NIMCET Mathematics Previous Year Questions (PYQs) – Page 94 of 95
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A professor has 24 text books on computer science and is concerned about their coverage of the topics (P) compilers, (Q) data structures and (R) Operating systems. The following data gives the number of books that contain material on these topics: $n(P) = 8, n(Q) = 13, n(R) = 13,
n(P \cap R) = 3, n(P \cap R) = 3, n(Q \cap R) = 3, n(Q \cap R) = 6, n(P \cap Q \cap R) = 2 $ where $n(x)$ is the cardinality of the set $x$. Then the number of text books that have no material on compilers is
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Solution
Given: Total books = 24, and books with compilers \(n(P)=8\).
Asked: Number of books with no material on compilers = \(|P'|\).
\[ |P'| = \text{Total} - |P| = 24 - 8 = \boxed{16}. \]
Note: The other intersection counts aren’t needed since “no compilers” depends only on \(n(P)\).
Let A and B be sets. $A\cap X=B\cap X=\phi$ and $A\cup X=B\cup X$ for some set X, relation between A & B
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Solution
Given: \(A\cap X=\varnothing,\; B\cap X=\varnothing\) and \(A\cup X = B\cup X\).
Since \(A\cap X=\varnothing\), the union is a disjoint union, so \[ A = (A\cup X)\setminus X. \] Similarly, \(B = (B\cup X)\setminus X.\)
But \(A\cup X = B\cup X\). Hence \[ A = (A\cup X)\setminus X = (B\cup X)\setminus X = B. \]
Conclusion: \(\boxed{A=B}\).
If $a_1,a_2,\ldots,a_n$ are in A.P. and $a_1=0$ then the value of
$\left(\dfrac{a_3}{a_2}+\dfrac{a_4}{a_3}+\cdots+\dfrac{a_n}{a_{n-1}}\right)-a_2\left(\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_{n-2}}\right)$
is equal to
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Solution
Evaluate the following definite integral:$\int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos^4 x}$
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Solution
$\int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos^4 x}$
$=\int_{0}^{\frac{\pi}{4}} \sec^4 xdx$
Now,
$\sec^4 x=\sec^2 x(1+\tan^2 x)$
Let $\tan x=t$
$\sec^2 xdx=dt$
Limits:
$x=0 \Rightarrow t=0$
$x=\frac{\pi}{4} \Rightarrow t=1$
So,
$\int_0^1 (1+t^2)dt$
$=\left[t+\frac{t^3}{3}\right]_0^1$
$=1+\frac{1}{3}$
$=\frac{4}{3}$
If cos x = tan y , cot y = tan z and cot z = tan x, then sinx =
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Solution
If a, b, c, d are in HP and arithmetic mean of ab, bc, cd is 9 then which of the following number is the value of ad?
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Solution
$a, b, c, d$ are in H.P.
$\Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ and $\frac{2}{c} = \frac{1}{b} + \frac{1}{d}$
But $ab + bc + cd = 27$:
$\Rightarrow \frac{1}{a}, \frac{1}{b}, \frac{1}{c}, \frac{1}{d}$ are in A.P.
$\Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ and $\frac{2}{c} = \frac{1}{b} + \frac{1}{d}$
Given A.M. of $ab, bc, cd$ is $9$:
$\frac{ab + bc + cd}{3} = 9$
$\Rightarrow ab + bc + cd = 27$
Now multiply:
$\frac{2}{b} = \frac{1}{a} + \frac{1}{c} $
$\Rightarrow 2ac = b(a + c)$
$\frac{2}{c} = \frac{1}{b} + \frac{1}{d} \Rightarrow 2bd = c(b + d)$
Multiply both:
$4abcd = bc(a + c)(b + d)$
Cancel $bc$:
$4ad = (a + c)(b + d)$
Expand RHS:
$4ad = ab + ad + bc + cd$
$\Rightarrow 3ad = ab + bc + cd$
But $ab + bc + cd = 27$:
$\Rightarrow 3ad = 27$
$\Rightarrow ad = 9$
$\boxed{9}$
The value of $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$ is
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Solution
We need to find: $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ$
Using sum-to-product formula on first two terms:
$\cos 20^\circ + \cos 100^\circ = 2\cos\left(\dfrac{20^\circ + 100^\circ}{2}\right)\cos\left(\dfrac{100^\circ - 20^\circ}{2}\right)$
$= 2\cos 60^\circ \cos 40^\circ$
$= 2 \times \dfrac{1}{2} \times \cos 40^\circ$
$= \cos 40^\circ$
So the expression becomes:
$\cos 40^\circ + \cos 140^\circ$
Again applying sum-to-product formula:
$= 2\cos\left(\dfrac{40^\circ + 140^\circ}{2}\right)\cos\left(\dfrac{140^\circ - 40^\circ}{2}\right)$
$= 2\cos 90^\circ \cos 50^\circ$
$= 2 \times 0 \times \cos 50^\circ$
$= 0$
$\therefore \boxed{\cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0}$
Using sum-to-product formula on first two terms:
$\cos 20^\circ + \cos 100^\circ = 2\cos\left(\dfrac{20^\circ + 100^\circ}{2}\right)\cos\left(\dfrac{100^\circ - 20^\circ}{2}\right)$
$= 2\cos 60^\circ \cos 40^\circ$
$= 2 \times \dfrac{1}{2} \times \cos 40^\circ$
$= \cos 40^\circ$
So the expression becomes:
$\cos 40^\circ + \cos 140^\circ$
Again applying sum-to-product formula:
$= 2\cos\left(\dfrac{40^\circ + 140^\circ}{2}\right)\cos\left(\dfrac{140^\circ - 40^\circ}{2}\right)$
$= 2\cos 90^\circ \cos 50^\circ$
$= 2 \times 0 \times \cos 50^\circ$
$= 0$
$\therefore \boxed{\cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0}$
A die is rolled twice independently. What is the probability that either the first die shows a number no less than 4 or the second die shows at least 4?
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Solution
Let
$A=$ first die shows a number no less than 4
So,
$A={4,5,6}$
$P(A)=\frac{3}{6}=\frac{1}{2}$
Let
$B=$ second die shows at least 4
So, $B={4,5,6}$
$P(B)=\frac{3}{6}=\frac{1}{2}$
Required probability:
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$=\frac{1}{2}+\frac{1}{2}-\frac{1}{2}\cdot\frac{1}{2}$
$=1-\frac{1}{4}$
$=\frac{3}{4}$
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