Mark the tick against the correct answer in the following:
Question: Mark the tick against the correct answer in the following: Range of $\operatorname{coses}^{-1} \times$ is A. $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ B. $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ C. $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}$ D. None of these Solution: To Find: The range of $\operatorname{cosec}^{-1}(x)$ Here,the inverse function is given by $y=\mathrm{f}^{-1}(x)$ The graph of the function $y=\operatorname{cosec}^{-1}(x)$ can be obtained from the graph of ...
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Question: Write a value of $\int \sqrt{x^{2}-9} d x$ Solution: we know that $\int \sqrt{x^{2}-a^{2}} d x=\frac{x \sqrt{x^{2}-a^{2}}}{2}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+c$ Given $\int \sqrt{x^{2}-9} d x$ $=\int \sqrt{x^{2}-3^{2}} d x$ $=\frac{x \sqrt{x^{2}-3^{2}}}{2}-\frac{3^{2}}{2} \log \left|x+\sqrt{x^{2}-3^{2}}\right|$ $=\frac{x \sqrt{x^{2}-9}}{2}-\frac{9}{2} \log \left|x+\sqrt{x^{2}-9}\right|+c$...
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Question: Mark the tick against the correct answer in the following: Range of $\sec ^{-1} \times$ is A. $\left[0, \frac{\pi}{2}\right]$ B. $[0, \pi]$ C. $[0, \pi]-\left\{\frac{\pi}{2}\right\}$ D. None of these Solution: To Find:The range of $\sec ^{-1}(x)$ Here,the inverse function is given by $\mathrm{y}=\mathrm{f}^{-1}(x)$ The graph of the function $y=\sec ^{-1}(x)$ can be obtained from the graph of $Y=\sec x$ by interchanging $x$ and $y$ axes.i.e, if $(a, b)$ is a point on $Y=\sec x$ then $(b...
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Question: Write a value of $\int \sqrt{9+x^{2}} d x$. Solution: we know that $\int \sqrt{x^{2}+a^{2}} d x=\frac{x \sqrt{x^{2}-a^{2}}}{2}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \mid$ Given $\int x^{2}+9$ $=\int x^{2}+3^{2}$ $=\frac{x \sqrt{x^{2}+3^{2}}}{2}+\frac{3^{2}}{2} \log \left|x+\sqrt{x^{2}+3^{2}}\right|$ $=\frac{x \sqrt{x^{2}+9}}{2}+\frac{9}{2} \log \left|x+\sqrt{x^{2}+9}\right|+c$...
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Question: $\left|\begin{array}{ccc}y+z z y \\ z z+x x \\ y x x+y\end{array}\right|=4 x y z$ Solution: Given, $\quad\left|\begin{array}{ccc}y+z z y \\ z z+x x \\ y x x+y\end{array}\right|$ [Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ ] $=\left|\begin{array}{ccc}2(y+z) z y \\ 2(z+x) z+x x \\ 2(y+x) x x+y\end{array}\right|$ $=2\left|\begin{array}{ccc}y+z z y \\ z+x z+x x \\ x+y x \dot{x}+y\end{array}\right|$ Now, [Applying $C_{1} \rightarrow C_{1}-C_{2}$ ] $=2\left|\begin{array}{ccc}y z y \\ 0 z...
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Question: Mark the tick against the correct answer in the following: Range of $\tan ^{-1} \mathrm{x}$ is A. $\left(0, \frac{\pi}{2}\right)$ B. $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ C. $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ D. None of these Solution: To Find: The range of $\tan ^{-1} \mathrm{x}$ Here,the inverse function is given by $y=f^{-1}(x)$ The graph of the function $y=\tan ^{-1}(x)$ can be obtained from the graph of $Y=\tan x$ by interchanging $x$ and $y$ axes.i.e, if $(a, b)$...
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Question: Write a value of $\int \sqrt{4-\mathrm{x}^{2}} \mathrm{dx}$. Solution: we know that $\int \sqrt{a^{2}-x^{2}} d x=\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{x^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c$ Given $\int \sqrt{4-x^{2}}$ $=\int \sqrt{2^{2}-x^{2}}$ $=\frac{x \sqrt{2^{2}-x^{2}}}{2}+\frac{x^{2}}{2} \sin ^{-1}\left(\frac{x}{2}\right)$ $=\frac{x \sqrt{4-x^{2}}}{2}+\frac{x^{2}}{2} \sin ^{-1}\left(\frac{x}{2}\right)+c$...
