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Differentiation question

2020 · 5 Sep · Shift 2 · Q33
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  5. /2020 · 5 Sep · Shift 2 · Q33

Differentiation question

2020 · 5 Sep · Shift 2 · Q33

JEE MainMathematicsDifferentiationMCQ+4 / −1
The derivative of tan⁡−1(1+x2−1x){\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)tan−1(x1+x2​−1​) with respect to tan⁡−1(2x1−x21−2x2){\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)tan−1(1−2x22x1−x2​​) at x =12{1 \over 2}21​ is :
  1. A
    233{{2\sqrt 3 } \over 3}323​​
  2. B
    235{{2\sqrt 3 } \over 5}523​​
  3. C
    310{{\sqrt 3 } \over {10}}103​​
  4. D
    312{{\sqrt 3 } \over {12}}123​​
View written solutionFree

Correct answer: C

Let y= an^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right),\qquad z= an^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right). We need dydz=dy/dxdz/dxat x=12.\frac{dy}{dz}=\frac{dy/dx}{dz/dx}\quad \text{at }x=\frac12.dzdy​=dz/dxdy/dx​at x=21​.

1. Simplify the first expression

Consider 1+x2−1x.\frac{\sqrt{1+x^2}-1}{x}.x1+x2​−1​. Rationalizing,

=\frac{x}{\sqrt{1+x^2}+1}.$$ Now use the identity $$\tan\left(\frac{\theta}{2}\right)=\frac{\sin\theta}{1+\cos\theta}.$$ If we put $x=\tan\theta$, then $$\sin\theta=\frac{x}{\sqrt{1+x^2}},\qquad \cos\theta=\frac{1}{\sqrt{1+x^2}}.$$ Hence $$\tan\left(\frac\theta2\right)=\frac{\sin\theta}{1+\cos\theta} =\frac{x/\sqrt{1+x^2}}{1+1/\sqrt{1+x^2}} =\frac{x}{\sqrt{1+x^2}+1}.

Therefore, 1+x2−1x=tan⁡(12tan⁡−1x).\frac{\sqrt{1+x^2}-1}{x}=\tan\left(\frac12\tan^{-1}x\right).x1+x2​−1​=tan(21​tan−1x). So, y=tan⁡−1(tan⁡(12tan⁡−1x))=12tan⁡−1x,y=\tan^{-1}\left(\tan\left(\frac12\tan^{-1}x\right)\right)=\frac12\tan^{-1}x,y=tan−1(tan(21​tan−1x))=21​tan−1x, for the relevant principal values.

Thus,

At x=12x=\frac12x=21​,

2. Simplify the second expression

Now z=tan⁡−1(2x1−x21−2x2).z=\tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right).z=tan−1(1−2x22x1−x2​​). Use the identity tan⁡2ϕ=2tan⁡ϕ1−tan⁡2ϕ.\tan 2\phi=\frac{2\tan\phi}{1-\tan^2\phi}.tan2ϕ=1−tan2ϕ2tanϕ​. Let tan⁡ϕ=x1−x2.\tan\phi=\frac{x}{\sqrt{1-x^2}}.tanϕ=1−x2​x​. Then tan⁡2ϕ=x21−x2,\tan^2\phi=\frac{x^2}{1-x^2},tan2ϕ=1−x2x2​, and

=2x/1−x21−x2−x21−x2=2x1−x21−2x2.=\frac{2x/\sqrt{1-x^2}}{1-x^2-x^2\over 1-x^2} =\frac{2x\sqrt{1-x^2}}{1-2x^2}.=1−x21−x2−x2​2x/1−x2​​=1−2x22x1−x2​​.

So the given quantity is tan⁡2ϕ\tan 2\phitan2ϕ where tan⁡ϕ=x1−x2\tan\phi=\dfrac{x}{\sqrt{1-x^2}}tanϕ=1−x2​x​.

But if x=sin⁡θx=\sin\thetax=sinθ, then x1−x2=sin⁡θcos⁡θ=tan⁡θ,\frac{x}{\sqrt{1-x^2}}=\frac{\sin\theta}{\cos\theta}=\tan\theta,1−x2​x​=cosθsinθ​=tanθ, so ϕ=θ=sin⁡−1x\phi=\theta=\sin^{-1}xϕ=θ=sin−1x (principal values). Hence 2x1−x21−2x2=tan⁡(2sin⁡−1x).\frac{2x\sqrt{1-x^2}}{1-2x^2}=\tan(2\sin^{-1}x).1−2x22x1−x2​​=tan(2sin−1x). Therefore, z=tan⁡−1(tan⁡(2sin⁡−1x)).z=\tan^{-1}(\tan(2\sin^{-1}x)).z=tan−1(tan(2sin−1x)). At x=12x=\frac12x=21​, 2sin⁡−1(12)=2⋅π6=π3,2\sin^{-1}\left(\frac12\right)=2\cdot \frac\pi6=\frac\pi3,2sin−1(21​)=2⋅6π​=3π​, which lies in the principal range of tan⁡−1\tan^{-1}tan−1, so locally z=2sin⁡−1x.z=2\sin^{-1}x.z=2sin−1x. Thus,

At x=12x=\frac12x=21​,

3. Compute dydz\dfrac{dy}{dz}dzdy​

Now,

=2543=25⋅34=310.=\frac{\frac25}{\frac{4}{\sqrt3}} =\frac25\cdot \frac{\sqrt3}{4} =\frac{\sqrt3}{10}.=3​4​52​​=52​⋅43​​=103​​.

4. Match with options

310\frac{\sqrt3}{10}103​​ corresponds to Option C.

5. Comparison with stored answer

Stored correct answer: C

Our derived answer also gives C, so it agrees.

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