5 Marks Questions

Question 15 Marks

Draw the rough sketch of the curve $y=20 \cos 2 x ;\left(\right.$ where $\left.\frac{\pi}{6} \leq x \leq \frac{\pi}{3}\right)$
Using integration, find the area of the region bounded by the curve $y = 20 \cos2x$ from the ordinates
$x=\frac{\pi}{6}$ to $x=\frac{\pi}{3}$ and the $x-$axis.

Answer

$y=20 \cos 2 x ;\left\{\frac{\pi}{6} \leq x \leq \frac{\pi}{3}\right\}$

$\text { Required area }=20 \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \cos 2 x d x+\left|20 \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cos 2 x d x\right|$
$=20\left[\frac{\sin 2 x}{2}\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}}+\left|20\left[\frac{\sin 2 x}{2}\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}\right|$
$=10\left(1-\frac{\sqrt{3}}{2}\right)+10\left(1-\frac{\sqrt{3}}{2}\right)=20\left(1-\frac{\sqrt{3}}{2}\right) \text { sq. units. }$

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Question 25 Marks

Find the image of the point (1,2,1) with respect to the line$\frac{x-3}{1}=\frac{y+1}{2}=\frac{z-1}{3}$ Also find the equation of the line joining the given point and its image.

Answer

Let P(1, 2, 1) be the given point and Lbe the foot of the perpendicular from P to the given line AB (as shown in the figure above).
Let's put $\frac{x-3}{1}=\frac{y+1}{2}=\frac{z-1}{3}=\lambda$. Then, $x=\lambda+3, y=2 \lambda-1, z=3 \lambda+1$
Let the coordinates of the point $L$ be $(\lambda+3,2 \lambda-1,3 \lambda+1)$.
So, direction ratios of $P L$ are $(\lambda+3-1,2 \lambda-1-2,3 \lambda+1-1)$ i.e., $(\lambda+2,2 \lambda-3,3 \lambda)$
Let Q(x_{1}, y_{1}, z_{1}) ? the image of P(1, 2, 1) with respect to the given line. Then, L is the mid-point of PQ.
Therefore, $\frac{1+x_1}{2}=\frac{23}{7}, \frac{2+y_1}{2}=-\frac{3}{7}, \frac{1+z_1}{2}=\frac{13}{7} \Rightarrow x_1=\frac{39}{7}, y_1=-\frac{20}{7}, z_1=\frac{19}{7}$
Hence, the image of the point $P ( 1 , 2 , 1 )$ with respect to the given line $Q\left(\frac{39}{7},-\frac{20}{7}, \frac{19}{7}\right)$.
The equation of the line joining $P(1,2,1)$ and $Q\left(\frac{39}{7},-\frac{20}{7}, \frac{19}{7}\right)$ is
$\frac{x-1}{32 / 7}=\frac{y-2}{-34 / 7}=\frac{z-1}{12 / 7} \Rightarrow \frac{x-1}{16}=\frac{y-2}{-17}=\frac{z-1}{6}$.

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Question 35 Marks

If $(x-a)^2+(y-b)^2=c^2$, for some $c>0$, prove that $\frac{\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2 y}{d x^2}}$ is a constant independent of $a$ and $b$.

Answer

Given relation is $(x-a)^2+(y-b)^2=c^2, c>0$.
Let $x - a = c \cos \theta$ and $y-b=c \sin \theta$.
Therefore, $\frac{d x}{d \theta}=-c \sin \theta$ And $\frac{d y}{d \theta}=c \cos \theta$
$\therefore \frac{d y}{d x}=-\cot \theta$
Differentiate both sides with respect to $\theta$, we get $\frac{d}{d \theta}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}(-\cot \theta)$$\text { Or, } \frac{d}{d x}\left(\frac{d y}{d x}\right) \frac{d x}{d \theta}=\operatorname{cosec}^2 \theta$
$\text { Or, } \frac{d^2 y}{d x^2}(-c \sin \theta)=\operatorname{cosec}^2 \theta$
$\frac{d^2 y}{d x^2}=-\frac{\operatorname{cosec}^3 \theta}{c}$
$\therefore \frac{\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2 y}{d x^2}}=\frac{c\left[1+\cot ^2 \theta\right]^{\frac{3}{2}}}{-\operatorname{cosec}^2 \theta}=\frac{-c\left(\operatorname{cosec}^2 \theta\right)^{\frac{3}{2}}}{\operatorname{cosec}^3 \theta}=-c,$
which is constant and is independent of $a$ and $b.$

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Question 45 Marks

If $f: R \rightarrow R$ is defined by $f(x)=|x|^3$, show that $f^{\prime \prime}(x)$ exists for all real $x$ and find it.