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Question: Mark the tick against the correct answer in the following: Range of $\cos ^{-1} x$ is A. $[0, \pi]$ B. $\left[0, \frac{\pi}{2}\right]$ C. $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ D. None of these Solution: To Find: The range of $\cos ^{-1} x$ Here,the inverse function is given by $y=f^{-1}(x)$ The graph of the function $y=\cos ^{-1}(x)$ can be obtained from the graph of $Y=\cos x$ by interchanging $x$ and $y$ axes.i.e, if $(a, b)$ is a point on $Y=\cos x$ then $(b, a)$ is the point...
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Question: Write a value of $\int \mathrm{e}^{\mathrm{ax}} \mid$ a $f(x)+f^{\prime}(x) \mid d x$. Solution: given $\int e^{a x}\left|a f(x)+f^{\prime}(x)\right| d x$ $=a \int e^{a x} f(x) d x+\int e^{a x} f^{\prime}(x) d x$ $=a\left[f(x) \frac{e^{a x}}{a}-\int f^{\prime}(x) \frac{e^{a x}}{a} d x\right]+\int e^{a x} f^{\prime}(x) d x$ $=f(x) e^{a x}+c$...
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Question: Write a value of $\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x$. Solution: given $\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x$ $=\int \frac{e^{x}}{x} d x-\int \frac{e^{x}}{x^{2}} d x$ $=\int \frac{e^{x}}{x} d x-\left[\frac{e^{x}}{x^{2}}-\int-\frac{e^{x}}{x}\right]+c$ $=-\frac{e^{x}}{x^{2}}+c$...
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Question: Mark the tick against the correct answer in the following: Range of $\sin ^{-1} x$ is A. $\left[0, \frac{\pi}{2}\right]$ B. $[0, \pi]$ C. $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ D. None of these Solution: To Find: The range of $\sin ^{-1} x$ Here,the inverse function is given by $y=f^{-1}(x)$ The graph of the function $y=\sin ^{-1}(x)$ can be obtained from the graph of $Y=\sin x$ by interchanging $x$ and $y$ axes.i.e, if $(a, b)$ is a point on $Y=\sin x$ then $(b, a)$ is The point...
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Question: Write a value of $\int e^{a x} \cos b x d x .5$ Solution: we know $\int f(x) g(x)=f(x) \int g(x)-\int f^{\prime}(x) \int g(x)$ Let $\int e^{a x} \cos b x d x=i$ Given that $\int e^{a x} \cos b x d x$ $i=\cos b x \int e^{a x}-\int-b \sin b x \int e^{a x}$ $i=\cos b x \frac{e^{a x}}{a}+\int b \sin b x \frac{e^{a x}}{a}$ $i=\cos b x \frac{e^{a x}}{a}+\frac{1}{a}\left[b \sin b x \frac{e^{a x}}{a}-\frac{b^{2}}{a} \int e^{a x} \cos b x d x\right]$ $\mathrm{I}=\cos b x \frac{e^{a x}}{a}+\frac...
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Question: $\left|\begin{array}{ccc}y^{2} z^{2} y z y+z \\ z^{2} x^{2} z x z+x \\ x^{2} y^{2} x y x+y\end{array}\right|=0$ Solution: From the given, [Multiplying $R_{1}, R_{2}, R_{3}$ by $x, y, z$ respectively] $=\frac{1}{x y z}\left|\begin{array}{ccc}x y^{2} z^{2} x y z x y+x z \\ x^{2} y z^{2} x y z y z+x y \\ x^{2} y^{2} z x y z x z+y z\end{array}\right|$ Next [Taking $(x y z)$ common from $C_{1}$ and $C_{2}$ ] $=\frac{1}{x y z}(x y z)^{2}\left|\begin{array}{lll}y z 1 x y+x z \\ x z 1 y z+x y ...
