NIMCET 2012 Previous Year Questions (PYQs) – Page 1 of 12
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If $H$ is the harmonic mean between $P$ and $Q$, then $\dfrac{H}{P} + \dfrac{H}{Q}$ is
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Solution
Harmonic mean of $P$ and $Q$ is
$H = \dfrac{2PQ}{P+Q}$
Now,
$\dfrac{H}{P} + \dfrac{H}{Q} = H\left(\dfrac{1}{P} + \dfrac{1}{Q}\right)
= H \cdot \dfrac{P+Q}{PQ}$
Substitute
$H = \dfrac{2PQ}{P+Q}$:
$\dfrac{H}{P} + \dfrac{H}{Q}
= \dfrac{2PQ}{P+Q} \cdot \dfrac{P+Q}{PQ}
= 2$
So the correct Answer is $2$.
The number of values of $k$ for which the system of equations
$(k+1)x + 8y = 4k$ and $kx + (k+3)y = 3k-1$ has infinitely many solutions, is
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Solution
$ (k+1)x + 8y = 4k $
$ kx + (k+3)y = 3k - 1 $
Since the system has infinitely many solutions,
$ \dfrac{k+1}{k} = \dfrac{8}{k+3} = \dfrac{4k}{3k-1} $
Taking 1st and 3rd ratio:
$ (k+1)(3k-1) = 4k^2 $
$ 3k^2 + 2k - 1 = 4k^2 $
$ k^2 - 2k + 1 = 0 $
$ (k-1)^2 = 0 $
$ \therefore k = 1 $
The sum
$^{20}C_8 + ^{20}C_9 + ^{21}C_{10} + ^{22}C_{11} - ^{23}C_{11}$
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Solution
Use Pascal identity:
$^{n}C_r + ^{n}C_{r+1} = ^{n+1}C_{r+1}$
So,
$^{20}C_8 + ^{20}C_9 = ^{21}C_9$
Then,
$^{21}C_9 + ^{21}C_{10} = ^{22}C_{10}$
Next,
$^{22}C_{10} + ^{22}C_{11} = ^{23}C_{11}$
Thus the expression becomes:
$^{23}C_{11} - ^{23}C_{11} = 0$
$\therefore$ the answer is 0.
The value of
$\cot^{-1}(21) + \cot^{-1}(13) + \cot^{-1}(-8)$
is:
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Solution
Use identity for inverse cotangent:
$\cot^{-1}a + \cot^{-1}b = \cot^{-1}\left(\dfrac{ab - 1}{a + b}\right)$
Apply to the first two terms:
$\cot^{-1}(21) + \cot^{-1}(13)$
$= \cot^{-1}\left(\dfrac{21 \cdot 13 - 1}{21 + 13}\right)$
$= \cot^{-1}\left(\dfrac{273 - 1}{34}\right)$
$= \cot^{-1}\left(\dfrac{272}{34}\right)$
$= \cot^{-1}(8)$
Now the expression becomes:
$\cot^{-1}(8) + \cot^{-1}(-8)$
But we know:
$\cot^{-1}(x) + \cot^{-1}(-x) = \pi$
So,
$\cot^{-1}(8) + \cot^{-1}(-8) = \pi$
$\therefore$ The value is $\pi$.
Normal to the curve
$y = x^3 - 3x + 2$
at the point $(2,4)$ is:
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Solution
Differentiate:
$\dfrac{dy}{dx} = 3x^2 - 3$
At $x = 2$:
$\dfrac{dy}{dx} = 3(4) - 3 = 12 - 3 = 9$
So slope of tangent $m_t = 9$
Slope of normal:
$m_n = -\dfrac{1}{9}$
Equation of normal at $(2,4)$:
$y - 4 = -\dfrac{1}{9}(x - 2)$
Multiply by 9:
$9y - 36 = -(x - 2)$
$9y - 36 = -x + 2$
$ x + 9y - 38 = 0$
Answer: (c) $x + 9y - 38 = 0$
The value of
$\displaystyle \lim_{n\to\infty} \frac{\pi}{n}\left[\sin\frac{\pi}{n}+\sin\frac{2\pi}{n}+\cdots+\sin\frac{(n-1)\pi}{n}\right]$
is:
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Solution
This is a Riemann sum.
