Write a value
Question: Write a value of $\int a^{x} e^{x} d x$ Solution: We know that a and e are constant so, $a^{x} e^{x}=(a e)^{x}$ $y=\int(a e)^{x} d x$ Use formula $\int c^{x}=\frac{c^{x}}{\log c}$ where $c$ is constant $y=\frac{(a e)^{x}}{\log (a e)}+c$ $y=\frac{a^{x} e^{x}}{\log a+1}+c$...
Read More →Write a value of $int a^{x} e^{x} d x$
Question: Write a value of $\int a^{x} e^{x} d x$ Solution: We know that a and e are constant so, $a^{x} e^{x}=(a e)^{x}$ $y=\int(a e)^{x} d x$ Use formula $\int c^{x}=\frac{c^{x}}{\log c}$ where $c$ is constant $y=\frac{(a e)^{x}}{\log (a e)}+c$ $y=\frac{a^{x} e^{x}}{\log a+1}+c$...
Read More →Prove the following
Question: If A = find A2+ 2A + 7I. Solution: Given, $A=\left[\begin{array}{ll}1 2 \\ 4 1\end{array}\right]$ $A^{2}=A \cdot A=\left[\begin{array}{ll}1 2 \\ 4 1\end{array}\right]\left[\begin{array}{ll}1 2 \\ 4 1\end{array}\right]=\left[\begin{array}{ll}1+8 2+2 \\ 4+4 8+1\end{array}\right]=\left[\begin{array}{ll}9 4 \\ 8 9\end{array}\right]$ $A^{2}+2 A+7 I=\left[\begin{array}{ll}9 4 \\ 8 9\end{array}\right]+\left[\begin{array}{ll}2 4 \\ 8 2\end{array}\right]+\left[\begin{array}{ll}7 0 \\ 0 7\end{ar...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\tan ^{-1} 1+\tan ^{-1} \frac{1}{3}=?$ A. $\tan ^{-1} \frac{4}{3}$ B. $\tan ^{-1} \frac{2}{3}$ C. $\tan ^{-1} 2$ D. $\tan ^{-1} 3$ Solution: To Find: The value of $\tan ^{-1} 1+\tan ^{-1} \frac{1}{3}$ Let, $x=\tan ^{-1} 1+\tan ^{-1} \frac{1}{3}$ Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\Rightarrow \tan ^{-1} 1+\tan ^{-1} \frac{1}{3}=\tan ^{-1}\left(\frac{1+\frac{1}{3}}{1 \frac{1}...
Read More →Write a value
Question: Write a value of $\int \log _{e} x d x$ Solution: $y=\int 1 \times \log _{e} x d x$ By using integration by parts Let, $\log _{e} x$ as Ist function and 1 as IInd function Use formula $\int I \times I I d x=I \int I I d x-\int\left(\frac{d}{d x} I\right)\left(\int I I d x\right) d x$ $y=\log _{e} x \int d x-\int\left(\frac{d}{d x} \log _{e} x\right)\left(\int d x\right) d x$ $y=\left(\log _{e} x\right) x-\int\left(\frac{1}{x}\right)(x) d x$ $y=x \log _{e} x-\int d x$ $y=x \log _{e} x-x...
Read More →Find the matrix A such that
Question: Find the matrix A such that $\left[\begin{array}{cc}2 -1 \\ 1 0 \\ -3 4\end{array}\right] \quad A=\left[\begin{array}{ccc}-1 -8 -10 \\ 1 -2 -5 \\ 9 22 15\end{array}\right]$ Solution: $A=\left[\begin{array}{lll}a b c \\ d e f\end{array}\right]$ $\left[\begin{array}{cc}2 -1 \\ 1 0 \\ -3 4\end{array}\right]\left[\begin{array}{lll}a b c \\ d e f\end{array}\right]=\left[\begin{array}{ccc}-1 -8 -10 \\ 1 -2 -5 \\ 9 22 15\end{array}\right]$ $\left[\begin{array}{ccc}2 a-d 2 b-e 2 c-f \\ a b c \...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\cot \left(\tan ^{-1} x+\cot ^{-1} x\right)=?$ A. 1 B. $\frac{1}{2}$ C. 0 D. none of these Solution: To Find: The value of $\cot \left(\tan ^{-1} x+\cot ^{-1} x\right)$ Let, $x=\cot \left(\tan ^{-1} x+\cot ^{-1} x\right)$ $\Rightarrow \mathrm{x}=\cot \left(\frac{\pi}{2}\right)\left(\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right)$ $\Rightarrow \mathrm{x}=0$...