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Question: Mark the tick against the correct answer in the following: $\cos ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}=?$ A. $\frac{\pi}{6}$ B. $\frac{\pi}{4}$ C. $\frac{\pi}{3}$ D. $\frac{3 \pi}{4}$ Solution: To Find: The value of $\cot ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}$ Now $\cot ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}$ can be written in terms of tan inverse as $\cot ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}=\tan ^{-1} \frac{1}{9}+\tan ^{...
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Question: Write a value of $\int \mathrm{e}^{2 x} \sin b x d x$ Solution: we know $\int f(x) g(x)=f(x) \int g(x)-\int f^{\prime}(x) \int g(x)$ Let $\int e^{a x} \sin b x d x=i$ Given that $\int e^{a x} \sin b x d x$ $i=\sin b x \int e^{a x}-\int b \cos b x \int e^{a x}$ $i=\sin b x \frac{e^{a x}}{a}-\int b \cos b x \frac{e^{a x}}{a}$ $i=\sin b x \frac{e^{a x}}{a}-\frac{1}{a}\left[b \cos b x \frac{e^{a x}}{a}-\frac{b^{2}}{a} \int e^{a x} \sin b x d x\right]$ $\mathrm{i}=\sin b x \frac{e^{a x}}{a}...
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Question: Mark the tick against the correct answer in the following: $\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}=?$ A. $\frac{13}{6}$ B. $\frac{17}{6}$ C. $\frac{19}{6}$ D. $\frac{23}{6}$ Solution: To Find: The value of $\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}$ Let $x=\cos ^{-1} \frac{4}{5}$ $\Rightarrow \cos x=\frac{4}{5}=\frac{\text { adjacent side }}{\text { hypotenuse }}$ By pythagorus theroem, (Hypotenuse ) $^{2}=$ (opposite side $)^{2}+(\tex...
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Question: Write a value of $\int \frac{1}{x(\log x)^{n}} d x$ Solution: Let, $\log x=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=\frac{1}{x}$ $\Rightarrow d t=\frac{1}{x} d x$ $y=\int \frac{1}{t^{n}} d t$ Use formula $\int \frac{1}{t^{n}} d t=\frac{t^{-n+1}}{-n+1}$ $y=\frac{t^{-n+1}}{-n+1}+c$ Again, put $t=\log x$ $y=\frac{(\log x)^{-n+1}}{-n+1}+c$...
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Question: $\left|\begin{array}{ccc}a-b-c 2 a 2 a \\ 2 b b-c-a 2 b \\ 2 c 2 c c-a-b\end{array}\right|$ Solution: Given, $\left|\begin{array}{ccc}a-b-c 2 a 2 a \\ 2 b b-c-a 2 b \\ 2 c 2 c c-a-b\end{array}\right|$ [Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ ] $=\left|\begin{array}{ccc}a+b+c a+b+c a+b+c \\ 2 b b-c-a 2 b \\ 2 c 2 c c-a-b\end{array}\right|$ [Taking $(a+b+c)$ common from the first row] $=(a+b+c)\left|\begin{array}{ccc}1 1 1 \\ 2 b b-c-a 2 b \\ 2 c 2 c c-a-b\end{array}\right|$ [Appl...
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Question: Mark the tick against the correct answer in the following: If $\tan ^{-1} 3 \mathrm{x}+\tan ^{-1} 2 \mathrm{x}=\frac{\pi}{4}$ then $\mathrm{x}=?$ A. $\frac{1}{2}$ or $-2$ B. $\frac{1}{3}$ or $-3$ C. $\frac{1}{4}$ or $-2$ D. $\frac{1}{6}$ or $-1$ Solution: Given: $\tan ^{-1} 3 x+\tan ^{-1} 2 x=\frac{\pi}{4}$ To Find: The value of $x$ Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\Rightarrow \tan ^{-1} 3 x+\tan ^{-1} 2 x=\tan ^{-1}\left(\frac{3 ...
