2 Marks Questions

Question 12 Marks

Find values of $k$ if area of triangle is $35$ square units having vertices as $(2, -6), (5, 4), (k, 4).$

Answer

Area of triangle$=35$ units
$\Rightarrow \frac{1}{2}\left[\begin{array}{ccc}2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1\end{array}\right]= \pm 35$ Expanding along row $I^{st},$
$\Rightarrow \frac{1}{2}[2(4-4)+6(5-k)+1(20-4 k)]= \pm 35$
$\Rightarrow \frac{1}{2}[30-6 k+20-4 k]= \pm 35$
$\Rightarrow \frac{1}{2}[50-10 k]= \pm 35$
$\Rightarrow 25-5 k= \pm 35$
$\Rightarrow 25-5 k=35 $ or $ 25-5 k=-35$
$\Rightarrow-5 k=10 $ or $ 5 k=60$
$\Rightarrow k=-2
$ or $ k=12$

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Question 22 Marks

Evaluate: $\int \frac{\left\{e^{\sin ^{-1} x}\right\}^2}{\sqrt{1-x^2}} d x$

Answer

Let $I =\int \frac{\left\{e^{\sin ^{-1} \alpha}\right\}^2}{\sqrt{1-x^2}} d x$
Also let sin $x = t$ then, we have
$ d \left(\sin ^{-1} x \right)= dt$
$\Rightarrow \frac{1}{\sqrt{1-x^2}} d x=d t$
$\Rightarrow d x=\sqrt{1-x^2} d t$
Putting $\sin ^{-1} x = t$ and $dx =\sqrt{1-x^2} d t$ in equation $(i),$ we get
$I=\int \frac{\left(e^t\right)^2}{\sqrt{1-x^2}} \times \sqrt{1-x^2} d t$
$=\int e^{2 t} d t$
$=\frac{e^2}{2}+c$
$=\frac{e^{2 \sin ^{-1} x}}{2}+c$
$\therefore I=\frac{\left\{e^{\sin }-1\right\}^2}{2}+c$

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Question 32 Marks

Show that $f ( x )=\cos ^2 x$ is a decreasing function on $\left(0, \frac{\pi}{2}\right)$.

Answer

Given: $f(x)=\cos ^2 x$
Theorem$:-$ Let $f$ be a differentiable real function defined on an open interval $(a,b).$
i. If $f^\ {\prime}(x);0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$
ii. If $f^\ {\prime}(x)<0$ for all,$x \in(a, b)$ then $f(x)$ is decreasing on $(a, b)$
For the value of $x$ obtained in $(ii) f(x)$ is increasing and for remaining points in its domain it is decreasing.
Here we have,
$ f ( x )=\cos ^2 x$
$\Rightarrow f(x)=\frac{d}{d x}\left(\cos ^2 x\right)$
$= f^\ {\prime}( x )=3 \cos x (-\sin x )$
$= f^\ {\prime}( x )=-2 \sin ( x ) \cos ( x )$
$= f^\ {\prime}( x )=-\sin 2 x ;$ as $\sin 2 A=2 \sin A \cos A $
Now, as given
$x \in\left(0, \frac{\pi}{2}\right)$
$=2 x \in(0, \pi)$
$=\operatorname{Sin}(2 x);0$
$=-\operatorname{Sin}(2 x)<0$
$\Rightarrow f^{\prime}(x)<0$
hence, it is the condition for $f(x)$ to be decreasing Thus, $f(x)$ is decreasing on interval $\left(0, \frac{\pi}{2}\right)$

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Question 42 Marks

A ladder $13 m$ long is leaning against a vertical wall. The bottom of the ladder is dragged away from the wall along the ground at the rate of $2 \ cm/\sec.$ How fast is the height on the wall decreasing when the foot of the ladder is $5 m$ away from the wall?

Answer

Let $AB$ be the ladder length of ladder is $5m$
i..e, $AB = 5$
$\& OB$ be the wall $OA$ be the ground.

