Jamia Millia Islamia MCA Differentiation Previous Year Question Paper and Solution with PDF

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Jamia Millia Islamia MCA

Jamia Millia Islamia MCA Previous Year Questions (PYQs)

Jamia Millia Islamia MCA Differentiation PYQ

Qus : 1 Jamia Millia Islamia MCA PYQ

If $y=\log(\tan x)$, then $\dfrac{dy}{dx}$ equals …

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2017 PYQ

Solution

$y'=\dfrac{\sec^2 x}{\tan x}=\dfrac{1}{\sin x\cos x}=2\csc2x$.

Qus : 2 Jamia Millia Islamia MCA PYQ

If $y=e^{2x}$, then $\dfrac{d^{2}y}{dx^{2}}\cdot\dfrac{d^{2}x}{dy^{2}}$ equals …

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2017 PYQ

Solution

$\dfrac{d^{2}y}{dx^{2}}=4e^{2x}=4y$; $x=\tfrac12\ln y\Rightarrow \dfrac{d^{2}x}{dy^{2}}=-\dfrac{1}{2y^{2}}$. Product $=4y\cdot\left(-\dfrac{1}{2y^{2}}\right)=-\dfrac{2}{y}=-2e^{-2x}$.

Qus : 3 Jamia Millia Islamia MCA PYQ

For $y = \sin x + \cos x - 5a$, what is the value of $\dfrac{dy}{dx}$?

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

$\dfrac{dy}{dx} = \dfrac{d}{dx}(\sin x + \cos x - 5a)$ $= \cos x - \sin x.$

Qus : 4 Jamia Millia Islamia MCA PYQ

If $y = \tan^{-1}\!\left(\dfrac{1 + \tan x}{1 - \tan x}\right)$, then $\dfrac{dy}{dx}$ is equal to …

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

We know $\tan(2x) = \dfrac{2\tan x}{1 - \tan^2 x}$. Here, $\dfrac{1 + \tan x}{1 - \tan x} = \tan\!\left(\dfrac{\pi}{4} + x\right)$. So, $y = \tan^{-1}(\tan(\dfrac{\pi}{4} + x)) = \dfrac{\pi}{4} + x$. Hence, $\dfrac{dy}{dx} = 1$.

Qus : 5 Jamia Millia Islamia MCA PYQ

If $y = \log(\tan \theta)$, then $\dfrac{dy}{d\theta}$ is equal to …

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

$y=\log(\tan\theta)\ \Rightarrow\ \frac{dy}{d\theta} =\frac{1}{\tan\theta}\cdot\sec^{2}\theta =\frac{\sec^{2}\theta}{\tan\theta} =\frac{1}{\sin\theta\cos\theta} =\sec\theta\,\csc\theta.$

Qus : 6 Jamia Millia Islamia MCA PYQ

If $y = x + e^{x}$, then $\dfrac{dx}{dy}$ is …

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

$\dfrac{dy}{dx}=1+e^{x}\ \Rightarrow\ \dfrac{dx}{dy}=\dfrac{1}{1+e^{x}}$.

Qus : 7 Jamia Millia Islamia MCA PYQ

If $y=(x^{x})^{x}$, then $\dfrac{dy}{dx}$ is …

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MCA 2016 PYQ

Solution

$(x^{x})^{x}=x^{x^{2}}=e^{x^{2}\ln x}$. $\Rightarrow \dfrac{y'}{y}=2x\ln x+x\ \Rightarrow\ y'=y(x+2x\ln x)=xy+2xy\log x$.

Qus : 8 Jamia Millia Islamia MCA PYQ

If a determinant of order $3\times3$ is formed using numbers $1$ or $-1$, then the minimum value of the determinant is:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

Possible determinant values with elements $\pm1$ range from $-8$ to $+8$. For minimum, consider alternating signs pattern (Hadamard form): $\begin{vmatrix} 1 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & -1 \end{vmatrix} = -4$.

Qus : 9 Jamia Millia Islamia MCA PYQ

If $m$ is the slope of tangent at any point on the curve $e^y = 1 + x^x$, then:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

Given $e^y = 1 + x^x$ Differentiate both sides: $e^y \frac{dy}{dx} = x^x (\ln x + 1)$ $\Rightarrow \frac{dy}{dx} = \dfrac{x^x(\ln x + 1)}{e^y}$ Since $e^y = 1 + x^x$, $\Rightarrow m = \dfrac{x^x(\ln x + 1)}{1 + x^x}$ For $x > 0$, this ratio always lies between $-2$ and $2$.

Qus : 10 Jamia Millia Islamia MCA PYQ

The derivative of $e^x$ is:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2025 PYQ

Solution

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

Qus : 11 Jamia Millia Islamia MCA PYQ

The second derivative of $x^3$ is:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2025 PYQ

Solution

$y = x^3$

First derivative: $\frac{dy}{dx} = 3x^2$

Second derivative: $\frac{d^2 y}{dx^2} = 6x$

Qus : 12 Jamia Millia Islamia MCA PYQ

The derivative of $ x^x $ is:

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Jamia Millia Islamia MCA Previous Year PYQ Jamia Millia Islamia MCA JAMIA MILLIA ISLAMIA MCA 2025 PYQ

Solution

Log rule

$ \log(ab) = \log a + \log b = 2 + 3 = 5 $

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