Jamia Millia Islamia MCA Differentiation Previous Year Questions (PYQs) – Page 1 of 2
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If $y=\log(\tan x)$, then $\dfrac{dy}{dx}$ equals …
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Solution
$y'=\dfrac{\sec^2 x}{\tan x}=\dfrac{1}{\sin x\cos x}=2\csc2x$.
If $y=e^{2x}$, then $\dfrac{d^{2}y}{dx^{2}}\cdot\dfrac{d^{2}x}{dy^{2}}$ equals …
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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 $=4y\cdot\left(-\dfrac{1}{2y^{2}}\right)=-\dfrac{2}{y}=-2e^{-2x}$.
For $y = \sin x + \cos x - 5a$, what is the value of $\dfrac{dy}{dx}$?
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Solution
$\dfrac{dy}{dx} = \dfrac{d}{dx}(\sin x + \cos x - 5a)$
$= \cos x - \sin x.$
If $y = \tan^{-1}\!\left(\dfrac{1 + \tan x}{1 - \tan x}\right)$, then $\dfrac{dy}{dx}$ is equal to …
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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, $\dfrac{dy}{dx} = 1$.
If $y = \log(\tan \theta)$, then $\dfrac{dy}{d\theta}$ is equal to …
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Solution
$y=\log(\tan\theta)\ \Rightarrow\ \frac{dy}{d\theta}
=\frac{1}{\tan\theta}\cdot\sec^{2}\theta
=\frac{\sec^{2}\theta}{\tan\theta}
=\frac{1}{\sin\theta\cos\theta}
=\sec\theta\,\csc\theta.$
If $y = x + e^{x}$, then $\dfrac{dx}{dy}$ is …
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Solution
$\dfrac{dy}{dx}=1+e^{x}\ \Rightarrow\ \dfrac{dx}{dy}=\dfrac{1}{1+e^{x}}$.
If $y=(x^{x})^{x}$, then $\dfrac{dy}{dx}$ is …
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Solution
$(x^{x})^{x}=x^{x^{2}}=e^{x^{2}\ln x}$.
$\Rightarrow \dfrac{y'}{y}=2x\ln x+x\ \Rightarrow\ y'=y(x+2x\ln x)=xy+2xy\log x$.
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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Solution
Possible determinant values with elements $\pm1$ range from $-8$ to $+8$.
For minimum, consider alternating signs pattern (Hadamard form):
$\begin{vmatrix}
1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{vmatrix} = -4$.
If $m$ is the slope of tangent at any point on the curve $e^y = 1 + x^x$, then:
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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 $2$.
Jamia Millia Islamia MCA
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Jamia Millia Islamia MCA
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