NIMCET 2021 Previous Year Questions (PYQs) – Page 4 of 12
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 120 questions
The area of the region bounded by x-axis and the curves defined by $y=tanx$, $-\frac{\pi}{3}\leq x\leq \frac{\pi}{3}$ and $y=cotx$, $-\frac{\pi}{6}\leq x\leq \frac{3\pi}{2}$ is
1
2
3
4
Solution
If $\log (1-x+x^2)={{a}}_1x+{{a}}_{2{}^{{}^{}}}{x}^2+{{{}{{a}}_3{x}^3+.\ldots.}}^{}$ then ${{a}}_3+{{a}}_6+{{a}}_9+.\ldots.$ is equal to
1
2
3
4
Solution
The number of common tangents to the circle $x^2+y^2=4$ and $x^2+y^2-6x-8y=24$ is
1
2
3
4
Solution
In a ΔABC, if $\tan ^2\frac{A}{2}+\tan ^2\frac{B}{2}+\tan ^2\frac{C}{2}=k$ , then k is always
1
2
3
4
Solution
Let $\vec{a}=2\widehat{i}\, +\widehat{j}\, +2\widehat{k}$ , $\vec{b}=\widehat{i}-\widehat{j}+2\widehat{k}$ and $\vec{c}=\widehat{i}+\widehat{j}-2\widehat{k}$ are are three vectors. Then, a vector in the plane of $\vec{a}$ and $\vec{c}$ whose projection on $\vec{b}$ is of magnitude $\frac{1}{\sqrt{6}}$ is
1
2
3
4
Solution
For what value of p, the polynomial $x^4-3x^3+2px^2-6$ is exactly divisible by $(x-1)$
1
2
3
4
Solution
Let the given polynomial be:
$P(x) = x^4 - 3x^3 + 2px^2 - 6$
By the Factor Theorem, if $P(x)$ is exactly divisible by $(x - 1)$, then:
$P(1) = 0$
Substitute $x = 1$ into $P(x)$:
$P(1) = (1)^4 - 3(1)^3 + 2p(1)^2 - 6 = 0$
$1 - 3 + 2p - 6 = 0$
$2p - 8 = 0$
$2p = 8$
$p = 4$
Thus, the value of $p$ is $4$.
If $F(\theta)=\begin{bmatrix}{\cos \theta} & {-\sin \theta} & {0} \\ {\sin \theta} & {\cos \theta} & {0} \\ {0} & {0} & {1}\end{bmatrix}$ , then $F(\theta)F(\alpha)$ is equal to
1
2
3
4
Solution
If $\frac{n!}{2!(n-2)!}$ and $\frac{n!}{4!(n-4)!}$ are in the ratio 2:1, then the value of n is
1
2
3
4
Solution
The locus of the point of intersection of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which meet right angles is
1
2
3
4
Solution
NIMCET
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
NIMCET
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
Commented Dec 24, 2021
3 Upvote 2 Downvote Reply
Your reply to this comment: