})^2},$ when $\text{x}\neq\frac{\pi}{2}=\lambda$ then f(x) will be continuous function at $\text{x}=\frac{\pi}{2},$ where $\lambda=$

$\frac{1}{8}$
$\frac{1}{4}$
$\frac{1}{2}$
none of these

✓

Answer

$\frac{1}{8}$

Solution:

If f(x) is continuous at $\text{x}=\frac{\pi}{2},$ then

$\lim\limits_{\text{x}\rightarrow\frac{\text{x}}{2}}\text{f}\text{(x)}=\text{f}\Big(\frac{\pi}{2}\Big)$

$\lim\frac{1-\sin\text{x}}{\text{x}\rightarrow\frac{\pi}{2}(\pi-2\text{x})^2}=\text{f}\Big(\frac{\pi}{2}\Big) \ ...(\text{i})$$$

suppose $\Big(\frac{\pi}{2}-\text{x}\Big)=\text{t},$ then

$\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{1-\sin\Big(\frac{\pi}{2}-\text{t}\Big)}{(2\text{t})^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$ [From eq.(i)]

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{1-\cos\text{t}}{(2\text{t})^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$

$\Rightarrow\frac{1}{4}\lim\limits_{t\rightarrow0}\begin{bmatrix}\frac{2\sin^2\Big(\frac{\text{t}}{2}\Big)}{\text{t}^2} \end{bmatrix}=\text{f}\Big(\frac{\pi}{2}\Big)$

$\Rightarrow \frac{1}{4}\lim\limits_{\text{t} \rightarrow0} \begin{bmatrix}\frac{\frac{2}{4}\sin^{2}\big(\frac{\text{t}}{2}\big)}{\frac{\text{t}^{2}}{4}} \end{bmatrix} = \text{f} \Big(\frac{\pi}{2}\Big)$

$\Rightarrow \frac{1}{8}\lim\limits_{\text{t} \rightarrow0} \begin{bmatrix}\frac{\frac{2}{4}\sin^{2}\big(\frac{\text{t}}{2}\big)}{\frac{\text{t}^{2}}{4}} \end{bmatrix} = \text{f} \Big(\frac{\pi}{2}\Big)$

$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{bmatrix}\frac{\sin\Big(\frac{\text{t}}{2}\Big)}{\frac{\text{t}}{2}} \end{bmatrix}^2=\text{f}\Big(\frac{\pi}{2}\Big)$

$\Rightarrow\text{f}\Big(\frac{\pi}{2}\Big)=\lambda=\frac{1}{8}$