Prove that
Question: $\left|\begin{array}{ccc}a+x y z \\ x a+y z \\ x y a+z\end{array}\right|$ Solution: Given, $\quad\left|\begin{array}{ccc}a+x y z \\ x a+y z \\ x y a+z\end{array}\right|$ [Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ ] $=\left|\begin{array}{ccc}a+x+y+z y z \\ a+x+y+z a+y z \\ a+x+y+z y a+z\end{array}\right|$ $=(a+x+y+z)\left|\begin{array}{ccc}1 y z \\ 1 a+y z \\ 1 y a+z\end{array}\right|$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ ] $=(a+x+y+z)\left|...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: If $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$ then $\left(\cos ^{-1} x+\cos ^{-1} y\right)=?$ A. $\frac{\pi}{6}$ B. $\frac{\pi}{3}$ C. $\pi$ D. $\frac{2 \pi}{3}$ Solution: Given: $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$ To Find: The value of $\cos ^{-1} x+\cos ^{-1} y$ Since we know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x$ Similarly $\cos ^{-1} y=\frac{\pi}{2}-\sin...
Read More →Prove the following
Question: $\left|\begin{array}{cc}x^{2}-x+1 x-1 \\ x+1 x+1\end{array}\right|$ Solution: Given, $\left|\begin{array}{cc}x^{2}-x+1 x-1 \\ x+1 x+1\end{array}\right|$ [Applying $C_{1} \rightarrow C_{1}-C_{2}$ ] $=\left|\begin{array}{cc}x^{2}-2 x+2 x-1 \\ 0 x+1\end{array}\right|$ = (x2 2x + 2) . (x + 1) (x 1) . 0 = x3 2x2+ 2x + x2 2x + 2 = x3 x2+ 2...
Read More →Write a value
Question: Write a value of $\int \frac{\sin \mathrm{x}}{\cos ^{3} \mathrm{x}} \mathrm{dx}$ Solution: Let, $\cos x=t$ Differentiating both sides with respect to $\mathrm{x}$ $\frac{d t}{d x}=-\sin x$ $\Rightarrow-d t=\sin x d x$ $y=\int \frac{-1}{t^{3}} d t$ Use formula $\int \frac{1}{t^{n}} d t=\frac{t^{-n+1}}{-n+1}$ $y=-\frac{t^{-2}}{-2}+c$ Again, put $t=\cos x$ $y=\frac{1}{2(\cos x)^{2}}+c$...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: If $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ then $x=?$ A. 1 B. $-1$ C. 0 D. $\frac{1}{2}$ Solution: To Find: The value of $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\Rightarrow \tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\tan ^{-1}\left(\frac{(1+x)+(1-x)}{1-(1+x)(1-x)}\right)$ $=\tan ^{-1}\left(\frac{2}{1-\left(1-x^{2}\right)}\right)$ $=\ta...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: If $\tan ^{-1} \mathrm{x}=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3}$ then $\mathrm{x}=?$ A. $\frac{1}{2}$ B. $\frac{1}{4}$ C. $\frac{1}{6}$ D. None of these Solution: To Find: The value of $\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3}$ Now, $\tan ^{-1} x=\tan ^{-1} 1-\tan ^{-1} \frac{1}{3}\left(\because \tan \frac{\pi}{4}=1\right)$ Since we know that $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ $\Rightar...
Read More →Find x, y, z if A = satisfies A¢ = A-1.
Question: Find x, y, z if A = satisfies A = A-1. $\left[\begin{array}{ccc}0 2 y z \\ x y -z \\ x -y z\end{array}\right]$ Solution: Matrix A is such that A = A-1 AA = I $\left[\begin{array}{ccc}0 2 y z \\ x y -z \\ x -y z\end{array}\right]\left[\begin{array}{ccc}0 x x \\ 2 y y -y \\ z -z z\end{array}\right]=\left[\begin{array}{ccc}1 0 0 \\ 0 1 0 \\ 0 0 1\end{array}\right]$ $\left[\begin{array}{ccc}4 y^{2}+z^{2} 2 y^{2}-z^{2} -2 y^{2}+z^{2} \\ 2 y^{2}-z^{2} x^{2}+y^{2}+z^{2} x^{2}-y^{2}-z^{2} \\ -...
Read More →Write a value
Question: Write a value of $\int \frac{1+\log x}{3+x \log x} d x$ Solution: Let, $x(\log x)=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=x \frac{1}{x}+\log x=1+\log x$ $\Rightarrow d t=(1+\log x) d x$ $y=\int \frac{1}{3+t} d t$ Use formula $\int \frac{1}{a+t} d t=\log (a+t)$ $y=\log (3+t)+c$ Again, put $t=x(\log x)$ $y=\log (3+x(\log x))+c$...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\sin \left[2 \sin ^{-1} \frac{4}{5}\right]$ A. $\frac{12}{25}$ B. $\frac{16}{25}$ C. $\frac{24}{25}$ D. None of these Solution: To Find: The value of $\sin \left(2 \sin ^{-1} \frac{4}{5}\right)$ Let, $x=\sin ^{-1} \frac{4}{5}$ $\Rightarrow \sin x=\frac{4}{5}$ We know that, $\cos x=\sqrt{1-\sin ^{2} x}$ $=\sqrt{1-\left(\frac{4}{5}\right)^{2}}$ $=\frac{3}{5}$ Now since, $x=\sin ^{-1} \frac{4}{5}$, hence $\sin \left(2 \sin ^{-1} ...