Read More →Write a value
Question: Write a value of $\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x$ Solution: We know that $1+\sin 2 x=\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=(\sin x+\cos x)^{2}$ $y=\int \frac{\sin x+\cos x}{\sqrt{(\sin x+\cos x)^{2}}} d x$ $y=\int \frac{(\sin x+\cos x)}{(\sin x+\cos x)} d x$ $y=\int d x$ Use formula $\int c d x=c x$, where $c$ is constant $y=x+c$...
Read More →Find the value of a, b, c and d, if
Question: Find the value of a, b, c and d, if Solution: Given, $3\left[\begin{array}{ll}a b \\ c d\end{array}\right]=\left[\begin{array}{cc}a 6 \\ -1 2 d\end{array}\right]+\left[\begin{array}{cc}4 a+b \\ c+d 3\end{array}\right]$ $\left[\begin{array}{ll}3 a 3 b \\ 3 c 3 d\end{array}\right]=\left[\begin{array}{cc}a+4 6+a+b \\ -1+c+d 2 d+3\end{array}\right]$ Now, 3a = a + 4 ⇒ a = 2 3b = 6 + a + b 3b b = 8 ⇒ b = 4 And, 3d = 3 + 2d ⇒ d = 3 And, 3c = c + d 1 2c = 3 1 = 2 ⇒ c = 1 Hence, a = 2, b = 4, c...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=?$ A. $\frac{\pi}{2}$ B. $\pi$ C. $\frac{3 \pi}{2}$ D. $\frac{2 \pi}{3}$ Solution: To Find: The value of $\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ Let, $x=\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ $\Rightarrow \mathrm{x}=-\tan ^{-1}(1)+\left(\pi-\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right)$ $\left(\because \tan ^{-1}(-\theta)=-\tan...
Read More →If , then find A2 – 5A – 14I.
Question: If , then find A2 5A 14I. Hence, obtain A3. $A=\left[\begin{array}{cc}3 -5 \\ -4 2\end{array}\right]$ Solution: Given, $A=\left[\begin{array}{cc}3 -5 \\ -4 2\end{array}\right]$$\ldots$ (i) Now, $A^{2}=A \cdot A=\left[\begin{array}{cc}3 -5 \\ -4 2\end{array}\right]\left[\begin{array}{cc}3 -5 \\ -4 2\end{array}\right]=\left[\begin{array}{cc}29 -25 \\ -20 24\end{array}\right]$ $A^{2}-5 A-14 I=\left[\begin{array}{cc}29 -25 \\ -20 24\end{array}\right]-\left[\begin{array}{cc}15 -25 \\ -20 10...
Read More →Write a value of
Question: Write a value of $\int \frac{\sec ^{2} x}{(5+\tan x)^{4}} d x$ Solution: Let, $\tan x=t$ Differentiating both side with respect to $x$ $\frac{d t}{d x}=(\sec x)^{2} \Rightarrow d t=\sec ^{2} x d x$ $y=\int \frac{d t}{(5+t)^{4}}$ Use formula $\int \frac{1}{(a+t)^{n}} d t=\frac{(a+t)^{-n+1}}{-n+1}$ $y=\frac{(5+t)^{-3}}{-3}+c$ Again, put $t=\tan x$ $y=-\frac{1}{3(5+\tan x)^{3}}+c$...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=?$ A. $\frac{\pi}{2}$ B. $\pi$ C. $\frac{3 \pi}{2}$ D. none of these Solution: To Find: The value of $\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$ Let, $x=\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$ $\Rightarrow x=-\sin ^{-1}\left(\frac{1}{2}\right)+2\left[\pi-\cos ^{-1}\le...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: If $\cot ^{-1}\left(\frac{-1}{5}\right)=x$ then $\sin x=?$ A. $\frac{1}{\sqrt{26}}$ B. $\frac{5}{\sqrt{26}}$ C. $\frac{1}{\sqrt{24}}$ D. none of these Solution: Given: $\cot ^{-1} \frac{-1}{5}=x$ To Find: The value of $\sin x$ Since, $x=\cot ^{-1} \frac{-1}{5}$ $\Rightarrow \cot x=\frac{-1}{5}=\frac{\text { adjacent side }}{\text { opposite side }}$ By pythagorus theroem, (Hypotenuse $)^{2}=(\text { opposite side })^{2}+(\text ...
