IITJEE Mains (Offline) 2016

Geometry Level 3

Consider f ( x ) = tan ⁡ − 1 ( 1 + sin ⁡ x 1 − sin ⁡ x ) , x f(x)=\tan^{-1}\left(\sqrt{\dfrac{1+\sin x}{1-\sin x}}\right),x belongs to ( 0 , π 2 ) \left(0,\dfrac{\pi}{2}\right) . A normal to y = f ( x ) y=f(x) at x = π 6 x=\dfrac{\pi}{6} also passes through the point __________ \text{\_\_\_\_\_\_\_\_\_\_} .

( π 6 , 0 ) (\frac{\pi}{6},0) ( 0 , 0 ) (0,0) ( 0 , 2 π 3 ) (0,\frac{2\pi}{3}) ( π 4 , 0 ) (\frac{\pi}{4},0)

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Tanishq Varshney
Apr 5, 2016

An easy problem

sin ⁡ x = cos ⁡ ( π 2 − x ) \sin x = \cos \left( \frac{\pi}{2}-x \right)

and using 1 + cos ⁡ 2 x = 2 cos ⁡ 2 x a n d 1 − cos ⁡ 2 x = 2 sin ⁡ 2 x \large{1+ \cos 2x=2 \cos^2 x~~and ~~ 1-\cos 2x=2 \sin^2 x}

we get f ( x ) = tan ⁡ − 1 ( cot ⁡ ( π 4 − x 2 ) ) \large{f(x)=\tan^{-1} \left( \cot \left( \frac{\pi}{4}-\frac{x}{2} \right) \right)}

⇒ f ( x ) = x 2 + π 4 \Rightarrow f(x)=\frac{x}{2}+\frac{\pi}{4}

equation of normal to this line (line perpendicular to it).

y = − 2 x + c y=-2x+c . This line passes through ( π 6 , π 3 ) \large{\left(\frac{\pi}{6},\frac{\pi}{3}\right)}

⇒ y = − 2 x + 2 π 3 \large{\Rightarrow y=-2x+\frac{2 \pi }{3}}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

can u provide easier expalnation tanishq. how can we say that line passes through (pi/6 ,pi/3)

rishabh dubey - 5 years, 2 months ago

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put the value of x x in f ( x ) f(x) to find y y .

Tanishq Varshney - 5 years, 2 months ago

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