4 Marks Questions

Question 14 Marks

Find the first principle the derivative of $x^3-27$.

Answer

Let, $\quad f(x)=x^3-27$
$\Rightarrow \frac{d f}{d x}=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$\begin{array}{l}=\lim _{h \rightarrow 0} \frac{\left\{(x+h)^3-27\right\}-\left(x^3-27\right)}{h} \\ =\lim _{h \rightarrow 0} \frac{x^3+3 x^2 h+3 x h^2+h^3-x^3}{h} \\ =\lim _{h \rightarrow 0} \frac{h\left(3 x^2+3 x h+h^2\right)}{h}\end{array}$
$\begin{array}{l}=\lim _{h \rightarrow 0}\left(3 x^2+3 x h+h^2\right) \quad(h \neq 0) \\ =3 x^2+3 \cdot x \cdot(0)+(0)^2 \\ =3 x^2 .\end{array}$

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Question 24 Marks

If $x=a(\cos 2 \theta+2 \theta \sin 2 \theta)$ and $y=a(\sin 2 \theta-2 \theta$ $\cos 2 \theta)$, find $\frac{d^2 y}{d x^2}$ at $\theta=\frac{\pi}{8}$

Answer

$x=a(\cos 2 \theta+2 \theta \sin 2 \theta)$
$\Rightarrow \quad \frac{d x}{d \theta}=a(-2 \sin 2 \theta+2 \sin 2 \theta$ $+\ 4 \theta \cos 2 \theta)$
$\Rightarrow \quad \frac{d x}{d \theta}=a(4 \theta \cos 2 \theta)\quad \ldots(1)$
$y=a(\sin 2 \theta-2 \theta \cos 2 \theta)$
$\Rightarrow \quad \frac{d y}{d \theta}=a(2 \cos 2 \theta+4 \theta \sin 2 \theta$ $-\ 2 \cos 2 \theta)$
$\Rightarrow \quad \frac{d y}{d \theta}=a(4 \theta \sin 2 \theta)\quad \ldots(2)$
Using (1) and (2),
$
\begin{array}{ll}
\Rightarrow \frac{d y}{d x}=\frac{a(4 \theta \sin 2 \theta)}{a(4 \theta \cos 2 \theta)} \\
\Rightarrow \frac{d y}{d x}=\frac{\sin 2 \theta}{\cos 2 \theta}=\tan 2 \theta
\end{array}
$
Differentiating again with respect to $x$, we get
$\Rightarrow \frac{d^2 y}{d x^2}=2 \sec ^2 2 \theta \cdot \frac{d \theta}{d x}$
$\Rightarrow \frac{d^2 y}{d x^2}=2 \sec ^2 2 \theta \cdot \frac{1}{a(4 \theta \cos 2 \theta)}$
$\left[\frac{d^2 y}{d x^2}\right]_{\theta=\frac{\pi}{8}}=2 \sec ^2 \frac{\pi}{4} \cdot \frac{1}{a\left(4 \frac{\pi}{8} \cos \frac{\pi}{4}\right)}$
$=\frac{8 \sqrt{2}}{\pi a}$.

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Question 34 Marks

If $y=a e^{2 x}+b e^{-x}$, then show that $\frac{d^2 y}{d x^2}-\frac{d y}{d x}-2 y=0$.

Answer

$y=a e^{2 x}+b e^{-x}\quad \ldots(1)$
$\frac{d y}{d x}=2 a e^{2 x}-b e^{-x}\quad \ldots(2)$
$\frac{d^2 y}{d x^2}=4 a e^{2 x}+b e^{-x}\quad \ldots(3)$
Putting the values on LHS
$
\begin{array}{l}
=\frac{d^2 y}{d x^2}-\frac{d y}{d x}-2 y \\
=\left(4 a e^{2 x}+b e^{-x}\right)-\left(2 a e^{2 x}-b e^{-x}\right)-2\left(a e^{2 x}+b e^{-x}\right) \\
=4 a e^{2 x}+b e^{-x}-2 a e^{2 x}+b e^{-x}-2 a e^{2 x}-2 b e^{-x} \\
=0
\end{array}
$

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Question 44 Marks

If $y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}$, prove that $(2 x y) \frac{d y}{d x}=\frac{x}{a}-\frac{a}{x}$.