Read More →Write a value
Question: Write a value of $\int \frac{\left(\tan ^{-1} x\right)^{3}}{1+x^{2}} d x$ Solution: Let, $\tan ^{-1} x=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=\frac{1}{1+x^{2}}$ $\Rightarrow d t=\frac{d x}{1+x^{2}}$ $y=\int t^{3} d t$ Use formula $\int t^{n} d t=\frac{t^{n+1}}{n+1}$ $y=\frac{t^{4}}{4}+c$ Again, put $t=\tan ^{-1} x$ $y=\frac{\left(\tan ^{-1} x\right)^{4}}{4}+c$...
Read More →If find a matrix C such
Question: If find a matrix C such that 3A + 5B + 2C is a null matrix. $A=\left[\begin{array}{cc}1 5 \\ 7 12\end{array}\right]$ and $B=\left[\begin{array}{ll}9 1 \\ 7 8\end{array}\right]$ Solution: Lets consider a matrix C, such that $C=\left[\begin{array}{ll}a b \\ c d\end{array}\right]$ $3 A+5 B+2 C=O$ $\left[\begin{array}{cc}3 15 \\ 21 36\end{array}\right]+\left[\begin{array}{cc}45 5 \\ 35 40\end{array}\right]+\left[\begin{array}{cc}2 a 2 b \\ 2 c 2 d\end{array}\right]=\left[\begin{array}{ll}0...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}=?$ A. $\frac{\pi}{3}$ B. $\frac{\pi}{4}$ C. $\frac{3 \pi}{4}$ D. $\frac{2 \pi}{3}$ Solution: To Find: The value of $\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$ Let, $x=\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$ $\Rightarrow \mathrm{x}=\tan ^{-1}\left\{2 \cos \left(2\left(\frac{\pi}{6}\right)\right)\right...
Read More →If then find values of
Question: If then find values of x, y, z and w. $\left[\begin{array}{cc}x y 4 \\ z+6 x+y\end{array}\right]=\left[\begin{array}{cc}8 w \\ 0 6\end{array}\right]$ Solution: In the given matrix equation, Comparing the corresponding elements, we get x + y = 6, xy = 8, z + 6 = 0 and w = 4 From the first two equations, we have (6 y) . y = 8 y2 6y + 8 = 0 (y 2) (y 4) = 0 y = 2 or y = 4 Hence, x = 4 and x = 2 Also, z + 6 = 0 z = -6 and w = 4 Therefore, x = 2, y = 4 or x = 4, y = 2, z = -6 and w = 4...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=?$ A. $\frac{1}{\sqrt{5}}$ B. $\frac{2}{\sqrt{5}}$ C. $\frac{1}{\sqrt{10}}$ D. $\frac{2}{\sqrt{10}}$ Solution: To Find: The value of $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$ Let $x=\cos ^{-1} \frac{4}{5}$ $\Rightarrow \cos x=\frac{4}{5}$ Therefore $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$ becomes $\sin \left(\frac{1}{2} x\right), i . e \sin \left(\...
