Solve for x :

Question: Solve for x : $\tan ^{-1} x=\sin ^{-1} \frac{1}{\sqrt{2}}$ Solution: To find: value of x Given: $\tan ^{-1} \mathrm{X}=\sin ^{-1} \frac{1}{\sqrt{2}}$ We know that $\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}$ Therefore, $\frac{\pi}{4}=\sin ^{-1} \frac{1}{\sqrt{2}}$ Substituting in the given equation, $\tan ^{-1} x=\frac{\pi}{4}$ $x=\tan \frac{\pi}{4}$ $\Rightarrow x=1$ Therefore, x = 1 is the required value of x....

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Find values of a and b if A = B,

Question: Find values of a and b if A = B, where $\mathrm{A}=\left[\begin{array}{cc}a+4 3 b \\ 8 -6\end{array}\right], \quad \mathrm{B}=\left[\begin{array}{cc}2 a+2 b^{2}+2 \\ 8 b^{2}-5 b\end{array}\right]$ Solution: Given, matrix A = matrix B Then their corresponding elements are equal. So, we have a11= b11; a + 4 = 2a + 2 ⇒ a = 2 a12= b12; 3b = b2+ 2 ⇒ b2 3b + 2 = 0 ⇒ b = 1, 2 a22= b22; -6 = b2 5b ⇒ b2 5b + 6 = 0 ⇒ b = 2, 3 Hence, a = 2 and b = 2 (common value)...

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Evaluate the following integral:

Question: Evaluate the following integral: $\int \frac{x+1}{x\left(1+x e^{x}\right)} d x$ Solution: Let, $\mathrm{I}=\int \frac{\mathrm{x}+1}{\mathrm{x}\left(1+\mathrm{xe}^{\mathrm{x}}\right)} \mathrm{dx}$ $\Rightarrow \quad I=\int \frac{(x+1)\left(1+x e^{x}-x e^{x}\right)}{x\left(1+x e^{x}\right)} d x$ $\Rightarrow, \quad I=\int \frac{(x+1)\left(1+x e^{x}\right)}{x\left(1+x e^{x}\right)} d x-\int \frac{(x+1)\left(x e^{x}\right)}{x\left(1+x e^{x}\right)} d x$ $\Rightarrow, \quad \mathrm{I}=\int ...

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Construct a 3 × 2 matrix whose elements

Question: Construct a 3 2 matrix whose elements are given by aij= ei.xsin jx Solution: Let A be a 3 x 2 matrix Such that, aij= ei.xsin jx; where where 1 i 3; 1 j 2 So, the terms are given as $a_{11}=e^{x} \sin x \quad a_{12}=e^{x} \sin 2 x$ $a_{21}=e^{2 x} \sin x \quad a_{22}=e^{2 x} \sin 2 x$ $a_{31}=e^{3 x} \sin x \quad a_{32}=e^{3 x} \sin 2 x$ Therefore, $A=\left[\begin{array}{ll}e^{x} \sin x e^{x} \sin 2 x \\ e^{2 x} \sin x e^{2 x} \sin 2 x \\ e^{3 x} \sin x e^{3 x} \sin 2 x\end{array}\right...

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Solve for x :

Question: Solve for x : $\cos \left(\sin ^{-1} x\right)=\frac{1}{2}$ Solution: To find: value of x Given: $\cos \left(\sin ^{-1} x\right)=\frac{1}{2}$ $\mathrm{LHS}=\cos \left(\sin ^{-1} \mathrm{x}\right)$ $=\cos \left(\cos ^{-1}\left(\sqrt{1-\mathrm{x}^{2}}\right)\right)$ $=\sqrt{1-\mathrm{x}^{2}}$ Therefore, $\sqrt{1-\mathrm{x}^{2}}=\frac{1}{2}$ Squaring both sides, $1-x^{2}=\frac{1}{4}$ $x^{2}=1-\frac{1}{4}$ $x^{2}=\frac{3}{4}$ $x=\pm \frac{\sqrt{3}}{2}$ Therefore, $x=\pm \frac{\sqrt{3}}{2}$ ...

