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Functions question

2019 · 12 Apr · Shift 1 · Q27
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Functions question

2019 · 12 Apr · Shift 1 · Q27

JEE MainMathematicsFunctionsMCQ+4 / −1
For x ∈\in∈ (0, 3/2), let f(x) = x\sqrt xx​, g(x) = tan x and h(x) =1−x21+x2{{1 - {x^2}} \over {1 + {x^2}}}1+x21−x2​. If ϕ\phiϕ(x) = ((hof)og)(x), then ϕ(π3)\phi \left( {{\pi \over 3}} \right)ϕ(3π​) is equal to :
  1. A
    tan⁡7π12\tan {{7\pi } \over {12}}tan127π​
  2. B
    tan⁡11π12\tan {{11\pi } \over {12}}tan1211π​
  3. C
    tan⁡π12\tan {\pi \over {12}}tan12π​
  4. D
    tan⁡5π12\tan {{5\pi } \over {12}}tan125π​
View written solutionFree

Correct answer: B

  1. Identify the compositions

Given: f(x)=x,g(x)=tan⁡x,h(x)=1−x21+x2f(x)=\sqrt{x},\qquad g(x)=\tan x,\qquad h(x)=\frac{1-x^2}{1+x^2}f(x)=x​,g(x)=tanx,h(x)=1+x21−x2​

and ϕ(x)=((h∘f)∘g)(x)=h(f(g(x))).\phi(x)=((h\circ f)\circ g)(x)=h(f(g(x))).ϕ(x)=((h∘f)∘g)(x)=h(f(g(x))).

So, ϕ(x)=h(tan⁡x).\phi(x)=h\big(\sqrt{\tan x}\big).ϕ(x)=h(tanx​).

  1. Compute ϕ(π3)\phi\left(\frac{\pi}{3}\right)ϕ(3π​)

First, g(π3)=tan⁡π3=3.g\left(\frac{\pi}{3}\right)=\tan\frac{\pi}{3}=\sqrt{3}.g(3π​)=tan3π​=3​.

Then, f(g(π3))=3=31/4.f\left(g\left(\frac{\pi}{3}\right)\right)=\sqrt{\sqrt{3}}=3^{1/4}.f(g(3π​))=3​​=31/4.

Now apply hhh: \phi\left(\frac{\pi}{3}\right)=h(3^{1/4})= rac{1-(3^{1/4})^2}{1+(3^{1/4})^2}.

Since (31/4)2=3,(3^{1/4})^2=\sqrt{3},(31/4)2=3​, we get ϕ(π3)=1−31+3.\phi\left(\frac{\pi}{3}\right)=\frac{1-\sqrt{3}}{1+\sqrt{3}}.ϕ(3π​)=1+3​1−3​​.

  1. Simplify the expression

Rationalize: \frac{1-\sqrt{3}}{1+\sqrt{3}}\cdot\frac{1-\sqrt{3}}{1-\sqrt{3}}= rac{(1-\sqrt{3})^2}{1-3}= rac{1-2\sqrt{3}+3}{-2}=-2+\sqrt{3}.

So, ϕ(π3)=3−2.\phi\left(\frac{\pi}{3}\right)=\sqrt{3}-2.ϕ(3π​)=3​−2.

  1. Match with tangent values

We use the standard value: tan⁡15∘=tan⁡π12=2−3.\tan 15^\circ=\tan\frac{\pi}{12}=2-\sqrt{3}.tan15∘=tan12π​=2−3​.

Therefore, 3−2=−(2−3)=−tan⁡π12.\sqrt{3}-2=-(2-\sqrt{3})=-\tan\frac{\pi}{12}.3​−2=−(2−3​)=−tan12π​.

Now, tan⁡(π−π12)=tan⁡11π12=−tan⁡π12=3−2.\tan\left(\pi-\frac{\pi}{12}\right)=\tan\frac{11\pi}{12}=-\tan\frac{\pi}{12}=\sqrt{3}-2.tan(π−12π​)=tan1211π​=−tan12π​=3​−2.

Hence, ϕ(π3)=tan⁡11π12.\phi\left(\frac{\pi}{3}\right)=\tan\frac{11\pi}{12}.ϕ(3π​)=tan1211π​.

  1. Check options
  • A: tan⁡7π12=tan⁡105∘\tan\frac{7\pi}{12}=\tan 105^\circtan127π​=tan105∘ — not equal
  • B: tan⁡11π12=tan⁡165∘\tan\frac{11\pi}{12}=\tan 165^\circtan1211π​=tan165∘ — correct
  • C: tan⁡π12=2−3\tan\frac{\pi}{12}=2-\sqrt{3}tan12π​=2−3​ — positive mismatch
  • D: tan⁡5π12=tan⁡75∘\tan\frac{5\pi}{12}=\tan 75^\circtan125π​=tan75∘ — not equal

Therefore, the correct option is B.

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