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Let, f(x)=\left ( \sin (\tan ^{-1}x)+\sin (\cot ^{-1}x) \right )^{2}-1, \left | x \right |>1.\; If\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left ( \sin ^{-1}(f(x)) \right ) and y(\sqrt{3})=\frac{\pi }{6}, then y(-\sqrt{3}) is equal to :
Option: 1 \frac{\pi }{3}
Option: 2 \frac{2\pi }{3}
Option: 3 -\frac{\pi }{6}
Option: 4 \frac{5\pi }{6}
 

Answers (1)

best_answer

 

 

 

Integration as Reverse Process of Differentiation -

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. 

For example,

\\\mathrm{\frac{d}{dx}(\sin x)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2 \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x \right )=e^x}

In the above example,  the function cos x is the derived function of sin x. We say that sin x is an anti derivative (or an integral) of cos x. Similarly, x2 and ex  are the anti derivatives (or integrals) of 2x and ex respectively.  

Also note that the derivative of a constant  (C) is zero. So we can write the above examples as:

\\\mathrm{\frac{d}{dx}(\sin x+c)=\cos x}\\\\\mathrm{\frac{d}{dx}\left ( x^2+c \right )=2x}\\\\\mathrm{\frac{d}{dx}\left ( e^x +c\right )=e^x}

 

Thus, anti derivatives (or integrals) of the above functions are not unique. Actually, there exist infinitely many anti derivatives of each of these functions which can be obtained by selecting C arbitrarily from the set of real numbers. 

For this reason C is referred to as arbitrary constant. In fact, C is the parameter by varying which one gets different anti derivatives (or integrals) of the given function. 

 

If the function F(x) is an antiderivative of f(x), then the expression F(x) + C is the indefinite integral of the function f(x) and is denoted by the symbol ∫ f(x) dx. 

By definition,

\\\mathrm{\int f(x)dx=F(x)+c,\;\;\;where\;\;F'(x)=f(x)\;\;and\;\;'c' \;is\;constant.}

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Trigonometric Identities -

Trigonometric Identities-

These identities are the equations that hold true regardless of the angle being chosen.

 

\\\mathrm{\sin^2\mathit{t}+\cos^2\mathit{t}=1}\\\mathrm{1+\tan^2\mathit{t}=\sec^2\mathit{t}}\\\mathrm1+{\cot^2\mathit{t}=\csc^2\mathit{t}}\\\mathrm{\tan \mathit{t}=\frac{\sin \mathit{t}}{\cos \mathit{t}},\;\;\cot \mathit{t}=\frac{\cos\mathit{t}}{\sin\mathit{t}}}

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Principal Value of function f-1 (f (x)) -

Principal Value of function f-1 (f (x))

 

\begin{array} {l}\mathrm{1.\;\;\sin^{-1}(\sin (\theta))=\theta} \quad\quad\quad \;\mathrm{for\;all\;\theta\in[-\pi/2,\pi/2] }\\\mathrm{2.\;\;\cos^{-1}(\cos(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in[0,\pi]}\\\mathrm{3.\;\;\tan^{-1}(\tan(\theta))=\theta} \;\;\;\quad\quad \mathrm{for\;all\;\theta\in(-\pi/2,\pi/2)}\\\mathrm{4.\;\;\cot^{-1}(\cot(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in(0,\pi)} \\\mathrm{5.\;\;\sec^{-1}(\sec(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in[0,\pi]-\left \{ \pi/2 \right \}}\\\mathrm{6.\;\;\csc^{-1}(\csc(\theta))=\theta} \quad\quad\quad \mathrm{for\;all\;\theta\in[-\pi/2,\pi/2]-\left \{ 0 \right \}}\end{array}

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\\y=\frac{1}{2} \sin ^{-1}(f(x))+c\\\\\begin{array}{l}{\text {Let } \theta=\tan ^{-1} x \Rightarrow \cot ^{-1} x=\frac{\pi}{2}-\theta} \\\\ {f(x)=\left\{\sin \theta+\sin \left(\frac{\pi}{2}-\theta\right)\right\}^{2}-1}\end{array}\\\\\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta-1=\sin2\theta

\\y=\frac{1}{2} \sin ^{-1}\left(\sin \left(2 \tan ^{-1} x\right)\right)+C\\\\\frac{\pi}{6}=\frac{1}{2} \sin ^{-1}\left(\sin \left(2 \frac{\pi}{3}\right)\right)+C\\\\\frac{\pi}{6}=\frac{1}{2} \frac{\pi}{3}+C\Rightarrow C=0\\\\y=\frac{1}{2} \sin ^{-1}\left(\sin \left(-2 \frac{\pi}{3}\right)\right)=-\frac{\pi}{6}

Correct Option (4)

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Kuldeep Maurya

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