3 Marks Question

Question 13 Marks

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

Answer

Given $
y \sqrt{1-x^2}=\sin ^{-1} x
$
differentiating both sides w.r.t. $x$$
\begin{aligned}
\frac{d y}{d x} \sqrt{1-x^2}+y \frac{1}{2 \sqrt{1-x^2}} & \times(-2 x)=\frac{1}{\sqrt{1-x^2}} \\
\sqrt{1-x^2} \frac{d y}{d x}-\frac{x y}{\sqrt{1-x^2}} & =\frac{1}{\sqrt{1-x^2}} \\
\left(1-x^2\right) \frac{d y}{d x}-x y & =1 \\
\left(1-x^2\right) \frac{d y}{d x} & =1+x y \\
\frac{d y}{d x} & =\frac{1+x y}{\left(1-x^2\right)}
\end{aligned}
$

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Question 23 Marks

If $y=3 \cos x-2 \sin x$, then prove that $\frac{d^2 y}{d x^2}-y=0$

Answer

$y =3 \cos x-2 \sin x$
$\Rightarrow \frac{d y}{d x} =3(-\sin x)-2(\cos x)$
$\Rightarrow \frac{d^2 y}{d x^2}=-3 \cos x-2(-\sin x)$
$\Rightarrow \frac{d^2 y}{d x^2}=-3 \cos x+2 \sin x$
$\Rightarrow \frac{d^2 y}{d x^2} =-(3 \cos x-2 \sin x)=-y$
$\Rightarrow \frac{d^2 y}{d x^2}+y=0 \text { }$
Hence proved.

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Question 33 Marks

Examine the continuity of function.$
f(x)=\left\{\begin{array}{ll}
1+x, & x \leq 3 \\
7-x, & x>3
\end{array} \text { at } x=3\right.
$

Answer

ATQ $\quad f(x)=\left\{\begin{array}{ll}1+x, & x \leq 3 \\ 7-x, & x>3\end{array}\right.$
at $x=3 \quad f(3)=1+3=4 ...(1)$$
\begin{aligned}
\text { R.H.L. } f(3+0) & =\lim _{h \rightarrow 0} f(3+h) \\
& =\lim _{h \rightarrow 0} 7-(3-h)=\lim _{h \rightarrow 0}(4-h)=4.....(2)
\end{aligned}
$
L.H.L. $f(3-0)=\lim _{h \rightarrow 0} f(3-h)$$
=\lim _{h \rightarrow 0}(1+3-h)=\lim _{h \rightarrow 0}(4-h)=4.....(3)
$
$
\because \quad f(3)=f(3+0)=f(3-0)=4
$
$\therefore$ function is continuous at $x=3$.

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Question 43 Marks

Differentiation w.r.t. $x$ of $\tan ^{-1}\left[\frac{\sin x+\cos x}{\cos x-\sin x}\right]$.

Answer

Suppose that $y=\tan ^{-1}\left[\frac{\sin x+\cos x}{\cos x-\sin x}\right]$$
\begin{aligned}
\Rightarrow \quad y & =\tan ^{-1}\left(\frac{\frac{\sin x+\cos x}{\cos x}}{\frac{\cos x-\sin x}{\cos x}}\right)=\tan ^{-1}\left(\frac{\tan x+1}{1-\tan x}\right) \\
\Rightarrow \quad y & =\tan ^{-1}\left(\frac{\tan \left(\frac{\pi}{4}\right)+\tan x}{1-\tan \left(\frac{\pi}{4}\right) \tan x}\right) \\
y & =\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+x\right)\right] \Rightarrow \frac{\pi}{4}+x \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{d}{d x}\left(\frac{\pi}{4}\right)+\frac{d}{d x}(x) \Rightarrow \frac{d y}{d x}=1 \quad \text {}
\end{aligned}
$

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Question 53 Marks

$f(x)=\left\{\begin{array}{c}5 x^2-4, \text { if } x \leq 1 \\ 4 x^2-3 x \text {, if } x>2\end{array}\right.$ Examine the continuity.

Answer

value of L.H.L. at $x=1$.$
\begin{aligned}
\lim _{x \rightarrow 1^{-}} f(x) & =\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}[5(1-h)-4] \\
& =\lim _{h \rightarrow 0}(5-5 h-4)=\lim _{h \rightarrow 0}(1-5 h)=1
\end{aligned}
$
value of R.H.L. at $x=1$$
\begin{aligned}
\lim _{x \rightarrow 1^{+}} f(x) & =\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}\left[4(1+h)^2-3(1+h)\right] \\
& =\lim _{h \rightarrow 0}\left(4+4 h^2+8 h-3-3 h\right) \\
& =\lim _{h \rightarrow 0}\left(4 h^2+5 h+1\right)=1
\end{aligned}
$value of function at $x=1$$
\begin{aligned}
& & f(x)=5 x-4 \\
\therefore & & f(1)=5 \times 1-4=5-4=1 \\
\because & & f(1)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} f(1+h)
\end{aligned}
$
Hence function is continuous at $x=1$ and except $x=$ 1 this function is also continuous due to universal continuous.

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Question 63 Marks

Show that function$
f(x)=\left\{\begin{array}{ll}
3-x, & \text { if } x<1 \\
2, & \text { if } x=1 \\
1+x, & \text { if } x>1
\end{array}\right.
$
is continuous at $x=1$.

