Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: If $x \neq 0$ then $\cos \left(\tan ^{-1} x+\cot ^{-1} x\right)=?$ A. $-1$ B. 1 C. 0 D. none of these Solution: Given: $x \neq 0$ To Find: The value of $\cos \left(\tan ^{-1} x+\cot ^{-1} x\right)$ Now, let $x=\cos \left(\tan ^{-1} x+\cot ^{-1} x\right)$ $\Rightarrow x=\cos \left(\frac{\pi}{2}\right)\left(\because \tan ^{-1} \theta+\cot ^{-1} \theta=\frac{\pi}{2}\right)$ $\Rightarrow x=0\left(\because \cos \left(\frac{\pi}{2}\r...
Read More →If then verify that:
Question: If then verify that: (i) (A) = A (ii) (AB) = BA (iii) (kA) = (kA). $A=\left[\begin{array}{ccc}0 -1 2 \\ 4 3 -4\end{array}\right]$ and $B=\left[\begin{array}{cc}4 0 \\ 1 3 \\ 2 6\end{array}\right]$ Solution: Given, $A=\left[\begin{array}{ccc}0 -1 2 \\ 4 3 -4\end{array}\right]$ and $B=\left[\begin{array}{ll}4 0 \\ 1 3 \\ 2 6\end{array}\right]$ (i) We have to verify that, $\left(A^{\prime}\right)^{\prime}=A$ So, $A^{\prime}=\left[\begin{array}{cc}0 4 \\ -1 3 \\ 2 -4\end{array}\right]$ And...
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Question: Evaluate the following integral: $\int \frac{x}{(x-3) \sqrt{x+1}} d x$ Solution: re-writing the given equation as $\int \frac{(x-3)+3}{(x-3) \sqrt{x+1}} d x$ $\int \frac{(x-3)}{(x-3) \sqrt{x+1}} d x+\int \frac{3}{(x-3) \sqrt{x+1}} d x$ For the first part using identity $\int x^{n} d x=\frac{x^{n+1}}{n+1}$ $2 \sqrt{x+1}+c$ For the second part assume $x+1=t^{2}$ $\mathrm{d} x=2 \mathrm{tdt}$ $\int \frac{2 d t}{\left(t^{2}-4\right)}$ Using identity $\int \frac{d z}{(z)^{2}-1}=\frac{1}{2} ...
Read More →If then verify that
Question: If then verify that A2+ A = A (A + I), where I is 3 3 unit matrix. $A=\left[\begin{array}{ccc}1 0 -1 \\ 2 1 3 \\ 0 1 1\end{array}\right]$ Solution: Given, $A=\left[\begin{array}{ccc}1 0 -1 \\ 2 1 3 \\ 0 1 1\end{array}\right]$ So, $A^{2}=A \cdot A$ Thus, $\quad=\left[\begin{array}{ccc}1 0 -1 \\ 2 1 3 \\ 0 1 1\end{array}\right]\left[\begin{array}{ccc}1 0 -1 \\ 2 1 3 \\ 0 1 1\end{array}\right]$ $=\left[\begin{array}{ccc}1+0+0 0+0-1 -1+0-1 \\ 2+2+0 0+1+3 -2+3+3 \\ 0+2+0 0+1+1 0+3+1\end{arr...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}}{(x-1) \sqrt{x+2}} d x$ Solution: re-writing the given equation as $\int \frac{\left(x^{2}-1\right)+1}{(x-1) \sqrt{x+2}} d x$ $\int \frac{\left(x^{2}-1\right)}{(x-1) \sqrt{x+2}} d x+\int \frac{1}{(x-1) \sqrt{x+2}} d x$ $\int \frac{(x+1)}{\sqrt{x+2}} d x+\int \frac{1}{(x-1) \sqrt{x+2}} d x$ $\int \frac{(1)}{\sqrt{x+2}} d x+\int \sqrt{x+2} d x+\int \frac{1}{(x-1) \sqrt{x+2}} d x$ For the first- and second-part using identity $\int x^{n} ...
