NIMCET 2027 2022 Previous Year Questions (PYQs) – Page 3 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
If $0{\lt}P(A){\lt}1$ and $0{\lt}P(B){\lt}1$ and $P(A\cap B)=P(A)P(B)$, then
1
2
3
4
Solution
If ${{a}}_1,{{a}}_2,\ldots,{{a}}_n$ are any real numbers and $n$ is
any positive integer, then
1
2
3
4
Solution
$$
\begin{aligned}
&\textbf{To prove: } \quad n\sum_{i=1}^{n} a_i^2 \;\ge\; \left(\sum_{i=1}^{n} a_i\right)^2 \\[6pt]
&\text{(RMS}\ge\text{AM)}:\quad
\sqrt{\frac{\sum_{i=1}^{n} a_i^2}{\,n\,}}
\;\ge\;
\frac{\sum_{i=1}^{n} a_i}{\,n\,} \\[6pt]
&\Rightarrow\;
\frac{\sum_{i=1}^{n} a_i^2}{\,n\,}
\;\ge\;
\left(\frac{\sum_{i=1}^{n} a_i}{\,n\,}\right)^2
\;\Rightarrow\;
n\sum_{i=1}^{n} a_i^2
\;\ge\;
\left(\sum_{i=1}^{n} a_i\right)^2. \\[6pt]
&\text{Equality iff } a_1=a_2=\cdots=a_n.
\end{aligned}
$$
If $D={\begin{vmatrix}{1} & 1 & {1} \\ 1 & {2+x} & {1} \\ {1} & {1} & {2+y}\end{vmatrix}}\, for\, x\ne0,\, y\ne0$ then D is
1
2
3
4
Solution
Area of the parallelogram formed by the
lines y=4x, y=4x+1, x+y=0 and x+y=1
1
2
3
4
Solution
A four-digit number is formed using the
digits 1, 2, 3, 4, 5 without repetition. The
probability that is divisible by 3 is
1
2
3
4
Solution
For $a\in R$ (the set of al real numbers), $a \ne 1$, $\lim _{{n}\rightarrow\infty}\frac{({1}^a+{2}^a+{\ldots+{n}^a})}{{(n+1)}^{a-1}\lbrack(na+1)+(na+2)+\ldots+(na+n)\rbrack}=\frac{1}{60}$ . Then one of the value of $a$ is
1
2
3
4
Solution
If $\vec{a}=\lambda \hat{i}+\hat{j}-2\hat{k}$ , $\vec{b}=\hat{i}+\lambda \hat{j}-2\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}+\hat{k}$ and $\begin{bmatrix}{\vec{a}} & {\vec{b}} & {\vec{c}} \end{bmatrix}=7$, then the values of the $\lambda$ are
1
2
3
4
Solution
NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
Your reply to this comment:
Commented Jun 17, 2023
1 Upvote 0 Downvote Reply
Your reply to this comment: