Jamia Millia Islamia 2016 Previous Year Questions (PYQs) – Page 4 of 10
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If $A$ be a set of cardinality $n$, then number of one-to-one onto functions from set $A$ to $A$ is …
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Solution
Number of bijective (one-one and onto) functions from a set of $n$ elements to itself $= n!$.
If $f$ is a function from a finite set $A$ having $10$ elements to a finite set $B$ having $5$ elements, then number of functions from $A$ to $B$ is …
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Solution
Each of $10$ elements in $A$ can be mapped to any of $5$ elements in $B$.
Hence total functions $= 5^{10}$.
If $A$ and $B$ are two sets, then $(A \cup B)' \cap B$ is equal to …
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Solution
$(A \cup B)'$ means elements not in $A$ or $B$.
Hence, $(A \cup B)' \cap B$ contains elements that are both in $B$ and not in $B$, i.e., none.
So result is $\phi$.
If $A = \{a, b, c\}$ and $B = \{a, b, d, e, f\}$ are two sets, then number of elements in $(A - B) \times (A \cap B)$ is …
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Solution
$A - B = \{c\}$ (only $c$ not in $B$).
$A \cap B = \{a, b\}$.
Then $(A - B) \times (A \cap B) = \{(c, a), (c, b)\}$ → 2 elements.
If $A$ and $B$ are two disjoint sets having $3$ and $5$ elements respectively, then power-set of $A \times (B - A)$ contains … elements.
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Solution
Since $A$ and $B$ are disjoint, $B - A = B$.
Then $A \times B$ has $3 \times 5 = 15$ ordered pairs.
Power-set of it will have $2^{15}$ elements.
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 $\sqrt{x + y} + \sqrt{y - x} = \sqrt{2}a$, then $\dfrac{d^2 y}{d x^2}$ is equal to …
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Solution
Differentiate both sides:
$\dfrac{1}{2\sqrt{x + y}}(1 + \dfrac{dy}{dx}) + \dfrac{1}{2\sqrt{y - x}}(\dfrac{dy}{dx} - 1) = 0$.
Simplify to get $\dfrac{dy}{dx} = \dfrac{\sqrt{y - x} - \sqrt{x + y}}{\sqrt{y - x} + \sqrt{x + y}}$.
Differentiate again and substitute from the given equation $\sqrt{x + y} + \sqrt{y - x} = \sqrt{2}a$, we get
$\dfrac{d^2y}{dx^2} = \dfrac{2}{a}$.
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
$(x^{x})^{x}=x^{x^{2}}=e^{x^{2}\ln x}$.
$\Rightarrow \dfrac{y'}{y}=2x\ln x+x\ \Rightarrow\ y'=y(x+2x\ln x)=xy+2xy\log x$.
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