4 Marks

Question 14 Marks

In a culture, the bacteria count is $1,00,000$. The number is increased by $10 \%$ in 2 hours. In how many hours will the count reach $2,00,000$, if the rate of growth of bacteria is proportional to the number present ?

Answer

Suppose, Number of Bacteria at time $t$ is $p$.
Here, $\frac{d p}{d t} \propto p$
$\therefore \frac{d p}{d t}=k p(\text { where }, k>0)$
[If function is decreasing then Take $k<0$ ]
$\therefore \frac{d p}{p}=k d t$
$\rightarrow$ Integrate both the sides,
$\therefore \int \frac{d p}{p}=k \int 1 d t$
$\begin{array}{l}\therefore \log |p|=k t+c \ldots(1) \\ \rightarrow \text { Now, initially } t=0 \text { when } p=1,00,000 \\ \therefore \log |1,00,000|=0+c \\ \therefore c=\log |1,00,000|\end{array}$
$\rightarrow$ Put the value of $c$ in equation (1),
$\begin{array}{l}
\therefore \log |p|=k t+\log |1,00,000| \\
\therefore \log |p|-\log |1,00,000|=k t \\
\therefore \log \left|\frac{p}{1,00,000}\right|=k t \ldots(2)
\end{array}$
In 2 hr , Number of bacteria increases at the rate of $10 \%$.
$\begin{array}{l}
\rightarrow t=2 hr \Rightarrow p=1,00,000+(10 \% \text { of } 1,00,000) \\
\therefore p=1,00,000+1,00,000\left(\frac{10}{100}\right) \\
\therefore p=1,00,000+10,000 \\
\therefore p=1,10,000
\end{array}$
$\begin{array}{l}
\therefore \log \left|\frac{1,10,000}{1,00,000}\right|=2 k \\
\therefore 2 k=\log \left(\frac{11}{10}\right) \\
\therefore k=\frac{1}{2} \log \left(\frac{11}{10}\right)
\end{array}$
$\rightarrow$ Put the value of $k$ in equation (2),
$\therefore \log \left|\frac{p}{1,00,000}\right|=\frac{1}{2} \log \left(\frac{11}{10}\right) t$
$\rightarrow$ Now, $p=2,00,000$ then $t=$ ?
$\begin{array}{l}
\therefore \log \left|\frac{2,00,000}{1,00,000}\right|=\frac{1}{2} \log \left(\frac{11}{10}\right) t \\
\therefore \log 2=\frac{1}{2} \log \left(\frac{11}{10}\right) t \\
\therefore 2 \log 2=\log \left(\frac{11}{10}\right) t \\
\therefore t=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)} \\
\therefore t=\frac{\log 4}{\log \left(\frac{11}{10}\right)} hr
\end{array}$
In $\frac{\log 4}{\log \left(\frac{11}{10}\right)} hr$ will the count reach $2,00,000$ bacteria.

View full question & answer→

Question 24 Marks

Evulate :
$\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x.$

Answer

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

View full question & answer→

Question 34 Marks

Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Answer

Rectangle $\square ABCD$ is insecribed in a $r$ radius circle.
Length of rectangle, $AB = CD =x(x \neq 0, x>0)$
Breadth, $BC = AD =y$
Now, $(2 r)^2=x^2+y^2$
$\therefore \quad 4 r^2=x^2+y^2 \ldots(1)$
Now, Area of rectangle, $A=$ Length $\times$ Breadth
$\begin{array}{l}
\therefore A=x y \\
\therefore A=x\left(\sqrt{4 r^2-x^2}\right) \quad(\because \text { From equation (1)) } \\
\rightarrow \text { Take, } f(x)=x^2\left(4 r^2-x^2\right) \\
\quad f(x)=4 r^2 x^2-x^4 \\
\therefore f^{\prime}(x)=8 r^2 x-4 x^3 \\
\therefore f^{\prime \prime}(x)=8 r^2-12 x^2
\end{array}$
$\rightarrow$ Now, for finding maximum area
$\begin{aligned}
f^{\prime}(x) & =0 \\
\therefore \quad 8 r^2 x-4 x^3 & =0 \\
\therefore \quad 4 x\left(2 r^2-x^2\right) & =0 \\
x \neq 0,2 r^2-x^2 & =0
\end{aligned}$
$\therefore \quad x^2=2 r^2 \Rightarrow x=\sqrt{2} r \ldots(2)$
Now, $f^{\prime \prime}(\sqrt{2} r)=8 r^2-12 x^2$
$\begin{array}{l}=8 r^2-12\left(2 r^2\right) \\ =8 r^2-24 r^2 \\ =-16 r^2<0\end{array}$
$\therefore f$ has maximum value.
$\rightarrow$ From equation (1),
$\begin{aligned}
& 4 r^2=x^2+y^2 \\
\therefore & 4 r^2=2 r^2+y^2 \\
\therefore & y^2=2 r^2 \\
\therefore & y^2=x^2(\because \text { From equation (2)) } \\
\therefore & x=y \\
\therefore & \text { Rectangle is square. }
\end{aligned}$

View full question & answer→

Question 44 Marks

If $y=e^{a \cos ^{-1} x}$, show that $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-a^2 y=0$, where $-1 \leq x \leq 1$