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Construct a2 × 2 matrix where

Question: Construct a2 2matrix where (i) aij= (i 2j)2/ 2 (ii) aij= |-2i + 3j| Solution: We have, A = [aij]22 (i) Such that, aij= (i 2j)2/ 2; where 1 i 2; 1 j 2 So, the terms of the matrix are $a_{11}=\frac{(1-2)^{2}}{2}=\frac{1}{2}$ $a_{12}=\frac{(1-2 \times 2)^{2}}{2}=\frac{9}{2}$ $a_{21}=\frac{(2-2 \times 1)^{2}}{2}=0$ $a_{22}=\frac{(2-2 \times 2)^{2}}{2}=2$ Therefore, $A=\left[\begin{array}{ll}\frac{1}{2} \frac{9}{2} \\ 0 2\end{array}\right]$ (ii) Here, aij= |-2i + 3j| So, the terms of the ma...

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Solve for x:

Question: Solve for x: $\sin ^{-1} \frac{8}{x}+\sin ^{-1} \frac{15}{x}=\frac{\pi}{2}$ Solution: To find: value of x Given: $\sin ^{-1} \frac{8}{x}+\sin ^{-1} \frac{15}{x}=\frac{\pi}{2}$ We know $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$ Let $\sin ^{-1} \frac{8}{x}=\mathrm{P}$ $\Rightarrow \sin P=\frac{8}{x}$ Therefore, $\cos \mathrm{P}=\frac{\sqrt{\mathrm{x}^{2}-64}}{\mathrm{x}}$ $P=\cos ^{-1} \frac{\sqrt{x^{2}-64}}{x}$ $\cos ^{-1} \frac{\sqrt{x^{2}-64}}{x}+\sin ^{-1} \frac{15}{x}=\frac{\pi}{2}$ ...

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Evaluate the following integral:

Question: Evaluate the following integral: $\int \frac{1}{\sin x+\sin 2 x} d x$ Solution: Let, $I=\int \frac{1}{\sin x+\sin 2 x} d x$ $\Rightarrow I=\int \frac{1}{\sin x+2 \sin x \cos x} d x$ Multiplying and dividing by $\sin x$ $\Rightarrow I=\int \frac{\sin x}{\sin ^{2} x+2 \sin ^{2} x \cdot \cos x} d x$ $\Rightarrow I=\int \frac{\sin \mathrm{x}}{1-\cos ^{2} \mathrm{x}+2\left(1-\cos ^{2} \mathrm{x}\right) \cos \mathrm{x}} \mathrm{dx}$ Let $\cos x=t,-\sin x d x=d t$ $\Rightarrow 1=A(1+t)(1+2 t)...

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In the matrix A = , write :

Question: In the matrix A = , write : (i) The order of the matrix A (ii) The number of elements (iii) Write elements a23, a31, a12 Solution: For the given matrix, (i) The order of the matrix A is 3 x 3. (ii) The number of elements of the matrix = 3 x 3 = 9 (iii) Elements: a23= x2 y, a31= 0, a12= 1...

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If a matrix has 28 elements,

Question: If a matrix has 28 elements, what are the possible orders it can have? What if it has 13 elements? Solution: For a given matrix of order m x n, it has mn elements, where m and n are natural numbers. Here we have, m x n = 28 (m, n) = {(1, 28), (2, 14), (4, 7), (7, 4), (14, 2), (28, 1)} So, the possible orders are 1 x 28, 2 x 14, 4 x 7, 7 x 4, 14 x 2, 28 x 1. Also, if it has 13 elements, then m x n = 13 (m, n) = {(1, 13), (13, 1)} Thus, the possible orders are 1 x 13, 13 x 1. $\left[\beg...

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Solve for x:

Question: Solve for x: $\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$ Solution: To find: value of x Formula Used: $2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ Given: $\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$ LHS $=\cos \left(2 \sin ^{-1} x\right)$ Let $\theta=\sin ^{-1} x$ So, $x=\sin \theta \ldots$ (1) $\mathrm{LHS}=\cos (2 \theta)$ $=1-2 \sin ^{2} \theta$ Substituting in the given equation, $1-2 \sin ^{2} \theta=\frac{1}{9}$ $2 \sin ^{2} \theta=\frac{8}{9}$ $\sin ^{2} \theta...

