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If f^{'}\left ( x \right )=\tan ^{-1}\left ( \sec x+\tan x \right ),-\frac{\pi }{2}<x<\frac{\pi }{2}, and f(0)=0, then f(1) is equal to : 

 

Option: 1

\frac{\pi +1}{4}
 


Option: 2

\frac{\pi +2}{4}


Option: 3

\frac{1}{4}

 


Option: 4

\frac{\pi -1}{4}


Answers (1)

best_answer

 

 

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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Derivative of the Polynomial Function -

Derivative of the Polynomial Function

Finding derivatives of functions by using the definition of the derivative can be a lengthy and, for certain functions, a rather challenging process.

In this section, we will learn rules for finding derivatives that allow us to bypass this process. Let’s begin with the basics.

\\\mathbf{1.}\;\;\;\;\mathrm{\mathbf{\frac{\mathit{d}}{\mathit{dx}}(constant)=0}}
\\f(x)=c\;\;and\;\;f(x+h)=c \\\\\begin{aligned}f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{c-c}{h} \\ &=\lim _{h \rightarrow 0} \frac{0}{h} \\ &=\lim _{h \rightarrow 0} 0=0 \end{aligned}

The rule for differentiating constant functions is called the constant rule. It states that the derivative of a constant function is zero; that is, since a constant function is a horizontal line, the slope, or the rate of change, of a constant function, is 0.

\\\mathbf{2.}\;\;\;\;\mathrm{\mathbf{\frac{\mathit{d}}{\mathit{dx}}(x^n)=nx^{n-1}}}

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f^{\prime}(x)=\tan ^{-1}(\sec x+\tan x)=\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right)

\\\Rightarrow \tan ^{-1}\left(\frac{1-\cos \left(\frac{\pi}{2}+x\right)}{\sin \left(\frac{\pi}{2}+x\right)}\right)=\tan ^{-1}\left(\frac{2 \sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}\right)\\=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right)=\frac{\pi}{4}+\frac{x}{2}\\ {\left(f^{\prime}(x)\right) =\frac{\pi}{4}+\frac{x}{2} } \\ {f(x)=\frac{\pi}{4} x+\frac{x^{2}}{4}+c}

{f(0)=c=0 \Rightarrow \quad f(x)=\frac{\pi}{4} x+\frac{x^{2}}{4}} \\ {\text { So } f(1)=\frac{\pi+1}{4}}

 

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Riya

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