Answer

value of $f(x)$ at $x=1$
$
f(1)=2
$
value of (LHL)
$
\begin{aligned}
\lim _{h \rightarrow 0} f(1-h) & =\lim _{h \rightarrow 0} 3-(1-h) \\
& =2

\end{aligned}
$
value of RHL
$
\begin{aligned}
\lim _{h \rightarrow 0} f(1+h) & =\lim _{h \rightarrow 0} 1+(1+h) \\
& =2
\end{aligned}
$
hence value of function $=$ L.H.L. $=$ R.H.L.
hence function is continuous at $x=1$

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Question 73 Marks

If function $f(x)=\left\{\begin{array}{cc}x^5 \sin \frac{1}{x}, & x \neq 0 \\ k & x=0\end{array}\right.$, is continuous at $x =0$, find the value of $k$.

Answer

Function at $x=0$
$
\begin{aligned}
f(x) & =k, \quad \text { when } x=0 \\
\therefore f(0) & =k
\end{aligned}
$
value of R.H.L.$
\begin{aligned}
\lim _{x \rightarrow 0^{+}} f(x) & =\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0}(0+h)^5 \sin \frac{1}{0+h} \\
& =\lim _{h \rightarrow 0} h^5 \sin \frac{1}{h}=\lim _{h \rightarrow 0} h^5 \times \lim _{h \rightarrow 0} \sin \frac{1}{h} \\
& =0 \times\{\text { any constant between }-1 \text { and } 1\} \\
& =0
\end{aligned}
$
is continuous at $x=0$
$
\begin{aligned}
\therefore & & f(0) & =\lim _{h \rightarrow 0} f(0+h) \\
\Rightarrow & & k & =0 \text {}
\end{aligned}
$

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Question 83 Marks

If $y= P e^{a x}+ Q e^{b x}$ then show that$
\frac{d^2 y}{d x^2}-(a+b) \frac{d y}{d x}+a b y=0
$

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Question 93 Marks

Differentiate w.r.t. $x$ of $y$$
=x+\frac{1}{x+\frac{1}{x+\frac{1}{x+\ldots \ldots \infty}}}
$

Answer

$\begin{array}{lr}\text { Taking } & y=x+\frac{1}{y} \\ \Rightarrow & y^2=x y+1 \\ \Rightarrow & y^2-x y=1\end{array}$
now differentiating w.r.t. $x$
$\Rightarrow \quad \frac{d}{d x}\left(y^2\right)-\frac{d}{d x}(x y)=\frac{d}{d x}(1)$
$\Rightarrow 2 y \frac{d y}{d x}-\left[x \cdot \frac{d y}{d x}+y \cdot 1\right]=0$
$\Rightarrow \quad 2 y \frac{d y}{d x}-\frac{x d y}{d x}-y=0$
$\Rightarrow \quad \frac{d y}{d x}(2 y-x)=y$
$\Rightarrow \quad \frac{d y}{d x}=\frac{y}{2 y-x}$

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Question 103 Marks

Find differentiation the following function w.r.t. $x$.$
\log (\sqrt{x-1}+\sqrt{x+1})
$

Answer

Suppose that
$
y=\log (\sqrt{x-1}+\sqrt{x+1})
$
now differentiate w.r.t. $x$$
\frac{d y}{d x}=\frac{1}{(\sqrt{x-1}+\sqrt{x+1})} \times \frac{d}{d x}(\sqrt{x-1}+\sqrt{x+1})
$
$
\begin{aligned}
\frac{d y}{d x} & =\frac{1}{(\sqrt{x-1}+\sqrt{x+1})} \times\left\{\frac{1}{2 \sqrt{x-1}}+\frac{1}{2 \sqrt{x+1}}\right\} \\
& =\frac{1}{(\sqrt{x-1}+\sqrt{x+1})} \times \frac{(\sqrt{x+1}+\sqrt{x-1})}{2 \sqrt{x-1} \times \sqrt{x+1}} \\
& =\frac{1}{2 \sqrt{x^2-1}}
\end{aligned}
$

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Question 113 Marks

If function $f(x)=\left\{\begin{array}{cc}\frac{x^2-2 x-3}{x+1}, & : x \neq 1 \\ \lambda & : x=-1\end{array}\right.$ is continuous at $x = - 1$ then find value of $\lambda$.

Answer

at $x=-1$
$
f(x)=\lambda \quad \therefore \quad f(-1)=\lambda
$
value of R.H.L. at $x=-1$$
\begin{aligned}
\lim _{h \rightarrow 0} f(-1+h) & =\lim _{h \rightarrow 0}\left[\frac{(-1+h)^2-2(-1+h)-3}{-1+h+1}\right] \\
& =\lim _{h \rightarrow 0}\left[\frac{1-2 h+h^2+2-2 h-3}{h}\right] \\
& =\lim _{h \rightarrow 0} \frac{h^2-4 h}{h} \\
& =\lim _{h \rightarrow 0}(h-4)=-4
\end{aligned}
$
$\therefore$ function is continuous at $x=-1$
$\therefore \quad f(-1)=\lim _{h \rightarrow 0} f(-1+h)$
$\Rightarrow \quad \lambda=-4 \quad \therefore \lambda=-4$

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Question 123 Marks

Prove that $f(x)=x-|x|, x \in R$ is continuous at $x = 0$.

Answer

Function is written as after removing modulus :
$f(x)=\{0, \text { if } x \geq 0 \because|x|=x,$ when $x \geq 02 x,$ if $ x<0 \because|x|=-x,$ when $ x<0$
value of $\text{L.H.L.}$
$\lim _{h \rightarrow 0} f(0-h) =\lim _{h \rightarrow 0} 2(0-h)=\lim _{h \rightarrow 0}-(2 h)$
$ =-2 \times 0=0$
value of $(\text{R.H.L.) } f(0+0)=0$
value of $f(0)=0$
$\because \lim _{h \rightarrow 0} f(0-h)$
$=\lim _{h \rightarrow 0} f(0+h)=f(0)$
hence function is continuous at $x=0$.

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