2 Marks

Question 12 Marks

The radius of a circle is increasing uniformly at the rate of $3 cm / s$. Find the rate at which the area of the circle is increasing when the radius is 10 cm .

Answer

Let the area of circle be A and radius $r$.
According to question,$
\begin{aligned}
\frac{d r}{d t} & =3 cm / s \\
A & =\pi r^2 \\
\frac{d A}{d t} & =2 \pi r \frac{d r}{d t} \\
r & =10 cm, \frac{d r}{d t}=3 cm / s \\
\text { Putting } \quad \frac{d A}{d t} & =\frac{2 \pi \times 10 \times 3}{60 \pi cm^2 / s}
\end{aligned}
$

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Question 22 Marks

Find the maximum and minimum value of function $f(x)=\sin 2 x+5$.

Answer

$
\begin{array}{rlrl}
f(x) & =\sin 2 x+5 \\
\because & -1 \leq \sin 2 x \leq 1 \quad \forall x \in R \\
\therefore & -1+5 \leq \sin 2 x+5 \leq 1+5, \quad \forall x \in R \\
\Rightarrow& 4 \leq \sin 2 x+5 \leq 6 \quad \forall x \in R \\
\Rightarrow & 4 \leq f(x) \leq 6 \forall x \in R
\end{array}
$
hence maximum value is 6 and minimum value is 4 .

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Question 32 Marks

Find the interval where function $f(x)=x^3-3 x$ is decreasing.

Answer

$
\begin{aligned}
f(x) & =x^3-3 x \\
f^{\prime}(x) & =3 x^2-3=3\left(x^2-1\right)
\end{aligned}
$
For decreasing $f^{\prime}(x) \leq 0$
$\begin{aligned} \Rightarrow & & 3\left(x^2-1\right) & \leq 0 \\ \Rightarrow & & x^2 & \leq 1 \\ \Rightarrow & & -1 & < x<1 \\ \Rightarrow & & x & \in(-1,1)\end{aligned}$

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Question 42 Marks

What is the maximum value of $a \sin x+b \cos x$ ?

Answer

Suppose
$
\begin{aligned}
y & =a \sin x+b \sin x \\
\frac{d y}{d x} & =a \cos x-b \sin x
\end{aligned}
$
For, maxima and minima $\frac{d y}{d x}=0$$
\begin{array}{l}
\therefore a \cos x-b \sin x=0 \\
\tan x=\frac{a}{b} \\
\frac{d^2 y}{d x^2}=-a \sin x-b \cos x \\
=-(a \sin x+b \cos x)=-ve \\
\therefore \text { maximum value }=\frac{a \times a}{\sqrt{a^2+b^2}}+\frac{b \times b}{\sqrt{a^2+b^2}} \\
=\frac{a^2+b^2}{\sqrt{a^2+b^2}}=\sqrt{a^2+b^2}
\end{array}
$

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Question 52 Marks

For function $y=f(x)$ if $\frac{d y}{d x}=6(x-2)(x-3)$ then find the value of $x$ for the maximum value of $y$.

Answer

For maxima and minima $\frac{d y}{d x}=0$
$
\begin{array}{rlrl}
\Rightarrow 0 & =6(x-2)(x-3) \\
\text { or } x=2,3 \\
\frac{d y}{d x} & =6(x-2)(x-3)=6\left(x^2-5 x+6\right) \\
\frac{d^2 y}{d x^2} & =6(2 x-5)=12 x-30 \\
\left(\frac{d^2 y}{d x^2}\right)_{\text {at } x=2} & =12 \times 2-30=-6<0
\end{array}
$
hence $y$ is maximum at $x=2$.

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Question 62 Marks

Find the minimum value of a, function $f(x)=x^2$ $+a x+10$ is increasing in $[3,6]$.

Answer

Given,
$
\begin{array}{rr}
& f(x)=x^2+a x+10 \\
\text { So, } & f^{\prime}(x)=2 x+a \\
\text { because } & x \in[3,6] \Rightarrow 3 \leq x \leq 6 \\
\Rightarrow & 6 \leq 2 x \leq 12 \\
\Rightarrow & 6+a \leq 2 x+a \leq 12+a \\
\Rightarrow & 6+a \leq f^{\prime}(x) \leq 12+a
\end{array}
$
If $f(x)$ is increasing then $f^{\prime}(x)>0$$
\Rightarrow \quad 6+a>0 \Rightarrow a>-6
$
So, minimum value of a is -6 .

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Maths 2 Marks

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