Read More →If AB = BA for any two square matrices,
Question: If AB = BA for any two square matrices, prove by mathematical induction that (AB)n= AnBn. Solution: Let P(n) : (AB)n= AnBn So, P(1) : (AB)1= A1B1 AB = AB So, P(1) is true. Let P(n) is true for some k N Now, $(A B)^{k+1}=(A B)^{k}(A B)$(using (i)) $=A^{k} R^{k}(A B)$ $=A^{k} B^{k-1}(B A) B$ $=A^{k} B^{k-1}(A B) B \quad($ as given $A B=B A)$ $=A^{k} B^{k-1} A B^{2}$ $=A^{k} B^{k-2}(B A) B^{2}$ $=A^{k} B^{k-2} A B B^{2}$ $=A^{k} B^{k-2} A B^{3}$ .......... .......... $=A^{k+1} B^{k+1}$ He...
Read More →Write a value
Question: Write a value of $\int \frac{a^{x}}{3+a^{x}} d x$ Solution: Let, $3+a^{x}=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=a^{x} \log a$ $\Rightarrow \frac{d t}{\log a}=a^{x} d x$ $y=\int \frac{1}{(\log a) t} d t$ Use formula $\int \frac{1}{t} d t=\log t$ $y=\frac{\log t}{\log a}+c$ Again, put $t=3+a^{x}$ $y=\frac{\log \left(3+a^{x}\right)}{\log a}+c$...
Read More →If A, B are square matrices of same order
Question: If A, B are square matrices of same order and B is a skew-symmetric matrix, show that ABA is skew symmetric. Solution: Given, A and B are square matrices such that B is a skew-symmetric matrix So, B = -B Now, we have to prove that ABA is a skew-symmetric matrix. (ABA) = AB (A) [Since, (AB) = BA] = A (-B)A = -ABA Hence, ABA is a skew-symmetric matrix....
Read More →If A is square matrix such
Question: If A is square matrix such that A2= A, show that (I + A)3= 7A + I. Solution: We know that, A . I = I . A So, A and I are commutative. Thus, we can expand (I + A)3like real numbers expansion. So, (I + A)3= I3+ 3I2A + 3IA2+ A3 = I + 3IA + 3A2+ AA2(As In= I, n N) = I + 3A + 3A + AA = I + 3A + 3A + A2= I + 3A + 3A + A = I + 7A...
Read More →Write a value
Question: Write a value of $\int \frac{\sin 2 x}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x$ Solution: We know that $\cos ^{2} x=1-\sin ^{2} x$ $\left(a^{2} \sin ^{2} x+b^{2} \cos ^{2} x\right)=a^{2} \sin ^{2} x+b^{2}\left(1-\sin ^{2} x\right)$ $=\left(a^{2}-b^{2}\right) \sin ^{2} x+b^{2}$ $y=\int \frac{\sin 2 x}{\left(a^{2}-b^{2}\right)(\sin x)^{2}+b^{2}} d x$ Let, $\sin ^{2} x=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=2 \sin x \cos x$ $=\sin 2 x$ $\Rightarrow d t=\sin 2 ...
Read More →Write a value We know that $cos ^{2} x=1-sin ^{2} x$
Question: Write a value of $\int \frac{\sin 2 x}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x$ Solution: We know that $\cos ^{2} x=1-\sin ^{2} x$ $\left(a^{2} \sin ^{2} x+b^{2} \cos ^{2} x\right)=a^{2} \sin ^{2} x+b^{2}\left(1-\sin ^{2} x\right)$ $=\left(a^{2}-b^{2}\right) \sin ^{2} x+b^{2}$ $y=\int \frac{\sin 2 x}{\left(a^{2}-b^{2}\right)(\sin x)^{2}+b^{2}} d x$ Let, $\sin ^{2} x=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=2 \sin x \cos x$ $=\sin 2 x$ $\Rightarrow d t=\sin 2 ...
Read More →If P (x) = then show that
Question: If P (x) = then show that P (x) . P (y) = P (x + y) = P (y) . P (x). $\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]$ Solution: Given, $P(x)=\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]$ So, $\quad P(y)=\left[\begin{array}{cc}\cos y \sin y \\ -\sin y \cos y\end{array}\right]$ Now, $P(x) \cdot P(y)=\left[\begin{array}{cc}\cos x \sin x \\ -\sin x \cos x\end{array}\right]\left[\begin{array}{cc}\cos y \sin y \\ -\sin y \cos y\end{array}...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\sin \left[2 \tan ^{-1} \frac{5}{8}\right]$ A. $\frac{25}{64}$ B. $\frac{80}{89}$ C. $\frac{75}{128}$ D. none of these Solution: To Find: The value of $\sin \left(2 \tan ^{-1} \frac{5}{8}\right)$ Let, $x=\sin \left(2 \tan ^{-1} \frac{5}{8}\right)$ We know that $2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ $\Rightarrow x=\sin \left(\sin ^{-1}\left(\frac{2\left(\frac{5}{8}\right)}{1+\left(\frac{5}{8}\right)^{2}}\ri...
