NIMCET 2009 Previous Year Questions (PYQs) – Page 1 of 12
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If
$ \theta = \tan^{-1}\dfrac{1}{1+2} + \tan^{-1}\dfrac{1}{1+2\cdot3} + \tan^{-1}\dfrac{1}{1+3\cdot4} + \ldots + \tan^{-1}\dfrac{1}{1+n(n+1)} $,
then $\tan\theta$ is equal to:
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Solution
We use identity
$\tan^{-1}a - \tan^{-1}b = \tan^{-1}\dfrac{a-b}{1+ab}$
Here each term satisfies
$\tan^{-1}\dfrac{1}{1+k(k+1)} = \tan^{-1}(k+1) - \tan^{-1}(k)$
So the series telescopes:
$\theta = \tan^{-1}(n+1) - \tan^{-1}(1)$
Thus
$\tan\theta = \dfrac{(n+1)-1}{1+(n+1)(1)} = \dfrac{n}{n+2}$
If
$(1 + x - 2x^2)^6 = 1 + a_1 x + a_2 x^2 + \ldots + a_{12} x^{12}$,
then the value of $a_2 + a_4 + a_6 + \ldots + a_{12}$ is:
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Solution
Even–power coefficient sum = $\dfrac{f(1) + f(-1)}{2}$
$f(1) = (1 + 1 - 2)^6 = 0^6 = 0$
$f(-1) = (1 - 1 - 2)^6 = (-2)^6 = 64$
Required sum $= \dfrac{0 + 64}{2} = 32$
A square with side $a$ is revolved about its centre through $45^\circ$.
What is the area common to both the squares?
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Solution
The overlapping area of a square rotated by $45^\circ$ is:
Common area $= 2(\sqrt{2}-1)a^2$
A and B throw a die in succession to win a bet with A starting first.
Whoever throws ‘1’ first wins Rs. 110.
What are the respective expectations of A and B?
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Solution
Let $P_A$ = probability A wins.
$P_A = \dfrac{1}{6} + \dfrac{5}{6}\cdot\dfrac{5}{6} P_A$
Solving:
$P_A = \dfrac{6}{11}$
$P_B = \dfrac{5}{11}$
Expected money:
$E_A = 110 \cdot \dfrac{6}{11} = 60$
$E_B = 110 \cdot \dfrac{5}{11} = 50$
How many different paths in the $xy$-plane are there from $(1,3)$ to $(5,6)$,
if a path proceeds one step at a time either right (R) or upward (U)?
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Solution
Right steps = $5-1 = 4$
Up steps = $6-3 = 3$
Total steps = $7$
Number of paths = $\binom{7}{3} = 35$
If the distance of $(x,y)$ from the origin is defined as
$d(x,y) = \max(|x|,|y|)$,
then the locus of points where $d(x,y)=1$ is:
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Solution
$\max(|x|,|y|)=1$ describes a square with vertices $(\pm1,\pm1)$.
Side length = $2$
Area = $4$
If
$\sin^{-1}x + \cos^{-1}(1-x) = \sin^{-1}(1-x)$
then $x$ satisfies the equation:
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Solution
Simplifying the given inverse–trigonometric equation gives:
$2x^2 - 3x = 0$
A and B are independent witnesses.
Probability A speaks the truth = $x$,
Probability B speaks the truth = $y$.
If both agree on a statement, the probability that the statement is true is:
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Solution
Both agree and the statement is true: Probability = $xy$
Both agree but the statement is false Probability = $(1-x)(1-y)$
So required probability:
$\dfrac{xy}{xy + (1-x)(1-y)}$
If $A$ is a $3\times 3$ matrix with $\det(A)=3$, then $\det(\operatorname{adj}A)$ is:
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Solution
For an $n \times n$ matrix:
$\det(\operatorname{adj}A) = (\det A)^{,n-1}$
Here:
$n = 3$, $\det A = 3$
So,
$\det(\operatorname{adj}A) = 3^{3-1} = 3^2 = 9$
A set contains $(2n+1)$ elements. If the number of subsets that contain at most $n$ elements is $4096$, then the value of $n$ is:
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Solution
For odd number of elements,
$\displaystyle \sum_{k=0}^{n} \binom{2n+1}{k} = 2^{2n}$
Given:
$2^{2n} = 4096 = 2^{12}$
So,
$2n = 12 \Rightarrow n = 6$
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