Jamia Millia Islamia 2016 Previous Year Questions (PYQs) – Page 1 of 10
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$\displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{x - \frac{\pi}{4}}$ is equal to …
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Solution
his is the derivative of $\sin x-\cos x$ at $x=\frac{\pi}{4}$.
$\,(\sin x-\cos x)'=\cos x+\sin x \Rightarrow \cos\frac{\pi}{4}+\sin\frac{\pi}{4}
=\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}=\sqrt2.$
$\displaystyle \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$ is equal to …
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Solution
$\sqrt{x}$ at $x$:
$\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}.$
If $\displaystyle \lim_{x \to a} \frac{x^{\alpha} - x^{a}}{x - a} = -1$, then $\alpha$ is equal to …
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Solution
Limit $\Rightarrow \dfrac{d}{dx}x^{\alpha}=\alpha x^{\alpha-1}$.
At $x=a$, $\alpha a^{\alpha-1}=-1$.
If $a=1$, $\alpha=-1$.
$\displaystyle \lim_{x\to a}\frac{x^{10}-a^{10}}{x^{2}-a^{2}}$ is equal to …
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Solution
Using L’Hôpital’s rule or factorization,
$\dfrac{10a^{9}}{2a}=5a^{8}$.
$\displaystyle \lim_{x\to1}\frac{x+x^{2}+\ldots+x^{10}-10}{5x-5}$ is equal to …
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Solution
Let $f(x)=x+x^{2}+\ldots+x^{10}-10$.
Then $f'(x)=1+2x+3x^{2}+\ldots+10x^{9}$.
At $x=1$, $f'(1)=1+2+3+\ldots+10=55$.
Hence limit $=\dfrac{f'(1)}{5}=\dfrac{55}{5}=11$.
$\displaystyle \int_{1}^{x}(1+\log t)^{2}\,dt$ is equal to …
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Solution
Let $y=1+\log t \Rightarrow t=e^{y-1}$, $dt=e^{y-1}dy$.
After simplification:
$x((1+\log x)^{2}-2(1+\log x)+2)-1$.
If $x>0$, then $\displaystyle \int |x|^{3} dx$ is equal to …
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Solution
For $x>0$, $|x|=x$, so $\int x^{3}dx=\dfrac{x^{4}}{4}+C$.
$\displaystyle \int_{0}^{\frac{\pi}{4}}\sec^{2}x\sin x\,dx=a+\sqrt{2}$, find $a$
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Solution
Let $u=\tan x$, then $\sin x=\dfrac{u}{\sqrt{1+u^{2}}}$ and $du=\sec^{2}x\,dx$.
$\int_{0}^{1}\dfrac{u}{\sqrt{1+u^{2}}}du=\left[\sqrt{1+u^{2}}\right]_{0}^{1}=\sqrt2-1$.
Hence $a=-1$.
$\displaystyle \int_{0}^{1}\frac{x}{(1-x)^{1/2}}dx$ is equal to …
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Solution
Let $u=1-x \Rightarrow du=-dx$.
Integral $=\int_{0}^{1}(u^{-1/2}-u^{1/2})du=\left[2u^{1/2}-\frac{2}{3}u^{3/2}\right]_{0}^{1}=2-\frac{2}{3}=\frac{4}{3}$.
$\displaystyle \int \sqrt{x}e^{\sqrt{x}}dx$ is equal to …
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Solution
Let $t=\sqrt{x}$, $dx=2t\,dt$.
Then $\int 2t^{2}e^{t}dt=2e^{t}(t^{2}-2t+2)+C$.
Substitute $t=\sqrt{x}$ ⇒ $(2x-4\sqrt{x}+4)e^{\sqrt{x}}+C$.
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