2 Marks Questions

Question 12 Marks

A person standing at $0\ (0,0,0)$ is watching an aeroplane which is at the coordinate point $4\ (4,0,3)$. At the same time he saw a bird at the coordinate point $B\ (0,0,1)$. Find the angles which $\overrightarrow{B A}$ makes with the $x,y$ and $z$ axes.

Answer

$ \overrightarrow{B A}=\overrightarrow{O A}-\overrightarrow{O B}=(4 \hat{\imath}+3 \hat{k})-\hat{k}=4 \hat{\imath}+2 \hat{k}$
$\widehat{B A}=\frac{4}{2 \sqrt{5}} \hat{\imath}+\frac{2}{2 \sqrt{5}} \hat{k}=\frac{2}{\sqrt{5}} \hat{\imath}+\frac{1}{\sqrt{5}} \hat{k} $
So, the angles made by the vector $\overrightarrow{B A}$ with the $x , y$ and the $z$ axes are respectively $\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right), \frac{\pi}{2}, \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)$.

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Question 22 Marks

Differentiate the following function with respect to $x:x :(\cos x)^x ;\left(\right.$ where $\left.x \in\left(0, \frac{\pi}{2}\right)\right)$

Answer

Let $y =(\cos x)^x$.
Then, $y =e^{x \log _e \cos x}$ On differentiating both sides with respect to $x$,
we get $\frac{d y}{d x}=e^{x \log _e \cos x} \frac{d}{d x}\left(x \log _e \cos x\right)$
$ \Rightarrow \frac{d y}{d x}=(\cos x)^x\left\{\log _e \cos x \frac{d}{d x}(x)+x \frac{d}{d x}\left(\log _e \cos x\right)\right\}$
$\Rightarrow \frac{d y}{d x}=(\cos x)^x\left\{\log _e \cos x+x \cdot \frac{1}{\cos x}(-\sin x)\right\} $
$\Rightarrow \frac{d y}{d x}=(\cos x)^x\left(\log _e \cos x-x \tan x\right)$

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Question 32 Marks

Find the derivative of $\tan^{-1}x$ with respect to $\log.x; ($where $x \in (1,\infty)).$

Answer

$y=\tan ^{-1} x$ and $z=\log _e x$
Then $\frac{d y}{d x}=\frac{1}{1+x^2}$
and $\frac{d_z}{d x}=\frac{1}{x}$
$\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}$
So $, =\frac{\frac{1}{1+x^2}}{\frac{1}{x}}=\frac{x}{1+x^2}$

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Question 42 Marks

The two co-initial adjacent sides of a parallelogram are $2 \hat{\imath}-4 \hat{\jmath}-5 \hat{k}$ and $2 \hat{\imath}+2 \hat{\jmath}+3 \hat{k}$ Find its diagonals and use them to find the area of the parallelogram.

Answer

$\overrightarrow{d_1}=\vec{a}+\vec{b}=4 \hat{\imath}-2 \hat{\jmath}-2 \hat{k}, \quad \overrightarrow{d_2}=\vec{a}-\vec{b}=-6 \hat{\jmath}-8 \hat{k}$
Area of the parallelogram $=\frac{1}{2}\left|\overrightarrow{d_1} \times \overrightarrow{d_2}\right|=\frac{1}{2}\left|\begin{array}{ccc}\hat{\imath} & \hat{\jmath} & \hat{k} \\ 4 & -2 & -2 \\ 0 & -6 & -8\end{array}\right|=2|\hat{\imath}+8 \hat{\jmath}-6 \hat{k}|$
Area of the parallelogram $=2 \sqrt{101}$ sq. units.

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Question 52 Marks

If vectors $\vec{a}=2 \hat{ i }+2 \hat{\jmath}+3 \hat{ k }, \vec{b}=-\hat{ i }+2 \hat{\jmath}+\hat{ k }$ and $\vec{c}=3 \hat{ i }+\hat{ j }$ are such that $\vec{b}+\lambda \vec{c}$ is perpendicular to $\vec{a}$, then find the value of $\lambda$.

Answer

We have $\overrightarrow{ b }+\lambda \overrightarrow{ c }=(-1+3 \lambda) \hat{ i }+(2+\lambda) \hat{\jmath}+\hat{ k }$ $(\vec{b}+\lambda \vec{c}) \cdot \vec{a}=0 \Rightarrow 2(-1+3 \lambda)+2(2+\lambda)+3=0$
$\lambda=-\frac{5}{8}$

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Question 62 Marks

The cost (in rupees) of producing x items in factory, each day is given by
$C(x)=0.00013 x^3+0.002 x^2+5 x+2200$
Find the marginal cost when 150 items are produced.

Answer

The marginal cost function is $C ^{\prime}( x )= 0 . 0 0 0 3 9 x ^{ 2 }+ 0 . 0 0 4 x + 5$.
C'(150) = ₹14.375 .

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Question 72 Marks

If $\cot ^{-1}(3 x+5)>\frac{\pi}{4}$, then find the range of the values of $x$

Answer

$\cot ^{-1}(3 x+5)>\frac{\pi}{4}=\cot ^{-1} 1$
$\Rightarrow 3 x +5<1 \ ($ as $\cot ^{-1} x$ is strictly decreasing function in its domain$)$
$ \Rightarrow 3 x<-4$
$\Rightarrow x<-\frac{4}{3}$
$\therefore x \in\left(-\infty,-\frac{4}{3}\right)$

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