4 Marks

Question 14 Marks

Show that the differrential equation $2 y e^{\frac{x}{y}} d x+\left(y-2 x e^{\frac{x}{y}}\right) d y=0$ is homogeneous and find its particular solution, given that $x=0$ when $y=1$.

Answer

The given differential equation can be written as:
$\frac{d x}{d y}=\frac{2 x(e)^{\frac{x}{y}}-y}{2 y(e)^{\frac{x}{y}}} \ldots(1)$
Let,
$\begin{aligned} F (x, y) & =\frac{2 x(e)^{\frac{x}{y}}-y}{2 y(e)^{\frac{x}{y}}} \\ \therefore \quad F(\lambda x, \lambda y) & =\frac{\left(2 x(e)^{\frac{x}{y}}-y\right)}{\left(2 y(e)^{\frac{x}{y}}\right)} \\ & =\lambda^0 F(x, y)\end{aligned}$
Thus, $F (x, y)$ is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.
To solve it, we make the substitution
$x=v y \ldots(2)$
Differentiating equation (2) with respect to $y$, we get,
$\frac{d x}{d y}=v+y \frac{d v}{d y}$
Substituting the value of $x$ and $\frac{d x}{d y}$ in equation (1),
we get,
$\begin{aligned} v+y \frac{d v}{d y} & =\frac{2 v e^v-1}{2 e^v} \\ \therefore \quad y \frac{d v}{d y} & =\frac{2 v e^v-1}{2 e^v}-v\end{aligned}$
$\begin{array}{ll}\therefore & y \frac{d v}{d y}=-\frac{1}{2 e^v} \\ \therefore & 2 e^v d v=-\frac{d y}{y}\end{array}$
$\begin{array}{rlrl} \therefore \int 2 e^v d v =-\int \frac{d y}{y} \\ \therefore 2 e^v =-\log |y|+c\end{array}$
and replacing $v$ by $\frac{x}{y}$, we get,
$2(e)^{\frac{x}{y}}+\log |y|+c \ldots(3)$
Substituting $x=0$ and $y=1$ in equation (3), we get
$2 e^0+\log |1|=c \Rightarrow c=2$
Substituting the value of $c$ in equation (3), we get
$2(e)^{\frac{x}{y}}+\log |y|=2$
which is the particular solution of the given differential equation.

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Question 24 Marks

By using the properties of definite integral evaluate:
$\int_0^{\frac{x}{4}} \log (1+\tan x) d x.$

Answer

$I=\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x \ldots(1)$
By property (6), $x=\frac{\pi}{4}-x$
$\begin{aligned}
I & =\int_0^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \\
& =\int_0^{\frac{\pi}{4}} \log \left[1+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \cdot \tan x}\right] d x
\end{aligned}$
$=\int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x$
$\begin{aligned} I & =\int_0^{\frac{\pi}{4}} \log \left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right) d x \\ & =\int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) d x \\ & =\int_0^{\frac{\pi}{4}}(\log (2)-\log (1+\tan x)) d x\end{aligned}$
$\begin{array}{l} I =\log 2 \int_0^{\frac{\pi}{4}} 1 d x-\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x \\ I=\log 2[x]_0^{\frac{\pi}{4}}- I \quad(\because \text { From equation (1)) } \\ 2 I=\log 2\left(\frac{\pi}{4}-0\right) \\ \therefore I =\frac{\pi}{8} \log 2\end{array}$

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Question 34 Marks

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan ^{-1}(\sqrt{2})$

Answer

Suppose, radius of cone is $r$ height is $h$ and slant height is $l$.
$\therefore l^2=h^2+r^2 \ldots(1)$
Suppose, semi-vertical angle is $\alpha$.
$\begin{aligned}
\therefore \tan \alpha & =\frac{r}{h} \\
\therefore \quad \alpha & =\tan ^{-1}\left(\frac{r}{h}\right)
\end{aligned}$
$\begin{aligned} \rightarrow \text { Volume of cone } V & =\frac{1}{3} \pi r^2 h \\ & =\frac{1}{3} \pi\left(l^2-h^2\right) h\end{aligned}$
( $\because$ From equation (1))
$\therefore$ $f(h)=\frac{\pi}{3}\left(l^2 h-h^3\right)$
$f^{\prime}(h)=\frac{\pi}{3}\left(l^2-3 h^2\right)$
$\therefore$ $f^{\prime \prime}(h)=\frac{\pi}{3}(-6 h)$
$\therefore$ $f^{\prime \prime}(h)=-2 \pi h<0$
$\therefore f$ has minimum value
$\rightarrow$ For finding maximum volume of cone,
$f^{\prime}(h)=0$
$\therefore \quad \frac{\pi}{3}\left(l^2-3 h^2\right)=0$
$\begin{aligned} & \therefore l^2-3 h^2 =0 \\ & \therefore h^2+r^2-3 h^2=0 \\ & \therefore r^2-2 h^2 =0 \\ & \therefore r^2 =2 h^2 \\ & \therefore r =\sqrt{ } 2 h \\ & \therefore \frac{r}{h} =\sqrt{2}\end{aligned}$
$\therefore$ Semi-vertical angle $=\tan ^{-1}\left(\frac{r}{h}\right)$
$=\tan ^{-1}(\sqrt{2})$

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Question 44 Marks

If $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$, find $\left(\frac{d^2 y}{d x^2}\right)_{t=\frac{x}{4}}$.

