Jamia Millia Islamia 2016 Previous Year Questions (PYQs) – Page 1 of 10
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
$\displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{x - \frac{\pi}{4}}$ is equal to …
1
2
3
4
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 …
1
2
3
4
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 …
1
2
3
4
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 …
1
2
3
4
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 …
1
2
3
4
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 …
1
2
3
4
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 …
1
2
3
4
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 $a$
1
2
3
4
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 …
1
2
3
4
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 …
1
2
3
4
Solution
Let $t=\sqrt{x}$, $dx=2t\,dt$.
Then $\int 2t^{2}e^{t}dt=2e^{t}(t^{2}-2t+2)+C$.
Substitute $t=\sqrt{x}$ ⇒ $(2x-4\sqrt{x}+4)e^{\sqrt{x}}+C$.
Jamia Millia Islamia
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Jamia Millia Islamia
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
