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Trignometry

New kid on the Block

 Joined: 20 Mar 2012 Post: 20
19 May 2012 09:28:58 IST
1 People liked this
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f(x)=?cos(sinx) + sin^-1(1+x^2/2x)
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Trigonometry

plzzz give the answer fast u all great minds

New kid on the Block

Joined: 20 Mar 2012
Posts: 20
19 May 2012 09:37:21 IST
0 people liked this

sorry i forgot you have to find domain and range

Blazing goIITian

Joined: 19 Jan 2008
Posts: 1070
19 May 2012 23:41:10 IST
1 people liked this

$\hspace{-16}Here \bf{f(x)=\cos(\sin x)+\sin^{-1}\left(\frac{1+x^2}{2x}\right)}\\\\\\ First We will Calculate Domain of \bf{f(x)}\\\\\\ Domain of \bf{\cos(\sin x)} is \bf{x\in \mathbb{R}}\\\\\\ bcz \bf{-1\leq \sin x\leq 1\forall x\in \mathbb{R}}\\\\\\ So \bf{\cos(\sin x)} is Defined for all \bf{x\in\mathbb{R}}\\\\\\ Now \bf{\sin^{-1}\left(\frac{1+x^2}{2x}\right)} is Defined when \bf{\left| \frac{1+x^2}{2x}\right|\leq 1}\\\\\\ So \bf{\frac{1+x^2}{2\mid x \mid}\leq 1\Leftrightarrow 1+x^2\leq 2\mid x \mid}\\\\\\ \bf{1+x^2-2\mid x \mid\leq 0\Leftrightarrow \left(1+\mid x \mid\right)^2\leq 0}\\\\\\ But Square \bf{\geq 0}. So \bf{\left(1+\mid x \mid\right)^2=0\Leftrightarrow 1-\mid x \mid =0}\\\\\\ So \bf{x=\left\{-1\;,1\right\}}\\\\\\ So Domain of \bf{f(x)=\cos(\sin x)+\sin^{-1}\left(\frac{1+x^2}{2x}\right)} is \bf{x=\left\{-1\;,1\right\}}$

$\hspace{-16}Similarly We will find Range of \bf{f(x)=\cos(\sin x)+\sin^{-1}\left(\frac{1+x^2}{2x}\right)}\\\\\\ Put \bf{x=-1}, We Get \bf{f(-1)=\cos(-\sin 1)+\sin^{-1}(-1)}\\\\\\ So \bf{f(-1)=\cos(\sin 1)-\frac{\pi}{2}}\\\\\\ Put \bf{x=1}, We Get \bf{f(1)=\cos(\sin 1)+\sin^{-1}(1)}\\\\\\ So \bf{f(1)=\cos(\sin 1)+\frac{\pi}{2}}\\\\\\ So Range of \bf{f(x)} is \bf{\left\{\cos(\sin 1)-\frac{\pi}{2}\;,\cos(\sin 1)+\frac{\pi}{2}\right\}}$

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