Answer

$y=e^{a \cos ^{-1} x}$
Differentiate w.r.t. $x$,
$\begin{aligned}
& \frac{d y}{d x}=e^{a \cos ^{-1} x} a \cdot\left(\frac{-1}{\sqrt{1-x^2}}\right) \\
\therefore & \sqrt{1-x^2} \frac{d y}{d x}=-a e^{a \cos ^{-1} x}
\end{aligned}$
Squaring both the sides,
$\begin{aligned}
& \left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=a^2\left[e^{a \cos ^{-1} x}\right]^2 \\
\therefore & \left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=a^2 y^2
\end{aligned}$
Differentiate again w.r.t. $x$,
$\therefore\left(1-x^2\right) 2 \frac{d y}{d x} \frac{d^2 y}{d x}+\left(\frac{d y}{d x}\right)^2(-2 x)=a^2 \cdot 2 y \frac{d y}{d x}$
Now, each terms divide by $2 \frac{d y}{d x} \neq 0$
$\begin{array}{l}
\therefore\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=a^2 y \\
\therefore\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-a^2 y=0
\end{array}$

View full question & answer→

Question 54 Marks

Solve system of linear equation, using matrix method :
$\begin{aligned}
x+y+z & =6 \\
y+3 z & =11 \\
x+z & =2 y
\end{aligned}$

Answer

Let first, second and third numbers be denoted by $x, y$ and $z$, respectively.
Then, according to given conditions, we have
$\begin{array}{l}
x+y+z=6 \\
y+3 z=11 \\
x+z=2 y \text { or } x-2 y+z=0
\end{array}$
This system can be written as $AX = B$, where,
$A =\left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1\end{array}\right], X =\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B =\left[\begin{array}{c}6 \\ 11 \\ 0\end{array}\right]$
Hence, $| A |=1(1+6)-1(0-3)+1(0-1)=9 \neq 0$.
Now we find adj A.
$\begin{array}{l}
A_{11}=1(1+6)=7, \\
A_{12}=-(0-3)=3, \\
A_{13}=-1 \\
A_{21}=-(1+2)=-3,
\end{array}$
$\begin{array}{l}A_{22}=0 \\ A_{23}=-(-2-1)=3 \\ A_{31}=(3-1)=2, A_{32}=-(3-0)=-3, \\ A_{33}=(1-0)=1\end{array}$
Hence, adj $A =\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$
Thus, $\quad A ^{-1}=\frac{1}{|A|} \operatorname{adj}( A )$
$=\frac{1}{9}\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]$
Since $\quad X=A^{-1} B$
$X =\frac{1}{9}\left[\begin{array}{ccc}7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1\end{array}\right]\left[\begin{array}{c}6 \\ 11 \\ 0\end{array}\right]$
or $\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{9}\left[\begin{array}{c}42-33+0 \\ 18+0+0 \\ -6+33+0\end{array}\right]$
$=\frac{1}{9}\left[\begin{array}{c}9 \\ 18 \\ 27\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
Thus, $\quad x=1, y=2, z=3$.

View full question & answer→

Question 64 Marks

If $A =\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$, then, prove that $A^3-6 A^2+7 A+2 I=O$ and hence find $A^{-1}$.

Answer

$\begin{aligned} A ^2 & = A \cdot A \\ & =\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right] \\ & =\left[\begin{array}{lll}1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9\end{array}\right]=\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]\end{aligned}$
$\begin{aligned} A ^3 & = A ^2 \cdot A \\ & =\left[\begin{array}{lll}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right] \\ & =\left[\begin{array}{ccc}5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39\end{array}\right] \\ & =\left[\begin{array}{lll}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]\end{aligned}$
$\text { Now, L.H.S. }=A^3-6 A^2+7 A+2 I$
$\begin{array}{l}
=\left[\begin{array}{lll}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right]-6\left[\begin{array}{lll}
5 & 0 & 8 \\
2 & 4 & 5 \\
8 & 0 & 13
\end{array}\right]+7\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]+2\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
=\left[\begin{array}{lll}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right]+\left[\begin{array}{ccc}
-30 & 0 & -48 \\
-12 & -24 & -30 \\
-48 & 0 & -78
\end{array}\right]+\left[\begin{array}{ccc}
7 & 0 & 14 \\
0 & 14 & 7 \\
14 & 0 & 21
\end{array}\right]+\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right]
\end{array}$
$\begin{array}{l}=\left[\begin{array}{ccc}21-30+7+2 & 0+0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0+0+0+0 & 55-78+21+2\end{array}\right] \\ =\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]= O =\text { R.H.S. }\end{array}$
$A^3-6 A^2+7 A+2 I=0$
Multiplying both sides by $A ^{-1}$,
$\begin{array}{l}
\therefore A^{-1}\left(A^3-6 A^2+7 A+2 I\right)=0 \cdot A^{-1} \\
\therefore A^2-6 A+7 I+2 A^{-1}=0 \\
\therefore 2 A^{-1}=6 A-A^2-7 I
\end{array}$
$\begin{array}{l}\therefore 2 A^{-1}=6\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]-\left[\begin{array}{lll}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]-7\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \\ \therefore 2 A^{-1}=\left[\begin{array}{rrr}6 & 0 & 12 \\ 0 & 12 & 6 \\ 12 & 0 & 18\end{array}\right]+\left[\begin{array}{rrr}-5 & 0 & -8 \\ -2 & -4 & -5 \\ -8 & 0 & -13\end{array}\right]+\left[\begin{array}{rrr}-7 & 0 & 0 \\ 0 & -7 & 0 \\ 0 & 0 & -7\end{array}\right]\end{array}$
$=\left[\begin{array}{rrr}-6 & 0 & 4 \\ -2 & 1 & 1 \\ 4 & 0 & -2\end{array}\right]$
$\therefore \quad A^{-1}=\left[\begin{array}{rrr}-3 & 0 & 2 \\ -1 & \frac{1}{2} & \frac{1}{2} \\ 2 & 0 & -1\end{array}\right]$

View full question & answer→

Maths 4 Marks

Take a timed test