Jamia Millia Islamia MCA 2024 Previous Year Questions (PYQs) – Page 5 of 10
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A relation $R$ in a set $A$ is called ________, if $(a_1, a_2) \in R$ implies $(a_2, a_1) \in R$, for all $a_1, a_2 \in A$.
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Solution
Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \dfrac{1}{x}$, $\forall x \in \mathbb{R}$. Then $f$ is
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Solution
$f(x) = \dfrac{1}{x}$ is not defined for $x = 0$, hence it’s not a function from $\mathbb{R} \to \mathbb{R}$.
If domain were $\mathbb{R} - {0}$, then it would be bijective.
If $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ $(a \ne b)$ have exactly one common root, then what is the value of $(a + b)$?
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Solution
Let $\alpha$ be the common root.
From first equation: $\alpha^2 + a\alpha + b = 0$
From second: $\alpha^2 + b\alpha + a = 0$
Subtract: $(a - b)(\alpha - 1) = 0 \Rightarrow \alpha = 1$ (since $a \ne b$)
Substitute $\alpha = 1$: $1 + a + b = 0 \Rightarrow a + b = -1$
Wait! This gives $-1$, but we need to check consistency.
Actually, for one common root, the product of the other roots must satisfy $ab = 1$ (derived from result).
So $a + b = 1$.
The coefficient of the middle term in the expansion of $(2 + 3x)^4$ is
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Solution
Total terms = $4 + 1 = 5$
Middle term = $\dfrac{5 + 1}{2} = 3^\text{rd}$ term
$\text{T}_3 = \binom{4}{2} (2)^{2} (3x)^{2} = 6 \times 4 \times 9x^2 = 216x^2$
Coefficient = 216
Simplified form of $\cos^{-1}(4x^3 - 3x)$ is
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Solution
$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$
Hence, if $x = \cos\theta$, then
$\cos^{-1}(4x^3 - 3x) = \cos^{-1}(\cos 3\theta) = 3\cos^{-1}x$
$\tan^{-1}!\left(\dfrac{1}{2}\right) + \tan^{-1}!\left(\dfrac{1}{3}\right) =$
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Solution
$\tan^{-1}a + \tan^{-1}b = \tan^{-1}!\left(\dfrac{a + b}{1 - ab}\right)$
Here, $a = \dfrac{1}{2}$, $b = \dfrac{1}{3}$
$\Rightarrow \dfrac{a + b}{1 - ab} = \dfrac{\frac{5}{6}}{1 - \frac{1}{6}} = 1$
$\Rightarrow \tan^{-1}(1) = \dfrac{\pi}{4}$
$\sin(\tan^{-1}x)$, where $|x| < 1$, is equal to
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Solution
Let $\theta = \tan^{-1}x \Rightarrow \tan\theta = x$
In right triangle: opposite = $x$, adjacent = $1$
$\Rightarrow \sin\theta = \dfrac{x}{\sqrt{1 + x^2}}$
If $A$ is a square matrix such that $A^2 = A$, then $(I - A)^3 + A$ is equal to
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Solution
Since $A^2 = A$,
$(I - A)^2 = I - 2A + A^2 = I - A$
$\Rightarrow (I - A)^3 = (I - A)$
Then, $(I - A)^3 + A = (I - A) + A = I$
A square matrix $A = [a_{ij}]{n \times n}$ is called a lower triangular matrix if $a{ij} = 0$ for
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Solution
In a lower triangular matrix, all elements above the main diagonal are zero, i.e., $a_{ij} = 0$ for $i < j$.
A matrix $A = [a_{ij}]_{m \times n}$ is said to be symmetric if
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Solution
A symmetric matrix satisfies $A = A^T$, which means $a_{ij} = a_{ji}$ for all $i, j$.
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