Read More →Write a value
Question: Write a value of $\int \frac{\cos x}{\sin x \log \sin x} d x$ Solution: Let $\log (\sin x)=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=\frac{\cos x}{\sin x} \Rightarrow d t=\frac{\cos x}{\sin x} d x$ $y=\int \frac{1}{t} d t$ Use formula $\int \frac{1}{t} d t=\log t$ $y=\log t+c$ Again, put $t=\log (\sin x)$ $y=\log (\log (\sin x))+c$...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\cos \left(2 \tan ^{-1} \frac{1}{2}\right)=?$ A. $\frac{3}{5}$ B. $\frac{4}{5}$ C. $\frac{7}{8}$ D. none of these Solution: To Find: The value of $\cos \left(2 \tan ^{-1} \frac{1}{2}\right)$ Let, $x=\cos \left(2 \tan ^{-1} \frac{1}{2}\right)$ $\Rightarrow x=\cos \left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2}\right)$ Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\Rightarrow \tan ^...
Read More →Write a value
Question: Write a value of $\int\left(e^{x \log _{4} a}+e^{a \log _{4} x}\right) d x$ Solution: We know that by using property of logarithm $e^{x \log _{e} a}=e^{\log _{e} a^{x}}=a^{x}$ and $e^{a \log _{e} x}=e^{\log _{e} x^{a}}=x^{a}$ $y=\int a^{x}+x^{a} d x$ $y=\int a^{x} d x+\int x^{a} d x$ Use formula $\int a^{x} d x=\frac{a^{x}}{\log a}$ and $\int x^{a} d x=\frac{x^{a+1}}{a+1}$ $y=\frac{a^{x}}{\log a}+\frac{x^{a+1}}{a+1}+c$...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $2 \tan ^{-1} \frac{1}{3}=?$ A. $\tan ^{-1} \frac{3}{2}$ B. $\tan ^{-1} \frac{3}{4}$ C. $\tan ^{-1} \frac{4}{3}$ D. none of these Solution: To Find: The value of $2 \tan ^{-1} \frac{1}{3}$ i.e, $\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{3}$ Let, $x=\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{3}$ 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}{...
Read More →If the matrix is a skew symmetric matrix,
Question: If the matrix is a skew symmetric matrix, find the values of a, b and c. $\left[\begin{array}{ccc}0 a 3 \\ 2 b -1 \\ c 1 0\end{array}\right]$ Solution: Let $A=\left[\begin{array}{ccc}0 a 3 \\ 2 b -1 \\ c 1 0\end{array}\right]$ As, $A$ is skew - symmetric matrix. So, we have $A^{\prime}=-A$ Then, $\Rightarrow \quad\left[\begin{array}{ccc}0 2 c \\ a b 1 \\ 3 -1 0\end{array}\right]=-\left[\begin{array}{ccc}0 a 3 \\ 2 b -1 \\ c 1 0\end{array}\right]$ $\Rightarrow \quad\left[\begin{array}{c...
Read More →Write a value
Question: Write a value of $\int \mathrm{e}^{2 \mathrm{x}^{2}+\ln \mathrm{x}} \mathrm{dx}$ Solution: We know that $\mathrm{e}^{\mathrm{a}+\mathrm{b}}=\mathrm{e}^{\mathrm{a}} \mathrm{e}^{\mathrm{b}}$ $y=\int e^{2 x^{2}} e^{\ln x} d x$ $y=\int e^{2 x^{2}} x d x$ Let, $x^{2}=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=2 x$ $\Rightarrow \frac{1}{2} d t=x d x$ $y=\int \frac{1}{2} e^{2 t} d t$ Use formula $\int e^{a+b t}=\frac{e^{a+b t}}{b}$ $y=\frac{1}{2} \frac{e^{2 t}}{2}+c$ A...
Read More →Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=?$ A. $\frac{\pi}{3}$ B. $\frac{\pi}{4}$ C. $\frac{\pi}{2}$ D. $\frac{2 \pi}{3}$ Solution: To Find: The value of $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$ Let, $x=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$ 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{\frac{2}{...
Read More →If A = and A-1 = A’,
Question: If A = and A-1= A, find value of a. $\left[\begin{array}{cc}\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha\end{array}\right]$ Solution: $A=\left[\begin{array}{cc}\cos \alpha \sin \alpha \\ -\sin \alpha \cos \alpha\end{array}\right]$ and $A^{\prime}=\left[\begin{array}{cc}\cos \alpha -\sin \alpha \\ \sin \alpha \cos \alpha\end{array}\right]$ Also, $A^{-1}=A^{\prime}$ $A A^{-1}=A A^{\prime}$ $I=A A^{\prime}$ $\left[\begin{array}{ll}1 0 \\ 0 1\end{array}\right]=\left[\begin{array}{cc...
Read More →