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The value of

Question: The value of sin (2 tan-1(0.75)) is equal to (a) 0.75 (b) 1.5 (c) 0.96 (d) sin 1.5 Solution: (c) 0.96 We have, sin (2 tan-1(0.75)) $=\sin \left(2 \tan ^{-1} \frac{3}{4}\right)=\sin \left(\sin ^{-1} \frac{2 \cdot \frac{3}{4}}{1+\frac{9}{16}}\right)\left(\because 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}\right)$ $=\sin \left(\sin ^{-1} \frac{3 / 2}{25 / 16}\right)=\sin \left(\sin ^{-1} \frac{24}{25}\right)=\frac{24}{25}=0.96$...

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Evaluate the following integral:

Question: Evaluate the following integral: $\int \frac{1}{\sin x(3+2 \cos x)} d x$ Solution: Let, $\mathrm{I}=\int \frac{1}{\sin \mathrm{x}(3+2 \cos \mathrm{x})} \mathrm{dx}$ Multiplying and dividing by $\sin x$ $\therefore \mathrm{I}=\int \frac{\sin \mathrm{x}}{\sin ^{2} \mathrm{x}(3+2 \cos \mathrm{x})} \mathrm{dx}$ $\therefore \mathrm{I}=\int \frac{\sin \mathrm{x}}{\left(1-\cos ^{2} \mathrm{x}\right)(3+2 \cos \mathrm{x})} \mathrm{dx}$ Let $\cos x=t,-\sin x d x=d t$ So, $I=\int \frac{d t}{\left...

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If cos (sin-1 2/5 + cos-1 x) = 0,

Question: If cos (sin-12/5 + cos-1x) = 0, then x is equal to (a) 1/5 (b) 2/5 (c) 0 (d) 1 Solution: (b) 2/5 Given, cos (sin-12/5 + cos-1x) = 0 So, this can be rewritten as sin-12/5 + cos-1x = cos-10 sin-12/5 + cos-1x = /2 cos-1x = /2 sin-12/5 cos-1x =cos-12/5 [Since, cos-1x + sin-1x = /2] Hence, x = 2/5...

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Solve for x:

Question: Solve for x: $\cos \left(\sin ^{-1} x\right)=\frac{1}{9}$ Solution: To find: value of $x$ Given: $\cos \left(\sin ^{-1} x\right)=\frac{1}{9}$ LHS $=\cos \left(\sin ^{-1} x\right) \ldots$ (1) Let $\sin \theta=x$ Therefore $\theta=\sin ^{-1} \times \ldots$ (2) From the figure, $\cos \theta=\sqrt{1-x^{2}}$ $\Rightarrow \theta=\cos ^{-1} \sqrt{1-x^{2}} \ldots$ (3) From (2) and (3), $\sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}} \ldots$ (4) Substituting (4) in (1), we get $\mathrm{LHS}=\cos \left(...

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The domain of the function

Question: The domain of the function by f(x) = sin-1(x 1) is (a) [1, 2] (b) [-1, 1] (c) [0, 1] (d) none of these Solution: (a) [1, 2] We know that, sin-1x is defined for x [-1, 1] So, f(x) = sin-1(x 1) is defined if 0 (x 1) 1 0 x 1 1 1 x 2 Hence, x [1, 2]...

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The domain of the function

Question: The domain of the function cos-1 (2x 1) is (a) [0, 1] (b) [-1, 1] (c) [-1, 1] (d) [0, ] Solution: (a) [0, 1] Since, cos-1 x is defined for x [-1, 1] So, f(x) = cos-1 (2x 1) is defined if -1 2x 1 1 0 2x 2 Hence, 0 x 1...

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Evaluate the following integral:

Question: Evaluate the following integral: $\int \frac{1}{\cos x(5-4 \sin x)} d x$ Solution: Let, $I=\int \frac{d x}{\cos x(5-4 \sin x)}$ Multiplying and dividing by $\cos x$ $\operatorname{Let}, I=\int \frac{\cos x d x}{\cos ^{2} x(5-4 \sin x)}$ $\Rightarrow I=\int \frac{\cos x d x}{\left(1-\sin ^{2} x\right)(5-4 \sin x)}$ Let, $\sin x=t, \cos x d x=d t$ $\therefore I=\int \frac{d t}{\left(1-t^{2}\right)(5-4 t)}$ Now, let $\frac{1}{\left(1-\mathrm{t}^{2}\right)(5-4 \mathrm{t})}=\frac{\mathrm{A}...

