MCA NIMCET Differentiation Previous Year Question Paper and Solution with PDF

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Differentiation PYQ

Qus : 1 MCA NIMCET PYQ

$$ \frac{d^{2}x}{dy^{2}} \; = \; ? $$

$ (a) -\left(\dfrac{d^{2}y}{dx^{2}}\right)^{-1}\left(\dfrac{dy}{dx}\right)^{-3} $

$ (b) \left(\dfrac{d^{2}y}{dx^{2}}\right)\left(\dfrac{dy}{dx}\right)^{-2} $

$ (c) -\left(\dfrac{d^{2}y}{dx^{2}}\right)\left(\dfrac{dy}{dx}\right)^{-3} $

$ (d) \left(\dfrac{d^{2}y}{dx^{2}}\right)^{-1} $

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MCA NIMCET Mathematics PYQ MCA NIMCET Differentiation (Disha JEE) PYQ

Solution

Qus : 2 MCA NIMCET PYQ

If $y = f(x)$ is odd and differentiable on $(-\infty,\infty)$ such that $f'(3) = -2$, then $f'(-3)$ equals:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

If $f(x)$ is odd, then:

$f(-x) = -f(x)$

Differentiate both sides:

$f'(-x)(-1) = -f'(x)$

$f'(-x) = f'(x)$

So derivative is even.

Therefore:

$f'(-3) = f'(3) = -2$

Qus : 3 MCA NIMCET PYQ

Find $ \displaystyle \frac{d}{dx}\left( \sqrt{x} - \frac{5}{\sqrt{x}} \right) $

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

$ \displaystyle \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}},;; \frac{d}{dx}\left(\frac{5}{\sqrt{x}}\right)= 5\cdot\left(-\frac{1}{2}x^{-3/2}\right) $ So $ \displaystyle \frac{d}{dx}\left(\sqrt{x}-\frac{5}{\sqrt{x}}\right) = \frac{1}{2\sqrt{x}} + \frac{5}{2}x^{-3/2} $

Qus : 4 MCA NIMCET PYQ

If $ f(x)=\displaystyle \int_{0}^{x} t\sin t, dt $, then $f'(x)$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

$ f'(x) = x\sin x $

Qus : 5 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 6 MCA NIMCET PYQ

Let $f(x)$ be a polynomial function of second degree and $f(1) = f(–1)$. If $a, b, c$ are in A.P. then $f'(a), f'(b), f'(c)$ are in.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Incorrect question

Qus : 12 MCA NIMCET PYQ

If $y=\sin ^{-1}(\frac{{x}^2+1}{\sqrt[]{1+3{x}^2+{x}^4}}),\, (x>0),$ then $\frac{dy}{dx}$=

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 13 MCA NIMCET PYQ

Differential coefficient of $\log_{10} x$ with respect to $\log_{x} 10$ to

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 14 MCA NIMCET PYQ

If y = cos²x², find dy/dx

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Let $y = [\cos(x^2)]^2$. Using the chain rule:

\[ \frac{dy}{dx} = 2\cos(x^2)\cdot \frac{d}{dx}\big(\cos(x^2)\big) \] \[ = 2\cos(x^2)\cdot\big(-\sin(x^2)\cdot 2x\big) \] \[= -4x\,\cos(x^2)\sin(x^2). \] Using $\;2\sin u\cos u=\sin(2u)$ with $u=x^2$: \[ \frac{dy}{dx} = -2x\,\sin\!\big(2x^2\big). \]

Answer: ${\dfrac{dy}{dx}=-4x\,\sin\!\big(x^2\big) \cos (x^2)}$

Qus : 15 MCA NIMCET PYQ

The derivative of (x³ + e^x + 3^x + cotx) with respect to x is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

$\dfrac{d}{dx}(x^3) = 3x^2$

$\dfrac{d}{dx}(e^x) = e^x$

$\dfrac{d}{dx}(3x) = 3$

$\dfrac{d}{dx}(\cot x) = -cosec^2 x$

Therefore, total derivative $= 3x^2 + e^x + 3 - cosec^2 x$

Answer: $\boxed{3x^2 + e^x + 3 - cosec^2 x}$ ✅

Qus : 16 MCA NIMCET PYQ

Differentiate [- log(log x), x > 1] with respect to x

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Let $y=-\log(\log x)$

$\dfrac{dy}{dx}=-\dfrac{d}{dx}[\log(\log x)]$

$\dfrac{d}{dx}[\log(\log x)]=\dfrac{1}{\log x}\cdot\dfrac{1}{x}$

Therefore, $\dfrac{dy}{dx}=-\dfrac{1}{x\log x}$

Answer: $\boxed{-\dfrac{1}{x\log x}},\; x>1$ ✅

Qus : 17 MCA NIMCET PYQ

With the usual notation $\frac{d^{2}x}{dy^{2}}$

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

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