MAH CET MCA Inverse Trigonometrical Function Previous Year Questions (PYQs)
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Showing 2 questions
The valid range of $x$ for which $\cos^{-1}(2x-3)$ is defined is
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Solution
For $\cos^{-1}(,\cdot,)$ to be defined, $-1\le 2x-3\le 1$.
This gives $2\le 2x\le 4\Rightarrow 1\le x\le 2$.
If $\tan^{-1}\left(\dfrac{1-x}{1+x}\right)=\dfrac{1}{2}\tan^{-1}x$, then $x$ is
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Solution
Let $t=\tan\left(\dfrac{1}{2}\tan^{-1}x\right)=\dfrac{1-x}{1+x}$.
Using $\tan(2\theta)=\dfrac{2t}{1-t^2}=x$, we get $x=\dfrac{1-x^2}{2x}\Rightarrow 3x^2=1\Rightarrow x=\pm\dfrac{1}{\sqrt3}$.
From the equation, $x>0$.
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