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The value of

Question: The value of sin-1cos 33/5 is (a) 3/5 (b) -7/5 (c) /10 (d) -/10 Solution: (d) -/10 $\sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)=\sin ^{-1}\left(\cos \left(6 \pi+\frac{3 \pi}{5}\right)\right)=\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right)$ $=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right]$ $=\sin ^{-1}\left(\sin \left(-\frac{\pi}{10}\right)\right)$ $=-\frac{\pi}{10} \quad\left(\because \sin ^{-1}(\sin x)=x\right.$, for $\left.x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}...

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Solve for x:

Question: Solve for x: $\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3}$ Solution: To find: value of x Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y1$ Given: $\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3}$ $\mathrm{LHS}=\tan ^{-1}\left(\frac{2+\mathrm{x}+2-\mathrm{x}}{1-\{(2+\mathrm{x}) \times(2-\mathrm{x})\}}\right)$ $=\tan ^{-1} \frac{4}{1-\left(4-2 x+2 x-x^{2}\right)}$ $=\tan ^{-1} \frac{4}{x^{2}-3}$ Therefore, $\tan ^{-1} \frac{...

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Solve for x:

Question: Solve for x: $\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3}$ Solution: To find: value of x Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y1$ Given: $\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3}$ $\mathrm{LHS}=\tan ^{-1}\left(\frac{2+\mathrm{x}+2-\mathrm{x}}{1-\{(2+\mathrm{x}) \times(2-\mathrm{x})\}}\right)$ $=\tan ^{-1} \frac{4}{1-\left(4-2 x+2 x-x^{2}\right)}$ $=\tan ^{-1} \frac{4}{x^{2}-3}$ Therefore, $\tan ^{-1} \frac{...

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Evaluate the following integral:

Question: Evaluate the following integral: $\int \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x$ Solution: Let, $I=\int \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x$ Let $x^{2}=y$ Thus, $\frac{\mathrm{x}^{2}}{\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)\left(\mathrm{x}^{2}+\mathrm{b}^{2}\right)}=\frac{\mathrm{y}}{\left(\mathrm{y}+\mathrm{a}^{2}\right)\left(\mathrm{y}+\mathrm{b}^{2}\right)}$ Now, let $\frac{y}{\left(y+a^{2}\right)\left(y+b^{2}\right)}=\frac...

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Solve for x

Question: Solve for x: $\tan ^{-1}(\mathrm{x}+1)+\tan ^{-1}(\mathrm{x}-1)=\tan ^{-1} \frac{8}{31}$ Solution: To find: value of x Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y1$ Given: $\tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1} \frac{8}{31}$ $\mathrm{LHS}=\tan ^{-1}\left(\frac{\mathrm{x}+1+\mathrm{x}-1}{1-\{(\mathrm{x}+1) \times(\mathrm{x}-1)\}}\right)$ $=\tan ^{-1} \frac{2 x}{1-\left(x^{2}-x+x-1\right)}$ $=\tan ^{-1} \frac{2 x}{2-x^{2}}$ Therefore...

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If 3 tan-1 x + cot-1 x = π,

Question: If 3 tan-1x + cot-1x = , then x equals (a) 0 (b) 1 (c) -1 (d) Solution: (b) 1 Given, 3 tan-1x + cot-1x = 2 tan-1x + tan-1x + cot-1x = 2 tan-1x + /2 = (As tan-1+ cot-1= /2) 2 tan-1x = /2 tan-1x = /4 x = 1...

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Which of the following is the principal

Question: Which of the following is the principal value branch of cosec-1x? (a) (-/2, /2) (b) [0, ] {/2} (c) [-/2, /2] (d) [-/2, /2] {0} Solution: (d) [-/2, /2] {0} As the principal branch of cosec-1x is [-/2, /2] {0}....

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