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Question: $\left|\begin{array}{ccc}x+4 x x \\ x x+4 x \\ x x x+4\end{array}\right|$ Solution: Given, $\left|\begin{array}{ccc}x+4 x x \\ x x+4 x \\ x x x+4\end{array}\right|$ $=\left|\begin{array}{ccc}3 x+4 3 x+4 3 x+4 \\ x x+4 x \\ x x x+4\end{array}\right|$ [Applying $\left.R_{1} \rightarrow R_{1}+R_{2}+R_{3}\right]$ $=(3 x+4)\left|\begin{array}{ccc}1 1 1 \\ x x+4 x \\ x x x+4\end{array}\right|$ $=(3 x+4)\left|\begin{array}{ccc}0 0 1 \\ -4 4 x \\ 0 -4 x+4\end{array}\right|=16(3 x+4)$ [Applying...
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Question: Write a value of $\int \frac{\sin x-\cos x}{\sqrt{1+\sin 2 x}} d x$ Solution: We know that $1+\sin 2 x=\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x$ $=(\sin x+\cos x)^{2}$ $y=\int \frac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^{2}}} d x$ $y=\int \frac{(\sin x-\cos x)}{(\sin x+\cos x)} d x$ Let, $\sin x+\cos x=t$ Differentiating both sides with respect to $x$ $\frac{d t}{d x}=\cos x-\sin x$ $\Rightarrow-\mathrm{dt}=(\sin x-\cos x) d x$ $y=\int \frac{-1}{t} d t$ Use formula $\int \frac{1}{t}=\log...
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Question: Mark the tick against the correct answer in the following: If $\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8$ then $x=?$ A. $\frac{1}{3}$ B. $\frac{1}{5}$ C. 3 D. 5 Solution: Given: $\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8$ To Find: The value of $x$ Here $\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8$ can be written as $\tan ^{-1} x=\tan ^{-1} 8-\tan ^{-1} 3$ Since we know that $\tan ^{-1} x$ - $\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ $\tan ^{-1} x=\tan ^{-1} 8-\tan ^{-1} 3=\tan ^{-1}\...
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Question: $\left|\begin{array}{ccc}3 x -x+y -x+z \\ x-y 3 y z-y \\ x-z y-z 3 z\end{array}\right|$ Solution: Given, $\left|\begin{array}{ccc}3 x -x+y -x+z \\ x-y 3 y z-y \\ x-z y-z 3 z\end{array}\right|$ [Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ ] $=\left|\begin{array}{ccc}x+y+z -x+y -x+z \\ x+y+z 3 y z-y \\ x+y+z y-z 3 z\end{array}\right|$ [Taking $(x+y+z)$ common from column $C_{1}$ ] $=(x+y+z)\left|\begin{array}{ccc}1 -x+y -x+z \\ 1 3 y z-y \\ 1 y-z 3 z\end{array}\right|$ [Applying $R_{2...
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Question: $\left|\begin{array}{ccc}0 x y^{2} x z^{2} \\ x^{2} y 0 y z^{2} \\ x^{2} z z y^{2} 0\end{array}\right|$ Solution: Given, $\left|\begin{array}{ccc}0 x y^{2} x z^{2} \\ x^{2} y 0 y z^{2} \\ x^{2} z z y^{2} 0\end{array}\right|$ [Taking $x^{2}, y^{2}$ and $z^{2}$ common from $C_{1}, C_{2}$ and $C_{3}$, respectively] $=x^{2} y^{2} z^{2}\left|\begin{array}{lll}0 x x \\ y 0 y \\ z z 0\end{array}\right|$ [Applying $C_{2} \rightarrow C_{2}-C_{3}$ ] $=x^{2} y^{2} z^{2}\left|\begin{array}{ccc}0 0...
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Question: Mark the tick against the correct answer in the following: $\left(\tan ^{-1} 2+\tan ^{-1} 3\right)=?$ A. $\frac{-\pi}{4}$ B. $\frac{\pi}{4}$ C. $\frac{3 \pi}{4}$ D. $\pi$ Solution: To Find: The value of $\tan ^{-1} 2+\tan ^{-1} 3$ Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\Rightarrow \tan ^{-1} 2+\tan ^{-1} 3=\tan ^{-1}\left(\frac{2+3}{1-(2 \times 3)}\right)$ $=\tan ^{-1}\left(\frac{5}{-5}\right)$ $=\tan ^{-1}(-1)$ Since the principle valu...
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