Answer

Given, $y=\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}$
$\begin{array}{ll}\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}\right) \\ \Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{a}} \frac{d}{d x} \sqrt{x}+\sqrt{a} \frac{d}{d x} \frac{1}{\sqrt{x}} \\ \Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{a}} \cdot \frac{1}{2 \sqrt{x}}-\sqrt{a}\left(\frac{1}{2 x \sqrt{x}}\right)\end{array}$
$=\frac{1}{2 \sqrt{a x}}-\frac{\sqrt{a}}{2 x \sqrt{x}}$
$\therefore \quad$ L.H.S. $=(2 x y) \frac{d y}{d x}$
$\begin{array}{l}=2 x\left(\sqrt{\frac{x}{a}}+\sqrt{\frac{a}{x}}\right)\left(\frac{1}{2 \sqrt{a x}}-\frac{1}{2 x} \sqrt{\frac{a}{x}}\right) \\ =2 x\left(\frac{1}{2 a}-\frac{1}{2 x}+\frac{1}{2 x}-\frac{a}{2 x^2}\right) \\ =2 x\left(\frac{1}{2 a}-\frac{a}{2 x^2}\right) \\ =\frac{x}{a}-\frac{a}{x} \\ =\text { R.H.S. }\end{array}$

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Question 54 Marks

If $x \sqrt{1+y}+y \sqrt{1+x}=0$ and $x \neq y$ prove that $\frac{d y}{d x}=-\frac{1}{(x+1)^2}$.

Answer

Given, equation is, $x \sqrt{1+y}+y \sqrt{1+x}=0$
On squaring both sides, we get
$x^2(1+y)=y^2(1+x)$
$\begin{array}{lrl}\Rightarrow x^2+x^2 y =y^2+y^2 x \\ \Rightarrow x^2-y^2 =y^2 x-x^2 y \\ \Rightarrow (x+y)(x-y) =-x y(x-y)\end{array}$
$\begin{array}{lrl}\Rightarrow (x-y)(x+y)+x y(x-y) =0 \\ \Rightarrow (x-y)(x+y+x y) =0 \\ \therefore \text { Either } x-y =0 \text { or } x+y+x y=0\end{array}$
Now, $x-y=0 \text { or } x=y$
But it is given that $x \neq y$
So, it is contradiction
$\begin{array}{lr}\therefore x-y =0 \text { is rejected. } \\ \text { Now, consider } y+x y+x =0\end{array}$
or $y(1+x)=-x$
or $y=\frac{-x}{1+x}\quad \ldots(i)$
On differentiating both sides w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{(1+x) \frac{d}{d x}(-x)-(-x) \frac{d}{d x}(1+x)}{(1+x)^2}$
$\begin{array}{ll}\Rightarrow & \frac{d y}{d x}=\frac{(1+x)(-1)+x \times 1}{(1+x)^2} \\ \Rightarrow & \frac{d y}{d x}=\frac{-1-x+x}{(1+x)^2} \\ \Rightarrow & \frac{d y}{d x}=\frac{-1}{(1+x)^2} .\end{array}$
Hence Proved.

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Question 64 Marks

Find the value of $k, f$ or which
$
f(x)=\left\{\begin{array}{ccc}
\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}, & \text { if } & -1 \leq x<0 \\
\frac{2 x+1}{x-1}, & \text { if } & 0 \leq x<1
\end{array}\right.
$
is continuous at $x=0$.