Answer

$\begin{array}{l}
x=a(\cos t+t \sin t) \\
\therefore \frac{d x}{d t}=a(-\sin t+t \cos t+\sin t) \\
\therefore \frac{d x}{d t}=a t \cos t
\end{array}$
Now, $y=a(\sin t-t \cos t)$
$\begin{array}{l}
\therefore \frac{d y}{d t}=a[\cos t+t \sin t-\cos t] \\
\therefore \frac{d y}{d t}=a t \sin t \\
\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{t}}=\frac{a t \sin t}{a t \cos t}=t a n t\end{array}$
Now, differentiate again w.r.t. $x$,
$\begin{array}{l}
\frac{d^2 y}{d x^2}=\frac{d}{d t}\left(\frac{d y}{d x}\right) \cdot \frac{d t}{d x}=\frac{d}{d t}(\tan t) \cdot \frac{d t}{d x} \\
\frac{d^2 y}{d x^2}=\sec ^2 t \frac{d t}{d x}=\frac{\sec ^2 t}{a t \cos t} \\
\frac{d^2 y}{d x^2}=\frac{\sec ^3 t}{a t} ; 0 < t < \frac{\pi}{2}
\end{array}$

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Question 54 Marks

Solve system of linear equations, using matrix method :
$\begin{array}{l}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=6, \quad \frac{1}{y}+\frac{3}{z}=11, \\
\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=0
\end{array}$

Answer

$\begin{array}{l}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=6 \\ \frac{1}{y}+\frac{3}{z}=11 \\ \frac{1}{x}-\frac{2}{y}+\frac{1}{z}=0\end{array}$
$\Rightarrow$ The equation can be written as matrix form,
$\begin{array}{l}
{\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 & 3 \\
1 & -2 & 1
\end{array}\right]\left[\begin{array}{c}
\frac{1}{x} \\
\frac{1}{y} \\
\frac{1}{z}
\end{array}\right]=\left[\begin{array}{c}
6 \\
11 \\
0
\end{array}\right]} \\
\therefore A=\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 & 3 \\
1 & -2 & 1
\end{array}\right], X=\left[\begin{array}{c}
\frac{1}{x} \\
\frac{1}{y} \\
\frac{1}{z}
\end{array}\right], B=\left[\begin{array}{c}
6 \\
11 \\
0
\end{array}\right]
\end{array}$
$\begin{array}{l}
|A|=\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 & 3 \\
1 & -2 & 1
\end{array}\right] \\
=1(1+6)-1(0-3)+1(0-1) \\
=7+3-1 \\
=9 \neq 0 \\
\therefore \quad A^{-1} \text { exists. }
\end{array}$
$\operatorname{adj} A=\left[\begin{array}{rrr}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$
$\therefore \quad A^{-1}=\frac{1}{|A|}$ adj A
$=\frac{1}{9}\left[\begin{array}{rrr}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$
$\therefore$ $AX = B$
$\therefore$ $A ^{-1} AX = A ^{-1} B$
$\therefore$ $IX = A ^{-1} B$
$\therefore$ $X = A ^{-1} B$
$\begin{array}{l}=\frac{1}{9}\left[\begin{array}{rrr}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]\left[\begin{array}{c}6 \\ 11 \\ 0\end{array}\right] \\ =\frac{1}{9}\left[\begin{array}{rrr}42 & -33 & +0 \\ 18 & +0 & -0 \\ -6 & +33+0\end{array}\right] \\ =\frac{1}{9}\left[\begin{array}{c}9 \\ 18 \\ 27\end{array}\right]\end{array}$
$\left[\begin{array}{c}\frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z}\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
$\begin{array}{l}\therefore \quad \text { Solution : } \frac{1}{x}=1, \frac{1}{y}=2, \frac{1}{z}=3 \\ \quad \therefore x=1, y=\frac{1}{z}, z=\frac{1}{3}\end{array}$

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Question 64 Marks

Find $x$, if $[x-5-1]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=O$.

Answer

$\left[\begin{array}{lll}x & -5 & -1\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]= O$
$\therefore \quad\left[\begin{array}{lll}x+0-2 & 0-10+0 & 2 x-5-3\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=O$
$\therefore\left[\begin{array}{lll}x-2 & -10 & 2 x-8\end{array}\right]\left[\begin{array}{l}x \\ 4 \\ 1\end{array}\right]=O$
$\begin{array}{l}\therefore \quad[x(x-2)+(-10)(4)+(2 x-8)(1)]=O \\ \therefore \quad\left[x^2-2 x-40+2 x-8\right]=O \\ \therefore \quad\left[x^2-48\right]=[0] \\ \therefore \quad x^2-48=0 \\ \therefore \quad x^2=48 \\ \therefore \quad x= \pm \sqrt{48}= \pm 4 \sqrt{3}\end{array}$

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Maths 4 Marks

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