Suppose $OA = x OB = y$
Given that
The bottom of the ladder is pulled along the ground, away the wall at the rate of $2\ cm/s$
i.e., $\frac{d x}{d t}=2\ cm / sec ......(i)$
We need to calculate at which rate height of ladder on the wall.
Decreasing when foot of the ladder is $4 m$ away from the wall
i.e. we need to calculate $\frac{d y}{d t}$ when $x=4\ cm$
Wall $OB$ is perpendicular to the ground $OA$

Using Pythagoras theorem, we get
$(OB)^2+(OA)^2=(AB)^2$
$y^2+x^2=(5)^2$
$y^2+x^2=24 \ldots . \text { (ii) }$
Differentiating $w.r.t.$ time, we get
$\frac{d\left(y^2+x^2\right)}{d t}=\frac{d(25)}{d t}$
$\frac{d\left(y^2\right)}{d t}+\frac{d\left(x^2\right)}{d t}=0$
$\frac{d\left(y^2\right)}{d t} \times \frac{d y}{d y}+\frac{d\left(x^2\right)}{d t} \times \frac{d x}{d x}=0$
$2 y \times \frac{d y}{d t}+2 x \times \frac{d z}{d t}=0$
$2 y \times \frac{d y}{d z}+2 x \times(2)=0$
$2 y \frac{d y}{d t}+4 x=0$
$2 y \frac{d y}{d t}=-4 x$
$\frac{d y}{d t}=\frac{-4 x}{2 y}$
We need to find $\frac{d y}{d t}$ when $x =4\ cm$
$
\left.\frac{d y}{d t}\right|_{x=4}=\frac{-4 \times 4}{2 y}$
$\left.\frac{d y}{d t}\right|_{x=4}=\frac{-16}{2 y}
.....(iii)$
Finding value of $y$ From $(ii)$
$x^2+y^2=25$
Putting x = 4
$(4)^2+y^2=25$
$y^2=9$
$y=3$

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Question 52 Marks

Show that $f(x)=\frac{1}{1+x^2}$ is neither increasing nor decreasing on $R .$

Answer

$f(x)=\frac{1}{1+x^2}$
Let $x _1> x _2$
$\Rightarrow x_1^2>x_2^2$
$\Rightarrow 1+x_1^2>1+x_2^2$
$\Rightarrow \frac{1}{1+x_1^2}<\frac{1}{1+x_2^2}$
$\Rightarrow f\left(x_1\right) < f\left(x_2\right)$
$f(x)$
is decreasing on $[0, \infty)$
Case $2$
$ \Rightarrow x_1^2 < x_2^2$
$\Rightarrow 1+x_1^2 < 1+x_2^2$
$\Rightarrow \frac{1}{1+x_1^2} > \frac{1}{1+x_2^2}$
$\Rightarrow f\left(x_1\right)>f\left(x_2\right) $
So, $f(x)$ is increasing on $[0, \infty]$
Thus, $f(x)$ is neither increasing nor decreasing on $R$

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Question 62 Marks

$\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)=?$

Answer

$\tan ^{-1}\left(\tan \frac{3 \pi}{\pi}\right) \neq \frac{3 \pi}{4} \text { as } \frac{3 \pi}{4} \notin\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
$\because \tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)=\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{4}\right)\right]$
$=\tan ^{-1}\left[-\tan \left(\frac{\pi}{4}\right)\right]$
$=-\frac{\pi}{4}$

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Question 72 Marks

Evaluate$:- \tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left(\sin \left(-\frac{\pi}{2}\right)\right)$

Answer

$\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left(\sin \left(-\frac{\pi}{2}\right)\right)$
$=-\frac{\pi}{6}+\frac{\pi}{3}+\tan ^{-1}(-1)$
$=-\frac{\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4}$
$=-\frac{\pi}{12}$

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Mathematics 2 Marks Questions

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