Answer

Given,
$
f(x)=\left\{\begin{array}{ccc}
\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}, & \text { if } & -1 \leq x<0 \\
\frac{2 x+1}{x-1}, & \text { if } & 0 \leq x<1
\end{array}\right.
$
Since $f(x)$ is continuous at $x=0$,
i.e., $\quad L \cdot H \cdot L =f(0)= R \cdot H \cdot L$
or $\quad$ L.H.L $=f(0)$
or $\quad \lim _{x \rightarrow 0^{+}} f(x)=\frac{2(0)+1}{0-1}=\frac{1}{-1}$
or $\quad \lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}=\frac{1}{-1}$
or $\quad\lim _{x \rightarrow 0} \frac{(\sqrt{1+k x})^2-(\sqrt{1-k x})^2}{x(\sqrt{1+k x}+\sqrt{1-k x})}=-1$
or $\quad \lim _{x \rightarrow 0} \frac{1+k x-1+k x}{x(\sqrt{1+k x}+\sqrt{1-k x})}=-1$
or $\quad \lim _{x \rightarrow 0} \frac{2 k x}{x(\sqrt{1+k x}+\sqrt{1-k x})}=-1$
or $\quad \frac{2 k}{\sqrt{1+0}+\sqrt{1-0}}=-1$
or $\quad \frac{2 k}{1+1}=-1$
or $\quad \frac{2 k}{2}=-1$
$
\therefore \quad k=-1
$
Thus, $f(x)$ is continuous at $x=0$, if $k=-1$.

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Question 74 Marks

$\lim _{x \rightarrow 0}\left(\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right)=?$

Answer

$\lim _{x \rightarrow 0}\left(\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right)$
$=\lim _{x \rightarrow 0}\left\{\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}} \times \frac{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}\right.$$\left.\times \frac{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}\right\}$
$\begin{array}{l}=\lim _{x \rightarrow 0}\left\{\frac{\left(1+x^2\right)-(1+x)}{\left(1+x^3\right)-(1+x)} \times \frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\right\} \\ =\lim _{x \rightarrow 0}\left\{\frac{x(x-1)}{\left(x^3-x\right)} \times \frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\right\}\end{array}$
$\begin{array}{l}=\lim _{x \rightarrow 0} \frac{x(x-1)\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{x(x-1)(x+1)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)} \quad(x \neq 0) \\ =\lim _{x \rightarrow 0} \frac{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{(x+1)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}\end{array}$
$\begin{array}{l}=\frac{\left(\sqrt{1+0^3}+\sqrt{1+0}\right)}{(0+1)\left(\sqrt{1+0^2}+\sqrt{1+0}\right)} \\ =\frac{2}{1 \times 2} \\ =1\end{array}$

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Question 84 Marks

Find the value of ' $k$ ' if $\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}$.

Answer

Given, $\quad \lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}$
$
\Rightarrow \lim _{x \rightarrow 1} \frac{\left(x^2-1\right)\left(x^2+1\right)}{(x-1)}=\lim _{x \rightarrow k} \frac{(x-k)\left(x^2+x k+k^2\right)}{(x-k)(x+k)}
$
$\begin{array}{r}\Rightarrow \lim _{x \rightarrow 1} \frac{(x-1)\left(x+1\left(x^2+1\right)\right.}{(x-1)}=\lim _{x \rightarrow k} \frac{\left(x^2+x k+k^2\right)}{x+k} \quad(x \neq k)\end{array}$
$\Rightarrow \quad \lim _{x \rightarrow 1}(x+1)\left(x^2+1\right)=\frac{k^2+k^2+k^2}{k+k} \quad(x \neq 1)$
$\Rightarrow \quad(1+1)\left(1^2+1\right)=\frac{3 k^2}{2 k}$
$\Rightarrow \quad 4=\frac{3 k}{2}$
$\Rightarrow \quad k=\frac{8}{3}$

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Applied Maths 4 Marks Questions

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