2 Marks

Question 12 Marks

If function $F (x)=\frac{\sin (10 x)}{x}, x \neq 0$, is continuous at $x=0$. Then find the value of $F (0)$.

Answer

If function $F (x)=\frac{\sin (10 x)}{x}, x \neq 0$ is continuous. then$
\begin{aligned}
F(0) & =\lim _{x \rightarrow 0} F(x)=\lim _{x \rightarrow 0}\left[\frac{\sin (10 x)}{x}\right] \\
& =\lim _{x \rightarrow 0}\left[\frac{\sin (10 x)}{10 x} \times 10\right] \\
& =10\left[\lim _{x \rightarrow 0} \frac{\sin (10 x)}{(10 x)}\right] \\
& =10 \times 1=10
\end{aligned}
$

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Question 22 Marks

If $x=t^2, y=t^3$ then find $\frac{d^2 y}{d x^2}$.

Answer

$\begin{array}{l}x=t^2 \Rightarrow \frac{d x}{d t}=2 t \\ y=t^3 \Rightarrow \frac{d y}{d t}=3 t^2\end{array}$
then\[\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{3 t^2}{2 t}=\frac{3}{2} t\]
$\begin{aligned} \therefore \quad \frac{d^2 y}{d x^2} & =\frac{d}{d x}\left(\frac{3}{2} t\right)=\frac{3}{2} \cdot \frac{d t}{d x}=\frac{3}{2} \times \frac{1}{2 t} \\ & =\frac{3}{4 t}\end{aligned}$

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Question 32 Marks

If $y=\sin ^{-1} x$ then find $\frac{d^2 y}{d x^2}$.

Answer

$\begin{aligned} y & =\sin ^{-1} x \\ \frac{d y}{d x} & =\frac{1}{\sqrt{1-x^2}} \\ \text { and } \quad \frac{d^2 y}{d x^2} & =\left(-\frac{1}{2}\right)\left(1-x^2\right)^{-3 / 2}(0-2 x) \\ & =\frac{x}{\left(1-x^2\right) \sqrt{1-x^2}}\end{aligned}$

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Question 42 Marks

Find differentiation of $\log (1+\theta)$ w.r.t. $\sin ^{-1} \theta$.

Answer

Suppose $\quad y=\log (1+\theta)$ and $x=\sin ^{-1} \theta$
$\therefore \quad \frac{d y}{d \theta}=\frac{1}{1+\theta}$
and $\quad \frac{d x}{d \theta}=\frac{1}{\sqrt{1-\theta^2}}$
$\therefore \quad \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{1 / 1+\theta}{\frac{1}{\sqrt{1-\theta^2}}}$
$
=\sqrt{\frac{1-\theta}{1+\theta}}
$

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Question 52 Marks

If $y=\sqrt{\sin x+y}$, then find $\frac{d y}{d x}$.

Answer

$
\begin{aligned}
y & =\sqrt{\sin x+y} \\
y^2 & =\sin x+y \\
2 y \frac{d y}{d x} & =\cos x+\frac{d y}{d x} \\
(2 y-1) \frac{d y}{d x} & =\cos x \\
\therefore \quad \frac{d y}{d x} & =\frac{\cos x}{2 y-1} \text {}
\end{aligned}
$

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