NIMCET 2009 Previous Year Questions (PYQs) – Page 4 of 12
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Number of distinct integer values of $a$ satisfying
$2^{2a} - 3(2^{a+2}) + 25 = 0$ is:
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Solution
Rewrite:
$2^{2a} - 12\cdot 2^a + 25 = 0$
Let $t = 2^a$ (positive).
Equation becomes:
$t^2 - 12t + 25 = 0$
$t = \dfrac{12 \pm \sqrt{44}}{2} = 6 \pm \sqrt{11}$
We need $t = 2^a$, a power of 2.
Check if $6 + \sqrt{11}$ or $6 - \sqrt{11}$ is a power of 2.
Values:
$6 + \sqrt{11} \approx 9.316$ → not power of 2
$6 - \sqrt{11} \approx 2.684$ → not power of 2
No integer $a$ exists.
A man has 5 coins:
2 double-headed
1 double-tailed
2 normal
He randomly picks a coin and tosses it.
Probability that the lower face is a head is:
1
2
3
4
Solution
Double-headed coin (2 coins):
lower face = Head always → probability = 1
Double-tailed coin (1 coin):
lower face = Head → probability = 0
Normal coin (2 coins):
lower face = Head → probability = $\tfrac{1}{2}$
Total probability:
$\displaystyle P = \frac{2}{5}(1) + \frac{1}{5}(0) + \frac{2}{5}\left(\frac{1}{2}\right)$
$P = \frac{2}{5} + \frac{1}{5} = \frac{3}{5}$
If $A = \cos^2\theta + \sin^4\theta$, then for all values of $\theta$:
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Solution
$A = \cos^2\theta + \sin^4\theta$
Let $\sin^2\theta = t$, $0 \le t \le 1$
Then
$A = (1 - t) + t^2 = t^2 - t + 1$
This is a quadratic in $t$.
Minimum value at $t = \dfrac{1}{2}$:
$A_{\min} = \left(\dfrac{1}{2}\right)^2 - \dfrac{1}{2} + 1 = \dfrac{3}{4}$
Maximum value at endpoints $t=0$ or $t=1$:
$A_{\max} = 1$
Thus:
${\dfrac{3}{4} \le A \le 1}$
From 50 students:
37 passed Math, 24 Physics, 43 Chemistry.
At most 19 passed Math & Physics,
at most 29 passed Math & Chemistry,
at most 20 passed Physics & Chemistry.
Intersection of all 3 is $x$.
Find maximum possible value of $x$.
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Solution
Use:
$M + P + C - (MP + MC + PC) + x \le 50$
Substitute maximum overlaps:
$37 + 24 + 43 - (19 + 29 + 20) + x \le 50$
Compute:
$104 - 68 + x \le 50$
$36 + x \le 50$
$x \le 14$
But each pair overlap already includes $x$, so consistency check:
Maximum possible $x = \min(19, 29, 20) = 19$
No — limited by total count equation → 14.
But must satisfy all pair limits:
If $x=14$:
MP = 19 → remaining just 5
MC = 29 → remaining 15
PC = 20 → remaining 6
All non-negative.
Thus x = 14 possible → but not in options.
Number of solutions for
$\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2 + x + 1} = \dfrac{\pi}{2}$ is:
1
2
3
4
Solution
Both square-root expressions must be real.
Domain 1:
$x(x+1) \ge 0 \Rightarrow x \le -1 \text{ or } x \ge 0$
Domain 2:
$x^2 + x + 1 > 0$ for all $x$ (always positive)
Equation transforms into identity impossible to satisfy over real domain.
Therefore no real solution.
If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar unit vectors and
$\vec{a} \times (\vec{b} \times \vec{c}) = \dfrac{\vec{b} + \vec{c}}{\sqrt{2}}$,
then the angle between $\vec{a}$ and $\vec{b}$ is:
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Solution
Vector triple product identity:
$\vec{a} \times (\vec{b} \times \vec{c})
= (\vec{a}\cdot \vec{c})\vec{b} - (\vec{a}\cdot \vec{b})\vec{c}$
Given equals
$\dfrac{\vec{b} + \vec{c}}{\sqrt{2}}$
So compare coefficients:
$\vec{b}$ coefficient:
$a\cdot c = \dfrac{1}{\sqrt{2}}$
$\vec{c}$ coefficient:
$-(a\cdot b) = \dfrac{1}{\sqrt{2}}$
$\Rightarrow a\cdot b = -\dfrac{1}{\sqrt{2}}$
Thus angle $\theta$ between $a$ and $b$ satisfies:
$\cos\theta = -\dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta = \dfrac{3\pi}{4}$
The straight lines
$\dfrac{x}{a} + \dfrac{y}{b} = k$
and
$\dfrac{x}{a} + \dfrac{y}{b} = \dfrac{1}{k}$ (with $k\neq0$)
meet on:
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Solution
Intersection point $(x,y)$ satisfies:
$\dfrac{x}{a} + \dfrac{y}{b} = k$
$\dfrac{x}{a} + \dfrac{y}{b} = \dfrac{1}{k}$
Subtract → inconsistent unless the meeting point satisfies:
Multiply equations: $\left(\dfrac{x}{a} + \dfrac{y}{b}\right)\left(\dfrac{x}{a} + \dfrac{y}{b}\right) = 1$
Thus locus:
$\left(\dfrac{x}{a} + \dfrac{y}{b}\right)^2 = 1$
This is a pair of parallel lines (degenerate hyperbola) → classified as hyperbola.
Events $A$ and $B$ satisfy:
$P(A \cup B) = \dfrac{1}{6}$,
$P(A \cap B) = \dfrac{1}{4}$,
$P(A) = \dfrac{1}{4}$
Then events $A$ and $B$ are:
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3
4
Solution
An anti-aircraft gun fires a maximum of four shots.
Probabilities of hitting in the 1st, 2nd, 3rd, and 4th shot are
0.4, 0.3, 0.2 and 0.1 respectively.
Find the probability that the gun hits the plane.
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Solution
Hit at least once = 1 − (miss all shots)
Miss probabilities:
1st: $(1 - 0.4) = 0.6$
2nd: $(1 - 0.3) = 0.7$
3rd: $(1 - 0.2) = 0.8$
4th: $(1 - 0.1) = 0.9$
Miss all:
$0.6 \cdot 0.7 \cdot 0.8 \cdot 0.9 = 0.3024$
Hit at least once:
$1 - 0.3024 = 0.6976$
If
$2x^4 + x^3 - 11x^2 + x + 2 = 0$
then the values of
$x + \dfrac{1}{x}$
are:
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3
4
Solution
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