NIMCET 2027 2022 Previous Year Questions (PYQs) – Page 2 of 12
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If the volume of the parallelepiped whose adjacent edges are $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$, $\vec{b}=\hat{i}+\alpha \hat{j}+2\hat{k}$ and $\vec{c}=\hat{i}+2\hat{j}+\alpha \hat{k}$ is 15, then $\alpha$ is equal to
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Let $a$ be the distance between the lines $−2x + y = 2$ and $2x − y = 2$, and $b$ be the distance between the lines $4x − 3y= 5$ and
$6y − 8x = 1$, then
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If $cosec\theta-cot \theta=2$, then the value of $cosec\theta$ is
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Given $\,\csc\theta-\cot\theta=2\,$ and the identity $\;(\csc\theta-\cot\theta)(\csc\theta+\cot\theta)=1\;$, we get
$$\csc\theta+\cot\theta=\frac{1}{2}.$$
Adding the two equations:
$$(\csc\theta-\cot\theta)+(\csc\theta+\cot\theta) $$ $$=2+\frac{1}{2}$$ $$\;\Rightarrow\;2\,\csc\theta=\frac{5}{2}.$$
Hence
$$\csc\theta=\frac{5}{4}.$$
The solution of the equation ${4\cos }^2x+6{\sin }^2x=5$ are
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The function $f(x)=\begin{cases}{{(1+2x)}^{1/x}} & {,x\ne0} \\ {{e}^2} & {,x=0}\end{cases}$, is
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Let $\vec{a}=2\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}$ be another vector such that $\vec{a}.\vec{b}=14$ and $\vec{a} \times \vec{b}=3\hat{i}+\hat{j}-8\hat{k}$ the vector $\vec{b}$ =
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Which term of the series $\frac{\sqrt[]{5}}{3},\, \frac{\sqrt[]{5}}{4},\frac{1}{\sqrt[]{5}},\, ...$ is $\frac{\sqrt{5}}{13}$ ?
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Step 1: Rewrite all terms with same numerator
$\dfrac{\sqrt{5}}{3},\ \dfrac{\sqrt{5}}{4},\ \dfrac{\sqrt{5}}{5},\ \ldots$
(Since $\dfrac{1}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}$)
Step 2: Identify the AP in denominators
Denominators: $3,\ 4,\ 5,\ \ldots$
First term $a = 3$
Common difference $d = 1$
Step 3: Set up the equation
The $n^{th}$ term of the series $= \dfrac{\sqrt{5}}{a + (n-1)d}$
Given $n^{th}$ term $= \dfrac{\sqrt{5}}{13}$
So denominator of $n^{th}$ term $= 13$
Step 4: Solve for $n$
$a + (n-1)d = 13$
$3 + (n-1) \times 1 = 13$
$n - 1 = 10$
$n = 11$
Answer: $\dfrac{\sqrt{5}}{13}$ is the $\boxed{11^{th}}$ term of the series.
$\dfrac{\sqrt{5}}{3},\ \dfrac{\sqrt{5}}{4},\ \dfrac{\sqrt{5}}{5},\ \ldots$
(Since $\dfrac{1}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}$)
Step 2: Identify the AP in denominators
Denominators: $3,\ 4,\ 5,\ \ldots$
First term $a = 3$
Common difference $d = 1$
Step 3: Set up the equation
The $n^{th}$ term of the series $= \dfrac{\sqrt{5}}{a + (n-1)d}$
Given $n^{th}$ term $= \dfrac{\sqrt{5}}{13}$
So denominator of $n^{th}$ term $= 13$
Step 4: Solve for $n$
$a + (n-1)d = 13$
$3 + (n-1) \times 1 = 13$
$n - 1 = 10$
$n = 11$
Answer: $\dfrac{\sqrt{5}}{13}$ is the $\boxed{11^{th}}$ term of the series.
The first three moments of a distribution about 2 are 1, 16, -40 respectively. The mean and variance of the distribution are
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A survey is done among a population of 200 people who like either tea or coffee. It is found that 60% of the pop lation like tea and 72% of the population like coffee. Let $x$ be the number of people who like both tea & coffee. Let $m{\leq x\leq n}$, then choose the correct option.
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Total people = 200
People who like tea = \(60\% \times 200 = 120\)
People who like coffee = \(72\% \times 200 = 144\)
People who like tea = \(60\% \times 200 = 120\)
People who like coffee = \(72\% \times 200 = 144\)
Using the set formula: \[ |T \cup C| = |T| + |C| - |T \cap C| \] \[ 200 = 120 + 144 - x \quad \Rightarrow \quad x = 64 \]
Minimum possible intersection: \[ x \geq |T| + |C| - 200 = 64 \] Maximum possible intersection: \[ x \leq \min(120,144) = 120 \]
✅ Final Answer
The range of \(x\) is: 64 ≤ x ≤ 120 Hence, \(m = 64, \; n = 120\).
The value of $\cot \Bigg{(}{cosec}^{-1}\frac{5}{3}+{\tan }^{-1}\frac{2}{3}\Bigg{)}$ is
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Solution
NIMCET 2027
Online Test Series, Information About Examination,
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and More.
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NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
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