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Question: If verify that A (B + C) = (AB + AC). $\mathrm{A}=\left[\begin{array}{ll}2 1\end{array}\right], \quad \mathrm{B}=\left[\begin{array}{lll}5 3 4 \\ 8 7 6\end{array}\right]$ and $\mathrm{C}=\left[\begin{array}{ccc}-1 2 1 \\ 1 0 2\end{array}\right]$ Solution: Given, $A=\left[\begin{array}{ll}2 1\end{array}\right], B=\left[\begin{array}{lll}5 3 4 \\ 8 7 6\end{array}\right]$ and $C=\left[\begin{array}{ccc}-1 2 1 \\ 1 0 2\end{array}\right]$ Now, $A(B+C)=\left[\begin{array}{ll}2 1\end{array}\r...
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Question: Mark the tick against the correct answer in the following: The value of $\sin \left(\sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}\right)=?$ A. 0 B. 1 C. $-1$ D. none of these Solution: To Find: The value of $\sin \left(\sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}\right)$ Now, let $x=\sin \left(\sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}\right)$ $\Rightarrow x=\sin \left(\frac{\pi}{2}\right)\left(\because \sin ^{-1} \theta+\cos ^{-1} \theta=\frac{\pi}{2}\right)$ $\Rightarrow x=1\left...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x+1}{(x-1) \sqrt{x+2}} d x$ Solution: re-writing the given equation as $\int \frac{(x-1)+2}{(x-1) \sqrt{x+2}} d x$ Now splitting the integral in two parts $\int \frac{(x-1)}{(x-1) \sqrt{x+2}} d x+\int \frac{2}{(x-1) \sqrt{x+2}} d x$ For the first part using identity $\int x^{n} d x=\frac{x^{n+1}}{n+1}$ $2 \sqrt{x+2}$ For the second part assume $x+2=t^{2}$ $\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$ $\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$ $\int \fr...
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Question: Mark the tick against the correct answer in the following: $\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)=?$ A. 0 B. $\frac{2 \pi}{3}$ C. $\frac{\pi}{2}$ D. $\pi$ Solution: To Find: The value of $\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)$ Now, let $x=\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)$ $\Rightarrow x=\frac{\pi}{3}-\left(-\sin ^{-1}\left(\frac{1}{2}\right)\right)(\because \sin (-\theta)=-\sin (\theta)$ $\Rightarrow x=\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\l...
Read More →If: = A, find A.
Question: If: = A, find A. $\left[\begin{array}{lll}2 1 3\end{array}\right]\left[\begin{array}{ccc}-1 0 -1 \\ -1 1 0 \\ 0 1 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]$ Solution: Given, $\left[\begin{array}{lll}2 1 3\end{array}\right]_{1 \times 3}\left[\begin{array}{ccc}-1 0 -1 \\ -1 1 0 \\ 0 0 1\end{array}\right]_{3 \times 3}\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]_{3 \times 1}=A$ So, $\quad\left[\begin{array}{lll}2 1 3\end{array}\right]_{1 \times 3}\left...
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Question: Mark the tick against the correct answer in the following: The value of $\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ is A. $\frac{3 \pi}{4}$ B. $\frac{\pi}{4}$ C. $\frac{-\pi}{4}$ D. none of these Solution: To Find: The value of $\tan ^{-1}\left(\tan \left(\frac{3 \pi}{4}\right)\right)$ Now, let $x=\tan ^{-1}\left(\tan \left(\frac{3 \pi}{4}\right)\right)$ $\Rightarrow \tan x=\tan \left(\frac{3 \pi}{4}\right)$ Here range of principle value of $\tan$ is $\left[-\frac{\pi}{2}, \frac{\pi}...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{(x-1) \sqrt{2 x+3}} d x$ Solution: assume $2 x+3=t^{2}$ $\mathrm{d} \mathrm{x}=\mathrm{tdt}$ $\int \frac{d t}{\frac{t^{2}-3}{2}-1}$ $\int \frac{2 d t}{\left(t^{2}-5\right)}$ Using identity $\int \frac{\mathrm{dz}}{(\mathrm{z})^{2}-1}=\frac{1}{2} \log \left|\frac{z-1}{z+1}\right|+c$ $\frac{1}{\sqrt{5}} \log \left|\frac{t-\sqrt{5}}{t+\sqrt{5}}\right|+c$ $\frac{1}{\sqrt{5}} \log \left|\frac{\sqrt{(2 x+3)}-\sqrt{5}}{\sqrt{2 x+3}+\sqrt{5}}\rig...
