NIMCET Mathematics Previous Year Questions (PYQs) – Page 92 of 95
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Suppose that A and B are two events with probabilities $P(A) =\frac{1}{2} \, P(B)=\frac{1}{3}$ Then which of the
following is true?
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Solution
If $\prod ^n_{i=1}\tan ({{\alpha}}_i)=1\, \forall{{\alpha}}_i\, \in\Bigg{[}0,\, \frac{\pi}{2}\Bigg{]}$ where i=1,2,3,...,n. Then maximum value of $\prod ^n_{i=1}\sin ({{\alpha}}_i)$.
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Solution
If $\vec{a}, \vec{b}$ are vector such that $|\vec{a}+\vec{b}|=\sqrt{29}$ and
$\vec{a}\times (2\hat{i}+3\hat{j}+4\hat{k}) = (2\hat{i}+3\hat{j}+4\hat{k})\times \vec{b}$, then a possible value of $(\vec{a}+\vec{b})\cdot(-7\hat{i}+2\hat{j}+3\hat{k})$ is
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Solution
Let $A_k$ be the arithmetic mean of squares of $k$ natural numbers. If $\sum_{k=1}^{n}(6A_k-3k)=31$, find the value of $n$.
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Solution
$A_k=\frac{1^2+2^2+3^2+\cdots+k^2}{k}$
$A_k=\frac{\frac{k(k+1)(2k+1)}{6}}{k}$
$A_k=\frac{(k+1)(2k+1)}{6}$
Now,
$6A_k-3k=(k+1)(2k+1)-3k$
$=2k^2+3k+1-3k$
$=2k^2+1$
Given,
$\sum_{k=1}^{n}(6A_k-3k)=31$
$\sum_{k=1}^{n}(2k^2+1)=31$
For $n=3$,
$(2\cdot1^2+1)+(2\cdot2^2+1)+(2\cdot3^2+1)$
$=3+9+19=31$
Therefore,
$n=3$
If the volume of a parallelepiped whose adjacent edges are
a = 2i + 3j + 4k,
b = i + αj + 2k
c = i + 2k + αk
is 15, then α =
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Solution
The number of one-to-one functions from {1, 2, 3} to {1, 2, 3, 4, 5} is
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Solution
A speaks truth in 60% and B speaks the truth in 50% cases. In what percentage of cases they are likely incontradict each other while narrating some incident is
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Solution
They contradict when one tells the truth and the other lies.
Let \( p_A = 0.6 \) (A tells truth) and \( p_B = 0.5 \) (B tells truth).
\[ P(\text{contradict}) \;=\; P(A\ \text{true},\,B\ \text{lie}) + P(A\ \text{lie},\,B\ \text{true}) \;=\; p_A(1-p_B) + (1-p_A)p_B. \]
\[ =\; 0.6(1-0.5) + (1-0.6)\cdot 0.5 \;=\; 0.6\cdot 0.5 + 0.4\cdot 0.5 \;=\; 0.30 + 0.20 \;=\; 0.50. \]
Answer: \(50\%\).
Let $x_1,x_2,\ldots,x_n$ be $n$ observations such that $\sum x_i^2 = 400$ and $\sum x_i = 80$. Then a possible value of $n$ amongst the following is
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Solution
Given: $\sum x_i^2 = 400$ and $\sum x_i = 80$
Using the inequality:
$\sum x_i^2 \geq \dfrac{(\sum x_i)^2}{n}$
This follows from Cauchy-Schwarz inequality:
$n \cdot \sum x_i^2 \geq \left(\sum x_i\right)^2$
$\Rightarrow n \times 400 \geq (80)^2$
$\Rightarrow 400n \geq 6400$
$\Rightarrow n \geq \dfrac{6400}{400}$
$\Rightarrow n \geq 16$
Also, Variance $\geq 0$:
$\dfrac{\sum x_i^2}{n} - \left(\dfrac{\sum x_i}{n}\right)^2 \geq 0$
$\Rightarrow \dfrac{400}{n} - \dfrac{6400}{n^2} \geq 0$
$\Rightarrow \dfrac{400n - 6400}{n^2} \geq 0$
$\Rightarrow n \geq 16$
$\therefore$ A possible value of $n$ is $\boxed{n = 20}$
Using the inequality:
$\sum x_i^2 \geq \dfrac{(\sum x_i)^2}{n}$
This follows from Cauchy-Schwarz inequality:
$n \cdot \sum x_i^2 \geq \left(\sum x_i\right)^2$
$\Rightarrow n \times 400 \geq (80)^2$
$\Rightarrow 400n \geq 6400$
$\Rightarrow n \geq \dfrac{6400}{400}$
$\Rightarrow n \geq 16$
Also, Variance $\geq 0$:
$\dfrac{\sum x_i^2}{n} - \left(\dfrac{\sum x_i}{n}\right)^2 \geq 0$
$\Rightarrow \dfrac{400}{n} - \dfrac{6400}{n^2} \geq 0$
$\Rightarrow \dfrac{400n - 6400}{n^2} \geq 0$
$\Rightarrow n \geq 16$
$\therefore$ A possible value of $n$ is $\boxed{n = 20}$
$\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}$
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Solution
$\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3x}$
Using expansion:
$\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}$
$x^2+2\cos x-2=x^2+2\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)-2$
$=x^2+2-x^2+\frac{x^4}{12}-2$
$=\frac{x^4}{12}$
Also,
$\sin x\approx x$
So,
$x\sin^3x\approx x\cdot x^3=x^4$
Therefore,
$\lim_{x\to 0}\frac{x^2+2\cos x-2}{x\sin^3x}$
$=\frac{x^4/12}{x^4}$
$=\frac{1}{12}$
Solve the equation sin2 x - sinx - 2 = 0 for for x on
the interval 0 ≤ x < 2π
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Solution
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