Rewrite the expression:
$\displaystyle \frac{\pi}{n}\sum_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)$
Let
$x_k = \frac{k\pi}{n}$
Then spacing
$\Delta x = \frac{\pi}{n}$
As $n\to\infty$, this becomes:
$\displaystyle \int_{0}^{\pi} \sin x , dx$
Now evaluate:
$\displaystyle \int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi}$
$= (-\cos\pi) - (-\cos 0)$
$= -(-1) - (-1)$
$= 1 + 1 = 2$
$\therefore$ the value of the limit is 2.
The point on the curve
$y = 6x - x^2$
where the tangent is parallel to the x-axis is:
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Solution
Tangent ∥ x-axis ⇒ slope = 0
Differentiate:
$\dfrac{dy}{dx} = 6 - 2x$
Set derivative equal to zero:
$6 - 2x = 0$
$2x = 6$
$x = 3$
Now find $y$:
$y = 6(3) - (3)^2 = 18 - 9 = 9$
Therefore, the required point is $(3,9)$.
Answer: $(3,9)$
If
$ I_1 = \displaystyle \int_{0}^{1} 2^{x^2},dx,\quad
I_2 = \displaystyle \int_{0}^{1} 2^{x^3},dx,\quad
I_3 = \displaystyle \int_{1}^{2} 2^{x^2},dx,\quad
I_4 = \displaystyle \int_{1}^{2} 2^{x^3},dx,$
then
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Solution
On the interval $0 \le x \le 1$:
$ x^3 < x^2 \Rightarrow 2^{x^3} < 2^{x^2} $
So,
$ I_2 = \displaystyle \int_0^1 2^{x^3},dx < \int_0^1 2^{x^2}dx = I_1 $
Thus,
$ I_1 > I_2 $
On the interval $1 \le x \le 2$:
$ x^3 > x^2 \Rightarrow 2^{x^3} > 2^{x^2} $
So,
$ I_4 = \displaystyle \int_1^2 2^{x^3}dx > \int_1^2 2^{x^2}dx = I_3 $
Thus,
$ I_4 > I_3 $
The value of integral $\displaystyle \int_{0}^{\pi/2} \log \tan x dx$ is
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Solution
Let $I = \displaystyle \int_{0}^{\pi/2} \log \tan x dx$
Using property $\displaystyle \int_{0}^{\pi/2} f(x)dx = \int_{0}^{\pi/2} f\left(\dfrac{\pi}{2}-x\right)dx$
$I = \displaystyle \int_{0}^{\pi/2} \log \tan\left(\dfrac{\pi}{2}-x\right) dx$
$= \displaystyle \int_{0}^{\pi/2} \log \cot x dx$
$= \displaystyle \int_{0}^{\pi/2} \log\left(\dfrac{1}{\tan x}\right) dx$
$= \displaystyle \int_{0}^{\pi/2} [-\log \tan x] dx$
$I= -I$
$\Rightarrow I + I = 0 \Rightarrow 2I = 0 \Rightarrow I = 0$
Answer: $0$
A determinant is chosen at random from the set of all determinants of matrices of order 2 with elements 0 and 1 only.
The probability that the determinant chosen is non-zero is:
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Solution
Total $2\times2$ matrices with entries $0$ or $1$:
$2^4 = 16$
Determinant: $ad - bc$
Non-zero determinant occurs in two cases.
Case 1: $ad = 1$ and $bc = 0$
$a = 1, d = 1$
$(b,c)$ can be $(0,0), (0,1), (1,0)$
→ 3 matrices
Case 2: $bc = 1$ and $ad = 0$
$b = 1,c = 1$
$(a,d)$ can be $(0,0), (0,1), (1,0)$
→ 3 matrices
Total non-zero determinants = $3 + 3 = 6$
$P(\text{non-zero determinants})=\dfrac{6}{16}=\dfrac{3}{8}$
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