Read More →Prove the following
Question: If $\mathrm{P}=\left[\begin{array}{lll}x 0 0 \\ 0 y 0 \\ 0 0 z\end{array}\right]$ and $\mathrm{Q}=\left[\begin{array}{lll}a 0 0 \\ 0 b 0 \\ 0 0 c\end{array}\right]$, prove that $\mathrm{PQ}=\left[\begin{array}{ccc}x a 0 0 \\ 0 y b 0 \\ 0 0 z c\end{array}\right]=\mathrm{QP}$ Solution: Given, $P=\left[\begin{array}{ccc}x 0 0 \\ 0 y 0 \\ 0 0 z\end{array}\right]$ and $Q=\left[\begin{array}{ccc}a 0 0 \\ 0 b 0 \\ 0 0 c\end{array}\right]$ It's seen that both $\mathrm{P}$ and $\mathrm{Q}$ are ...
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Question: Mark the tick against the correct answer in the following: The value of $\operatorname{cosec}^{-1}\left(\operatorname{cosec} \frac{4 \pi}{3}\right)$ is A. $\frac{\pi}{3}$ B. $\frac{-\pi}{3}$ C. $\frac{2 \pi}{3}$ D. none of these Solution: To Find: The value of $\operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(\frac{4 \pi}{3}\right)\right)$ Now, let $x=\operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(\frac{4 \pi}{3}\right)\right)$ $\Rightarrow \operatorname{cosec} x=\ope...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{(x-1) \sqrt{x+2}} d x$ Solution: assume $x+2=t^{2}$ $\mathrm{d} \mathrm{x}=2 \mathrm{tdt}$ $\int \frac{2 d t}{\left(t^{2}-3\right)}$ Using identity $\int \frac{d z}{(z)^{2}-1}=\frac{1}{2} \log \left|\frac{z-1}{z+1}\right|+c$ $\frac{1}{\sqrt{3}} \log \left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+c$ $\frac{1}{\sqrt{3}} \log \left|\frac{\sqrt{(x+2)}-\sqrt{3}}{\sqrt{x+2}+\sqrt{3}}\right|+c$...
Read More →Prove the following
Question: If $A=\left[\begin{array}{cc}1 2 \\ -2 1\end{array}\right], \quad B=\left[\begin{array}{cc}2 3 \\ 3 -4\end{array}\right]$ and $\quad C=\left[\begin{array}{cc}1 0 \\ -1 0\end{array}\right]$ (i) (AB) C = A (BC) (ii) A (B + C) = AB + AC. Solution: Given, $A=\left[\begin{array}{cc}1 2 \\ -2 1\end{array}\right], B=\left[\begin{array}{cc}2 3 \\ 3 -4\end{array}\right]$ and $C=\left[\begin{array}{cc}1 0 \\ -1 0\end{array}\right]$ (i) $A B=\left[\begin{array}{cc}1 2 \\ -2 1\end{array}\right]\le...
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Question: Mark the tick against the correct answer in the following: The value of $\sec ^{-1}\left(\sec \frac{8 \pi}{5}\right)$ is A. $\frac{2 \pi}{5}$ B. $\frac{3 \pi}{5}$ C. $\frac{8 \pi}{5}$ D. none of these Solution: To Find: The value of $\sec ^{-1}\left(\sec \left(\frac{8 \pi}{5}\right)\right)$ Now, let $x=\sec ^{-1}\left(\sec \left(\frac{8 \pi}{5}\right)\right)$ $\Rightarrow \sec x=\sec \left(\frac{8 \pi}{5}\right)$ Here range of principle value of sec is $[0, \pi]$ $\Rightarrow x=\frac{8...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: The value of $\sec ^{-1}\left(\sec \frac{8 \pi}{5}\right)$ is A. $\frac{2 \pi}{5}$ B. $\frac{3 \pi}{5}$ C. $\frac{8 \pi}{5}$ D. none of these Solution: To Find: The value of $\sec ^{-1}\left(\sec \left(\frac{8 \pi}{5}\right)\right)$ Now, let $x=\sec ^{-1}\left(\sec \left(\frac{8 \pi}{5}\right)\right)$ $\Rightarrow \sec x=\sec \left(\frac{8 \pi}{5}\right)$ Here range of principle value of sec is $[0, \pi]$ $\Rightarrow x=\frac{8...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{\sin ^{4} x+\sin ^{2} x \cos ^{2} x+\cos ^{4} x} d x$ Solution: Re-writing the given equation as Multiplying $\sec ^{4} x$ in both numerator and denominator $\int \frac{\sec ^{4} x}{\tan ^{4} x+\tan ^{2} x+1} d x$ $=\int \frac{\left(\tan ^{2} x+1\right) \sec ^{2} x}{\tan ^{4} x+\tan ^{2} x+1} d x$ Assume $\tan x=t$ $\sec ^{2} x d x=d t$ $=\int \frac{\left(\mathrm{t}^{2}+1\right) \mathrm{dt}}{\mathrm{t}^{4}+\mathrm{t}^{2}+1}$ $=\int \frac{...