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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
If $\text{f(x)}=\begin{cases}\frac{1-\sin\text{x}}{(\pi-2\text{x}^2)}\times\frac{\log\sin\text{x}}{\log(1+\pi^2-4\pi\text{x}+4\text{x}^2)},&\text{x}\neq\frac{\pi}{2}\\\text{k},&\text{x}=\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then k =
  • A
    $-\frac{1}{16}$
  • B
    $-\frac{1}{32}$
  • ✓
    $-\frac{1}{64}$
  • D
    $-\frac{1}{28}$
✓

Answer

Correct option: C.
$-\frac{1}{64}$
if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ then

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

$\text{f }\frac{\pi}{2}-\text{x = t},$ then

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\text{f}\big(\frac{\pi}{2}-\text{t}\big)=\text{f}(\frac{\pi}{2})$

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\Bigg(\frac{1-\sin(\frac\pi{2}-\text{t}{})}{4\text{t}^2}\times\frac{\log\sin(\frac{\pi}{2})}{\log\big(1+\pi^2-4\pi\big(\frac{\pi}{2}-\text{t}\big)+4\big(\frac{\pi}{2}-\text{t}\big)^2\big)}\Bigg)=\text{k}$

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\Bigg(\frac{(1-\cos\text{t})}{4\text{t}^2}\times\frac{\log\cos\text{t}}{\log\big(1+\pi^2-2\pi^2+4\pi\text{t}+4\big(\frac{\pi^2}{4}+\text{t}^2-\pi\text{t}\big)}\Bigg)=\text{k}$

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{(1-\cos\text{t})}{4\text{t}^2}\times\frac{\log\cos\text{t}}{\log(1-\pi^2+4\pi\text{t})+(\pi^2+4\text{t}^2-4\pi\text{t}}\bigg)=\text{t}$

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{(1-\cos\text{t})}{4\text{t}^2}\times\frac{\log \cos\text{t}}{\log(1+4\text{t}^2)}\bigg)=\text{k}$

$\Rightarrow\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{2\sin^2\frac{\text{t}}{2}}{16\times\frac{\text{t}^2}{4}}\times\frac{\log\cos\text{t}}{\log(1+4\text{t}^2)}\bigg)=\text{k}$

$\Rightarrow\frac{2}{16}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{\bigg(\frac{\text{t}^2}{4}\bigg)}\times\frac{\log\cos\text{t}}{\bigg(\frac{4\text{t}^2\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{(\frac{\text{t}}{2})^2}\times\frac{\frac{\log\cos\text{t}}{4\text{t}^2}}{\bigg({\frac{\log(1+4\text{t}^2)}{4\text{t}}\bigg)}} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{(\frac{\text{t}}{2})}\times\frac{\frac{\log\sqrt{1-\sin^2\text{t}}}{4\text{t}^2}}{\bigg({\frac{\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)}} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{8}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{\big(\frac{\text{t}}{2}\big)^2}\times\frac{\bigg(\frac{\log(1-\sin^2\text{t})}{(8\text{t}^2)}\bigg)}{\bigg({\frac{\log(1+4^2\text{t})}{4\text{t}^2}}\bigg)} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{64}\lim\limits_{\text{t}\rightarrow0}\begin{pmatrix}\frac{\sin^2\frac{\text{t}}{2}}{\big(\frac{\text{t}}{2}\big)}\times\frac{\bigg(\frac{\log(1-\sin^2\text{t})}{\text{t}^2}\bigg)}{\bigg(\frac{\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{64}\begin{pmatrix}\lim\limits_{\text{t}\rightarrow0}\Bigg(\frac{\sin\frac{\text{t}}{2}}{\big({\frac{\text{t}}{2}\big)}}\Bigg)^2\times\frac{\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\log(1-\sin^2\text{t})}{\text{t}^2}\bigg)}{\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\log(1+4\text{t}^2)}{4\text{t}^2}\bigg)} \end{pmatrix}=\text{k}$

$\Rightarrow\frac{1}{64}\bigg(1\times\lim\limits_{\text{t}\rightarrow0}\frac{(-\sin^2\text{t})\log(1-\sin^2\text{t})}{\text{t}^2(-\sin^2\text{t})}\bigg)=\text{k}$

$\Rightarrow\frac{-1}{64}\bigg(\lim\limits_{\text{t}\rightarrow0}\frac{(\sin^2\text{t})\log(1-\sin^2\text{t})}{\text{t}^2(-\sin^2\text{t})}\bigg)=\text{k}$

$\Rightarrow\frac{-1}{64}\Bigg(\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\sin\text{t}}{\text{t}}\bigg)^2\lim\limits_{\text{t}\rightarrow0}\frac{\log(1-\sin^2\text{t})}{(-\sin^2\text{t})}\Bigg)=\text{k}$

$\Rightarrow\frac{-1}{64}\Bigg(\lim\limits_{\text{t}\rightarrow0}\bigg(\frac{\sin\text{t}}{\text{t}}\bigg)^2\lim\limits_{\text{t}\rightarrow0}\frac{\log(1-\sin^2\text{t})}{(-\sin^2\text{t})}\Bigg)=\text{k}$

$\Rightarrow\text{k}=\frac{-1}{64}$ $\bigg[\because\lim\limits_{\text{x}\rightarrow0}\frac{\log(1-\text{x})}{\text{x}}=1\bigg]$

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