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Day 174: MHT-CET Maths

5 questions. Try each one first, then open its answer.

  1. Question 1

    Differential Equations. Asked once: 13 May 2026 S1 (same type asked 10 times) See all 10 of this type

    86
    The solution of the differential equation y2−x2dydx=xydydxy^2 - x^2\dfrac{dy}{dx} = xy\dfrac{dy}{dx} is ...

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  2. Question 2

    Limits. Asked once: 15 May 2026 S1 (same type asked 10 times) See all 10 of this type

    71
    Let f(x)=12x−4x−3x+11−cos⁡2xf(x) = \dfrac{12^x - 4^x - 3^x + 1}{1 - \cos 2x}, for x≠0x \ne 0. If f(x)f(x) is continuous at x=0x = 0 and f(0)=(log⁡a)⋅(log⁡3)4kf(0) = \dfrac{(\log a) \cdot (\log 3)}{4k}, then the value of aa in terms of kk is:

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  3. Question 3

    Limits. Asked once: 14 May 2026 S2 (same type asked 10 times) See all 10 of this type

    72
    If f(x)=1−cos⁡(pπx2)x2f(x) = \dfrac{1 - \cos\left(\dfrac{p\pi x}{2}\right)}{x^2} is continuous at x=0x = 0 and f(0)=9π232f(0) = \dfrac{9\pi^2}{32}, then the value of pp is

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  4. Question 4

    Differentiation. Asked once: 18 Apr 2026 S1 (same type asked 10 times) See all 10 of this type

    68
    Let t∈(0,1)t \in (0, 1) and α∈(0,π4)\alpha \in \left(0, \dfrac{\pi}{4}\right). If x=csc⁡−1(1+t22t),y=cot⁡−1(1−t2t)x = \mathrm{\csc}^{-1}\left(\dfrac{1 + t^2}{2t}\right), y = \cot^{-1}\left(\dfrac{\sqrt{1 - t^2}}{t}\right) and dydx=f(t)\dfrac{dy}{dx} = f(t), then the value of f(tan⁡α)f(\tan\alpha) is

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  5. Question 5

    Differentiation. Asked once: 2 May 2023 S2 (same type asked 10 times) See all 10 of this type

    145
    Let f(θ)=sin⁡ ⁣(tan⁡−1 ⁣sin⁡θcos⁡2θ)f(\theta) = \sin\!\left(\tan^{-1}\!\frac{\sin\theta}{\sqrt{\cos2\theta}}\right) where −π4<θ<π4-\frac{\pi}{4}<\theta<\frac{\pi}{4}, then dd(tan⁡θ)(f(θ))\frac{d}{d(\tan\theta)}(f(\theta)) is

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