3 Marks

Question 13 Marks

Find the maximum value of $\sin \theta+\cos \theta$.

Answer

Given $\quad f(\theta)=\sin \theta+\cos \theta, \theta \in[0, \pi]$
$\therefore \quad f^{\prime}(\theta)=\cos \theta-\sin \theta$
$f^{\prime}(\theta)=0$
$\Rightarrow \quad \sin \theta-\cos \theta=0 \quad \Rightarrow \tan \theta=1$
$\therefore \quad \theta=\frac{\pi}{4}$
$\begin{array}{l}
\text { hence } f(0)=\sin 0+\cos 0=1 \\
f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2} \\
f(\pi)=\sin \pi+\cos \pi=0-1=-1 \\
\therefore \text { Absolute maximum value }=\sqrt{2}
\end{array}$

View full question & answer→

Question 23 Marks

A particle is moving in straight line such that on time $t$ distance ( $S$ ) from a fixed point is proportional to $n$ power of time. If at time $t$ its velocity $( V )$ and a acceleration then prove that :$
V^2=\frac{n a S}{(n-1)}
$

Answer

Given
$
\begin{array}{ll}
& \text { distance }(S) \propto(\text { time })^n \\
\therefore & S \propto t^n \\
\Rightarrow & S=k t^n \\
\Rightarrow & V=\frac{d S}{d t}=n k t^{n-1}
\end{array}
$
and acceleration (a) $=\frac{d V}{d t}=n(n-1) k \cdot t^{n-2}$
now, $\quad a \cdot S=n(n-1) k t^{n-2} \cdot$
In equation (2) multiplying by $\left(\frac{n}{n-1}\right)$$
\begin{array}{ll}
\Rightarrow & \frac{n}{(n-1)} a \cdot S=n^2 k^2 t^{2 n-2} \\
\Rightarrow & \frac{n \cdot a \cdot S}{(n-1)}=\left(n k t^{n-1}\right)^2
\end{array}
$
Put the value from (1)
$
\frac{n \cdot a \cdot S}{(n-1)}=V^2
$
hence$\quad$$
V^2=\frac{n a S}{n-1}
$
Hence proved.

View full question & answer→

Question 33 Marks

Find the interval for which function $f(x)=x^3-$ $3 x^2-24 x+5$ is increasing.

Answer

Given $\quad$$f(x)=x^3-3 x^2-24 x+5$
or$\quad$$y=x^3-3 x^2-24 x+5$
$\Rightarrow \quad \frac{d y}{d x}=3 x^2-6 x-24$
$\Rightarrow \quad \frac{d y}{d x}=3\left(x^2-2 x-8\right)$$
\begin{array}{ll}
\Rightarrow & \frac{d y}{d x}=3(x-4)(x+2) \\
\text { For increasing } & \frac{d y}{d x}>0
\end{array}
$
So, $\quad 3(x-4)(x+2)>0$$
\Rightarrow \quad(x-4)(x+2)>0 \quad[\because 3>0]
$
Case I, $x-4>0,(x+2)>0 \Rightarrow x>4, x>-2$
Case II, $x-4<0,(x+2)<0 \Rightarrow x<4, x<-2$$
x \in(-\infty,-2) \cup(4, \infty)
$

View full question & answer→

Question 43 Marks

In interval $[ 1 , 5]$ Find the absolute maximum and absolute minimum values of given function $f(x)=x^2-4 x+8$.

Answer

We know :$
\begin{aligned}
f(x) & =x^2-4 x+8 \\
\text { or } \quad f^{\prime}(x) & =2 x-4=2(x-2)
\end{aligned}
$
Put $f^{\prime}(x)=0$ we get $x=2$
We find the values of $f$ at $x=1, x=2$, and $x=5$.
now$
\begin{aligned}
\therefore f(x) & =x^2-4 x+8 \\
\therefore f(1) & =(1)^2-4 \times 1+8=5 \\
f(2) & =(2)^2-4 \times 2+8=4 \\
f(5) & =(5)^2-4 \times 5+8=13
\end{aligned}
$
Thus we find that in $[1,5]$ function at $x=5$
absolute maximum value is 13 and at $x=2$
absolute minimum value is 4 .