Answer

We have, $f(x)=|x|^3,\left\{\begin{array}{l}x^3, \text { if } x \geq 0 \\ (-x)^3=-x^3, \text { if } x<0\end{array}\right.$
Now, $( \ce{LHD}$ at $x=0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0^{-}}\left(\frac{-x^3-0}{x}\right)$
$= \lim _{x \rightarrow 0^{-}}\left(-x^2\right)=0$
$(\text { RHD}$ at $x =0) \lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0^{+}}\left(\frac{x^3-0}{x}\right)$
$=\lim _{x \rightarrow 0}\left(-x^2\right)=0$
$\therefore(\operatorname{LHD}$ of $f(x)$ at $x=0)=( \ce{RHD}$ of $f(x)$ at $x=0)$
So, $f(x)$ is differentiable at $x =0$ and the derivative of $f(x)$ is given by
$f^{\prime}(x)=\left\{\begin{array}{l}3 x^2, \text { if } x \geq 0 \\-3 x^2, \text { if } x<0\end{array}\right.$

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Question 55 Marks

Find the shortest distance between the lines $I_1$ and $l_2$
$\vec{r}=(-\hat{\imath}-\hat{\jmath}-\hat{k})+\lambda(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})$ and $\vec{r}=(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\mu(\hat{\imath}-2 \hat{\jmath}+\hat{k})$where $\lambda$ and $\mu$ are parameters.

Answer

Given that equation of lines are
$\vec{r} =(-\hat{\imath}-\hat{\jmath}-\hat{k})+\lambda(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})\ldots(i)$
$\vec{r} =(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\mu(\hat{\imath}-2 \hat{\jmath}+\hat{k})\ldots(ii)$
The given lines are non-parallel lines as vectors $7 \hat{\imath}-6 \hat{\jmath}+\hat{k}$ and $\hat{\imath}-2 \hat{\jmath}+\hat{k}$ are not parallel.
There is a unique line segment $P Q\ ( P$ lying on line $(i)$ and $Q$ on the other line $(i i) ),$ which is at right angles to both the lines $P Q$ is the shortest distance between the lines.
Hence, the shortest possible distance between the lines $=P Q$.
Let the position vector of the point $P$ lying on the line $r=(-\hat{\imath}-\hat{\jmath}-\hat{k})+\lambda(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})$ where ' $\lambda$ ' is a scalar, is $(7 \lambda-1) \hat{\imath}-(6 \lambda+1) \hat{\jmath}+(\lambda-1) \hat{k}$, for some $\lambda$ and the position vector of the point $Q$ lying on the line $r=(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\mu(\hat{\imath}-2 \hat{\jmath}+\hat{k})$ where $'\mu\ ^{\prime}$ is a scalar, is $(\mu+3) \hat{\imath}+(-2 \mu+5) \hat{\jmath}+(\mu+7) \hat{k}$, for some $\mu$.
Now, the vector
$\overline{P Q}=\overline{O Q}-\overline{O P}=(\mu+3-7 \lambda+1) \hat{\imath}+(-2 \mu+5+6 \lambda+1) \hat{\jmath}+(\mu+7-\lambda+1) \hat{k}$
i.e., $\overline{P Q}=(\mu-7 \lambda+4) \hat{\imath}+(-2 \mu+6 \lambda+6) \hat{\jmath}+(\mu-\lambda+8) \hat{k}; ($where $'O\ '$ is the origin$),$ is perpendicular to both the lines, so the vector $\overline{P Q}$ is perpendicular to both the vectors $7 \hat{\imath}-6 \hat{\jmath}+\hat{k}$ and $\hat{\imath}-2 \hat{\jmath}+\hat{k}$.
$\Rightarrow(\mu-7 \lambda+4) \cdot 7+(-2 \mu+6 \lambda+6) \cdot(-6)+(\mu-\lambda+8) \cdot 1=0$
$\ (\mu-7 \lambda+4) \cdot 1+(-2 \mu+6 \lambda+6) \cdot(-2)+(\mu-\lambda+8) \cdot 1=0$
$\Rightarrow 2 0 \mu- 8 6 \lambda= 0$
$\Rightarrow 1 0 \mu- 4 3 \lambda= 0 \ 6 \mu-20 \lambda=0 $
$\Rightarrow 3 \mu-10 \lambda=0$
On solving the above equations, we get $\mu=\lambda=0$
So, the position vector of the points $P$ and $Q$ are $-\hat{\imath}-\hat{\jmath}-\hat{k}$ and $3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k}$ respectively.
$\overline{P Q}=4 \hat{\imath}+6 \hat{\jmath}+8 \hat{k}$ and $|\overline{P Q}|=\sqrt{4^2+6^2+8^2}=\sqrt{116}=2 \sqrt{29}$ units.

