2 Marks Questions

Question 12 Marks

Evaluate the determinant $\left|\begin{array}{rr}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right|$

Answer

Let $A =\left|\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right|$
$\Rightarrow|A|=(a+i b)(a-i b)-(c+i d)(-c+i d)$
$=(a+i b)(a-i b)+(c+i d)(c-i d)$
$=a^2-i^2 b^2+c^2-i^2 d^2$
$=a^2-(-1) b^2+c^2-(-1) d^2$
$=a^2+b^2+c^2+d^2$
Thus $, |A|=a^2+b^2+c^2+d^2$

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Question 22 Marks

Evaluate: $\int \sec ^{\frac{4}{3}} x \csc ^{\frac{8}{3}} x d x$

Answer

Let $I =\int \sec ^{\frac{4}{3}} x \csc ^{\frac{8}{3}} x d x$.
Then, we have
$I=\int \frac{1}{\cos ^{4 / 3} x \sin ^{8 / 3} x} d x$
$=\int \cos ^{\frac{-4}{3}} x \sin ^{\frac{-8}{3}} x d x$
since $-\left(\frac{4}{3}+\frac{8}{3}\right)=-4$,
which is an even integer.
So, we divide both numerator and denominator by $\cos ^4 x$.
$\therefore \quad I=\int \frac{\sec ^4 x}{\tan ^{8 / 3} x} d x$
$=\int \frac{\left(1+\tan ^2 x\right)}{\tan ^{8 / 3} x} \sec ^2 x d x$
Put $\tan x=t$ and $\sec ^2=d t$, we get
$I=\int \frac{1+t^2}{t^{\frac{8}{3}}} d t$
$=\int\left(t^{\frac{-8}{3}}+t^{\frac{-2}{3}}\right) d t$
$=-\frac{3}{5} t^{\frac{-5}{3}}+3 t^{\frac{1}{3}}+c$
$\Rightarrow I=-\frac{3}{5} \tan ^{\frac{-5}{3}} x+3 \tan ^{\frac{1}{3}} x+C$

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Question 32 Marks

Find the point on the curve $y^2=8 x+3$ for which the $y-$coordinate change $8$ times more than coordinate of $x$.

Answer

$y^2=8 x+3 \ldots(i) ($given $)$
$\therefore 2 y \frac{d y}{d t}=8 \frac{d x}{d t}$
$\frac{d y}{d t}=8 \frac{d x}{d t} \ldots (ii) ($ given $)$
$\therefore 2 y \cdot 8 \frac{d x}{d t}=8 \frac{d x}{d t}$
$\Rightarrow y=\frac{8}{16}=\frac{1}{2}$
$\text { For } y=\frac{1}{2}$
From eq $(i), \left(\frac{1}{2}\right)^2=8 x+3$
or, $\frac{1}{4}-3=8 x$
or, $x=-\frac{11}{32}$
Hence, required point is $\left(-\frac{11}{32}, \frac{1}{2}\right)$.

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Question 42 Marks

Find the absolute maximum value and the absolute minimum value of the function
$f(x)=4 x-\frac{1}{2} x^2, x \in\left[-2, \frac{9}{2}\right]$

Answer

Given that $f(x)=4 x-\frac{1}{2} x^2, x \in\left[-2, \frac{9}{2}\right]$
$\Rightarrow f^{\prime}(x)=4-\frac{1}{2}(2 x)=4-x$
$\text { Now, } f(x)=0$
$\Rightarrow x=4$
Now, we evaluate the value of $f$ at critical point $x = 0$ and at end points of the interval
$f(4)=16-\frac{1}{2}(16)=16-8=8$
$f(-2)=-8-\frac{1}{2}(4)=-8-2=-10$
$f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}\left(\frac{9}{2}\right)^2$
$=18-\frac{81}{8}$
$=18-10.125$
$=7.875$
Therefore, the absolute maximum value of $f$ on $\left[-2, \frac{9}{2}\right]$ is $8$ occurring at $x=4$
And, the absolute minimum value of $f$ on $\left[-2, \frac{9}{2}\right]$ is $-10$ occurring at $x=-2$

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Question 52 Marks

Find the least value of a such that the function $f$ given by $f(x)=x^2+a x+1$ is strictly increasing on $(1, 2).$

Answer

$ f(x)=x^2+a x+1$
$\Rightarrow f^{\prime}(x)=2 x+a $
Since $f(x)$ is strictly increasing on $(1,2)$, therefore $f^{\prime}(x)=2 x+a;$
$0$ for all $x$ in $(1,2)$
$\therefore \text { On }(1,2) 1 < x < 2$
$\Rightarrow 2 < 2 x < 4$
$\Rightarrow 2 + a < 2 x + a < 4 + a$
$\therefore$Minimum value of $f^{\prime}(x)$ is $2+a$ and maximum value is $4+a$.
Since $f^{\prime}(x);0$ for all $x$ in $(1,2)$
$ \therefore 2+a;0$ and $4+a;0$
$\Rightarrow a;-2$ and $a;-4 $
Therefore least value of $a$ is $-2 .$
Which is the required solution.

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Question 62 Marks

$\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$

Answer

Let $\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=y$
$\Rightarrow \cos y=-\frac{1}{\sqrt{2}}$
$\Rightarrow \cos y=-\cos \frac{\pi}{4}$
$\Rightarrow \cos y=\cos \left(\pi-\frac{\pi}{4}\right)=\cos \frac{3 \pi}{4}$
Since, the principal value branch of $\cos ^{-1}$ is $[0, \pi]$.
Therefore, Principal value of $\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ is $\frac{3 \pi}{4}$.

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Question 72 Marks

Find the value of $\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left[\sin \left(\frac{-\pi}{2}\right)\right]$.

Answer

We have, $\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left[\sin \left(\frac{-\pi}{2}\right)\right]$.
$=\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cot ^{-1}\left(\cot \frac{\pi}{3}\right)+\tan ^{-1}(-1)$
$=\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{6}\right)\right]+\cot ^{-1}\left[\cot \left(\frac{\pi}{3}\right)\right]+\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{4}\right)\right]$
$=\tan ^{-1}\left(-\tan \frac{\pi}{6}\right)+\cot ^{-1}\left(\cot \frac{\pi}{3}\right)+\tan ^{-1}\left(-\tan \frac{\pi}{4}\right)\left[\begin{array}{c}\because \tan ^{-1}(\tan x)=x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \\ \cot ^{-1}(\cot x)=x, x \in(0, \pi) \\ \text { and } \tan ^{-1}(-x)=-\tan ^{-1} x\end{array}\right]$
$=-\frac{\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4}=\frac{-2 \pi+4 \pi-3 \pi}{12}$
$=\frac{-5 \pi+4 \pi}{12}=\frac{-\pi}{12}$

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Maths 2 Marks Questions

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