Read More →Find inverse, by elementary row operations (if possible),
Question: Find inverse, by elementary row operations (if possible), of the following matrices (i) $\left[\begin{array}{cc}1 3 \\ -5 7\end{array}\right]$ (ii) $\left[\begin{array}{cc}1 -3 \\ -2 6\end{array}\right]$ Solution: (i) Let $A=\left[\begin{array}{cc}1 3 \\ -5 7\end{array}\right]$ Now, $\left[\begin{array}{cc}1 3 \\ -5 7\end{array}\right]=\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right] A$ $\left[\begin{array}{cc}1 3 \\ 0 22\end{array}\right]=\left[\begin{array}{cc}1 0 \\ 5 1\end{array...
Read More →Write a value
Question: Write a value of $\int \frac{1}{1+2 \mathrm{e}^{\mathrm{x}}} \mathrm{dx}$ Solution: Take $e^{x}$ out from the denominator. $y=\int \frac{1}{e^{x}\left(e^{-x}+2\right)} d x$ $y=\int \frac{e^{-x}}{\left(e^{-x}+2\right)} d x$ Let, $e^{-x}+2=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=-e^{-x}$ $\Rightarrow-d t=e^{-x} d x$ $y=\int \frac{-d t}{t}$ Use formula $\int \frac{1}{t} d t=\ln t$ $Y=-\ln t+c$ Again, put $e^{-x}+2=t$ $Y=-\ln \left(e^{-x}+2\right)+c$ Note: Don't ...
Read More →Write a value of $int rac{1}{1+2 mathrm{e}^{mathrm{x}}} mathrm{dx}$
Question: Write a value of $\int \frac{1}{1+2 \mathrm{e}^{\mathrm{x}}} \mathrm{dx}$ Solution: Take $e^{x}$ out from the denominator. $y=\int \frac{1}{e^{x}\left(e^{-x}+2\right)} d x$ $y=\int \frac{e^{-x}}{\left(e^{-x}+2\right)} d x$ Let, $e^{-x}+2=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=-e^{-x}$ $\Rightarrow-d t=e^{-x} d x$ $y=\int \frac{-d t}{t}$ Use formula $\int \frac{1}{t} d t=\ln t$ $Y=-\ln t+c$ Again, put $e^{-x}+2=t$ $Y=-\ln \left(e^{-x}+2\right)+c$ Note: Don't ...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}=?$ A. 1 B. 0 C. $\frac{-1}{2}$ D. none of these Solution: To Find: The value of of $\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}$ Let, $x=\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}$ $\Rightarrow x=\sin \left\{\frac{\pi}{3}-\left(-\sin ^{-1} \frac{1}{2}\right)\right\}\left(\because \sin ^{-1}(-\theta)=-\sin ...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\cos \left(\tan ^{-1} \frac{3}{4}\right)=?$ A. $\frac{3}{5}$ B. $\frac{4}{5}$ C. $\frac{4}{9}$ D. none of these Solution: To Find: The value of $\cos \left(\tan ^{-1} \frac{3}{4}\right)$ Let $x=\tan ^{-1} \frac{3}{4}$ $\Rightarrow \tan x=\frac{3}{4}$ $\Rightarrow \tan x=\frac{3}{4}=\frac{\text { oppositeside }}{\text { adjacent side }}$ We know that by pythagorus theorem , (Hypotenuse $)^{2}=$ (opposite side $)^{2}+(\text { ad...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x}{\left(x^{2}+4\right) \sqrt{x^{2}+9}} d x$ Solution: assume $x^{2}+9=u^{2}$ $x d x=u d u$ $\int \frac{u d u}{\left(u^{2}-5\right) u}$ $\int \frac{d u}{\left(u^{2}-5\right)}$ Using identity $\int \frac{\mathrm{dz}}{(\mathrm{z})^{2}-1}=\frac{1}{2} \log \left|\frac{\mathrm{z}-1}{\mathrm{z}+1}\right|+\mathrm{c}$ $\frac{1}{2 \sqrt{5}} \log \left|\frac{u-\sqrt{5}}{u+\sqrt{5}}\right|+c$ Substituting $u=\sqrt{9+x^{2}}$ $\frac{1}{2 \sqrt{5}} \log \...
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