View full question & answer→

Question 53 Marks

Radius of a sphere measure 9 cm in which error is 0.02 cm . Find the approximate error in calculation of volume.

Answer

Suppose that radius of sphere is $r$ and the error measurement in it is $\Delta r$.
Here
$r=9 cm$ and $\Delta r=0.02 cm$
Volume of sphere $v=\frac{4}{3} \pi r^3$
or
$\begin{aligned}
\frac{d v}{d r} & =\frac{4}{3} \pi \times 3 r^2=4 \pi r^2 \\
d v & =\left(\frac{d v}{d r}\right) \Delta r \\
& =\left(4 \pi r^2\right) \Delta r
\end{aligned}$
Put the values
$\begin{array}{l}
=\left(4 \pi(9)^2\right) \times 0.02 \\
=4 \times 81 \times 0.02 \pi \\
=6.48 \pi cm^3
\end{array}$
hence the approximate error in calculation of volume $=6.48 \pi cm^3$

View full question & answer→

Question 63 Marks

Find the interval for which function $f(x)=2 \log (x-2)-x^2+4 x+1$ is increasing.

Answer

Here $f(x)=2 \log (x-2)-x^2+4 x+1$
$\Rightarrow \quad y=2 \log (x-2)-x^2+4 x+1$
$
\begin{array}{l}

\begin{aligned}
\Rightarrow \quad \frac{d y}{d x} & =2 \frac{2}{(x-2)}-2 x+4 \\
& =2\left[\frac{1}{x-2}-x+2\right]=2\left[\frac{1}{(x-2)}-(x-2)\right] \\
\Rightarrow \quad \frac{d y}{d x} & =2\left[\frac{1-(x-2)^2}{(x-2)}\right]=\frac{2(x-1)(3-x)}{(x-2)}
\end{aligned}
\end{array}
$because given function $f(x)$ is increasing
\[\begin{array}{l}\text { So, } \quad \frac{d y}{d x} > 0 \\\Rightarrow \frac{2(x-1)(3-x)}{(x-2)} > 0 \\\Rightarrow 2(x-1)(3-x) > 0 \\\Rightarrow (x-1)(3-x) > 0[\because 2>0] \\\Rightarrow 1< x < 3\end{array}\]
Hence interval is $1 < x < 3$.

View full question & answer→

Question 73 Marks

Find the interval for which $f(x)=x^4-2 x^2$ is increasing or decreasing.

Answer

Given $\quad f(x)=x^4-2 x^2$
So,$f^{\prime}(x)=4 x^3-4 x$
$\begin{array}{l}f^{\prime}(x)=4 x\left(x^2-1\right) \\ f^{\prime}(x)=4 x(x+1)(x-1)\end{array}$
For $f(x)$ is increasing $f^{\prime}(x) > 0$ $ \begin{array}{lr} \text { So, } 4 x(x+1)(x-1)>0 \\ \Rightarrow & x(x+1)(x-1) > 0 \\ \Rightarrow & -1< x <0 \text { or } x>1 \\ \Rightarrow & x \in(-1,0) \cup(1, \infty) \end{array} $

So, $f(x)$ is increasing in interval $(-1,0) \cup(1, \infty)$.
$\begin{array}{lc}f(x) & \text { is decreasing } f^{\prime}(x)<0 \\ \Rightarrow & 4 x(x+1)(x-1)< 0 \\ \Rightarrow & x(x+1)(x-1)< 0 \quad \therefore 4 > 0 \\ \Rightarrow & x< -1 \text { or } 0< x <1 \\ \Rightarrow & x \in(-\infty,-1) \cup(0,1)\end{array}$

So, $f(x)$ is decreasing in interval $(-\infty,-1) \cup(0,1)$

View full question & answer→

Maths 3 Marks

Test yourself on this topic