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Question 65 Marks

The equation of the path traversed by the ball headed by the footballer is$y=a x^2+b x+c ;($ where $0 \leq x \leq 14$ and $a, b, c \in R$ and $a \neq 0)$ (2,15), (4,25) and (14,15) Determine the values of a, b and c by solving the system of linear equations in a, b and c, using matrix method. Also find the equation of the path traversed by the ball.

Answer

$\begin{array}{l} y=a x^2+b x+c \\ 15=4 a+2 b+c \\ 25=16 a+4 b+c \\ 15=196 a+14 b+c \end{array}$
The set of equations can be represented in the matrix form as $A X = B$,
where $A=\left[\begin{array}{ccc}4 & 2 & 1 \\ 16 & 4 & 1 \\ 196 & 14 & 1\end{array}\right], X=\left[\begin{array}{l}a \\ b \\ c\end{array}\right]$ and $B=\left[\begin{array}{l}15 \\ 25 \\ 15\end{array}\right] \Rightarrow\left[\begin{array}{ccc}4 & 2 & 1 \\ 16 & 4 & 1 \\ 196 & 14 & 1\end{array}\right]\left[\begin{array}{l}a \\ b \\ c\end{array}\right]=\left[\begin{array}{l}15 \\ 25 \\ 15\end{array}\right]$ $ |A|=4(4-14)-2(16-196)+(224-784)=-40+360-560=-240 \neq 0 \text {. Hence } A ^{-1} $ exists. $ \begin{array}{l} \text { Now, } \operatorname{adj}(A)=\left[\begin{array}{ccc} -10 & 180 & -560 \\ 12 & -192 & 336 \\ -2 & 12 & -16 \end{array}\right]^T=\left[\begin{array}{ccc} -10 & 12 & -2 \\ 180 & -192 & 12 \\ -560 & 336 & -16 \end{array}\right] \\ \qquad\left[\begin{array}{l} a \\ b \\ c \end{array}\right]=-\frac{1}{240}\left[\begin{array}{ccc} -10 & 12 & -2 \\ 180 & -192 & 12 \\ -560 & 336 & -16 \end{array}\right]\left[\begin{array}{c} 15 \\ 25 \\ 15 \end{array}\right]=-\frac{5}{240}\left[\begin{array}{ccc} -10 & 12 & -2 \\ 180 & -192 & 12 \\ -560 & 336 & -16 \end{array}\right]\left[\begin{array}{l} 3 \\ 5 \\ 3 \end{array}\right]=-\frac{5}{240}\left[\begin{array}{c} 24 \\ -384 \\ -48 \end{array}\right] \\ \therefore a =-\frac{1}{ 1 }, b = 8 , c = 1 \end{array} $So, the equation becomes $y=-\frac{1}{2} x^2+8 x+1$

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MATHS 5 Marks Questions

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