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JEE Mathematics 2024 Question with Solution

If log⁡(e) y=3sin⁡−1x\log(e)\, y = 3\sin^{-1} x, then (1−x)2y′′−xy′(1 - x)^2 y'' - x y' at x=12x = \frac{1}{2} is equal to:

  • A

    9eπ/69e^{\pi/6}

  • B

    3eπ/63e^{\pi/6}

  • C

    3eπ/23e^{\pi/2}

  • D

    9eπ/29e^{\pi/2}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: ln⁡(y)=3sin⁡−1(x)\ln(y) = 3\sin^{-1}(x)

Find: (1−x2)y′′−xy′(1-x^2)y''-xy' at x=12x=\frac{1}{2}.

Differentiate both sides with respect to xx:

1yy′=3(11−x2)\frac{1}{y}y' = 3\left(\frac{1}{\sqrt{1-x^2}}\right)

So,

y′=3y1−x2y' = \frac{3y}{\sqrt{1-x^2}}

At x=12x=\frac{1}{2},

y=e3sin⁡−1(12)=eπ/2y = e^{3\sin^{-1}\left(\frac{1}{2}\right)} = e^{\pi/2}

and hence

y′=3eπ/21−(12)2=3eπ/232=23eπ/2y' = \frac{3e^{\pi/2}}{\sqrt{1-\left(\frac{1}{2}\right)^2}} = \frac{3e^{\pi/2}}{\frac{\sqrt{3}}{2}} = 2\sqrt{3}e^{\pi/2}

Differentiate again:

y′′=3(1−x2 y′−y⋅11−x2(−2x)1−x2)y'' = 3\left(\frac{\sqrt{1-x^2}\,y' - y\cdot \frac{1}{\sqrt{1-x^2}}(-2x)}{1-x^2}\right)

Therefore,

(1−x2)y′′=3(3y+xy1−x2)(1-x^2)y'' = 3\left(3y + \frac{xy}{\sqrt{1-x^2}}\right)

Now substitute x=12x=\frac{1}{2} and y=eπ/2y=e^{\pi/2}:

(1−x2)y′′∣x=12=3eπ/2(3+13)(1-x^2)y''\big|_{x=\frac{1}{2}} = 3e^{\pi/2}\left(3+\frac{1}{\sqrt{3}}\right)

Then,

(1−x2)y′′−xy′=3eπ/2(3+13)−12(23eπ/2)(1-x^2)y''-xy' = 3e^{\pi/2}\left(3+\frac{1}{\sqrt{3}}\right) - \frac{1}{2}\left(2\sqrt{3}e^{\pi/2}\right) =eπ/2(9+3−3)=9eπ/2= e^{\pi/2}(9+\sqrt{3}-\sqrt{3}) = 9e^{\pi/2}

Therefore, the correct option is D, and the required value is 9eπ/29e^{\pi/2}.

Using explicit form of y

Given: log⁡ey=3sin⁡−1x\log_e y = 3\sin^{-1}x, so y=e3sin⁡−1xy = e^{3\sin^{-1}x}.

Find: (1−x2)y′′−xy′(1-x^2)y''-xy' at x=12x=\frac{1}{2}.

Differentiate:

1yy′=3⋅11−x2\frac{1}{y}y' = 3\cdot\frac{1}{\sqrt{1-x^2}}

Hence,

y′=3y1−x2y' = \frac{3y}{\sqrt{1-x^2}}

Differentiate y′y' once more:

y′′=ddx(3y1−x2)y'' = \frac{d}{dx}\left(\frac{3y}{\sqrt{1-x^2}}\right)

Using the product or quotient form shown in the solution,

(1−x2)y′′=3(3y+xy1−x2)(1-x^2)y'' = 3\left(3y + \frac{xy}{\sqrt{1-x^2}}\right)

At x=12x=\frac{1}{2},

sin⁡−1(12)=π6\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}

so

y=e3⋅π/6=eπ/2y = e^{3\cdot\pi/6} = e^{\pi/2}

and

y′=3eπ/23/2=23eπ/2y' = \frac{3e^{\pi/2}}{\sqrt{3}/2} = 2\sqrt{3}e^{\pi/2}

Substitute into the target expression:

(1−x2)y′′−xy′=3eπ/2(3+13)−12(23eπ/2)(1-x^2)y''-xy' = 3e^{\pi/2}\left(3+\frac{1}{\sqrt{3}}\right) - \frac{1}{2}(2\sqrt{3}e^{\pi/2}) =9eπ/2= 9e^{\pi/2}

The second provided approach contains an internal inconsistency with eπ/6e^{\pi/6} in intermediate steps, but its final answer agrees with the correct derivation above.

Common mistakes

  • Using sin⁡−1(12)=π2\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2} is incorrect. The correct principal value is π6\frac{\pi}{6}, and only after multiplying by 33 do we get π2\frac{\pi}{2} in the exponent.

  • Differentiating ln⁡y\ln y as if it were just y′y' is wrong. One must use ddx(ln⁡y)=1yy′\frac{d}{dx}(\ln y)=\frac{1}{y}y' by the chain rule.

  • While differentiating 11−x2\frac{1}{\sqrt{1-x^2}}, students often miss the chain rule sign and factor of xx. Compute it carefully before forming y′′y''.

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