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: The value of $\cot ^{-1}\left(\cot \frac{5 \pi}{4}\right)$ is A. $\frac{\pi}{4}$ B. $\frac{-\pi}{4}$ C. $\frac{3 \pi}{4}$ D. none of these Solution: To Find: The value of $\cot ^{-1}\left(\cot \left(\frac{5 \pi}{4}\right)\right)$ Now, let $x=\cot ^{-1}\left(\cot \left(\frac{5 \pi}{4}\right)\right)$ $\Rightarrow \cot x=\cot \left(\frac{5 \pi}{4}\right)$ Here range of principle value of $\cot$ is $\left[-\frac{\pi}{2}, \frac{\pi}...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{x^{4}+3 x^{2}+1} d x$ Solution: re-writing the given equation as $\int \frac{\frac{1}{x^{2}}}{x^{2}+3+\frac{1}{x^{2}}} d x$ $\frac{1}{2} \int \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+3+\frac{1}{x^{2}}} d x$ $\frac{1}{2}\left[\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+5} d x-\int \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right)^{2}+1}\right]$ Assume $t=x-\frac{1}{x}$ and $z=x+\frac{1}{...
Read More →Prove the following
Question: If (i) (AB) C = A (BC) (ii) A (B + C) = AB + AC. Solution: Given, $A=\left[\begin{array}{cc}1 2 \\ -2 1\end{array}\right], B=\left[\begin{array}{cc}2 3 \\ 3 -4\end{array}\right]$ and $C=\left[\begin{array}{cc}1 0 \\ -1 0\end{array}\right]$ (i) $A B=\left[\begin{array}{cc}1 2 \\ -2 1\end{array}\right]\left[\begin{array}{cc}2 3 \\ 3 -4\end{array}\right]=\left[\begin{array}{cc}2+6 3-8 \\ -4+3 -6-4\end{array}\right]=\left[\begin{array}{cc}8 -5 \\ -1 -10\end{array}\right]$ and $(A B) C=\lef...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{(x-1)^{2}}{x^{4}+x^{2}+1} d x$ Solution: re-writing the given equation as $\int \frac{x^{2}-2 x+1}{x^{4}+x^{2}+1} d x$ $\int \frac{1-\frac{2}{x}+\frac{1}{x^{2}}}{x^{2}+1+\frac{1}{x^{2}}} d x$ $\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+3} d x-\int \frac{2 x}{x^{4}+x^{2}+1} d x$ Substituting $\mathrm{tas} \mathrm{x}-\frac{1}{\mathrm{x}}$ and $\mathrm{z}$ as $\mathrm{x}^{2}$ $\left(1+\frac{1}{x^{2}}\right) d x=d t$ and $2 x d...
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Question: Mark the tick against the correct answer in the following The value of $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$ is A. $\frac{7 \pi}{6}$ B. $\frac{5 \pi}{6}$ C. $\frac{\pi}{6}$ D. none of these Solution: To Find: The value of $\tan ^{-1}\left(\tan \left(\frac{7 \pi}{6}\right)\right)$ Now, let $x=\tan ^{-1}\left(\tan \left(\frac{7 \pi}{6}\right)\right)$ $\Rightarrow \tan x=\tan \left(\frac{7 \pi}{6}\right)$ Here range of principle value of $\tan$ is $\left[-\frac{\pi}{2}, \frac{\pi}...
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Question: Mark the tick against the correct answer in the following: The value of $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ is A. $\frac{13 \pi}{6}$ B. C. $\frac{5 \pi}{6}$ D. $\frac{\pi}{6} \frac{7 \pi}{6}$ Solution: To Find: The value of $\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$ Now, let $x=\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$ $\Rightarrow \cos x=\cos \left(\frac{13 \pi}{6}\right)$ Here, range of principle value of $\cos$ is $[0, \pi]$ $\Rightarrow ...
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