Jamia Millia Islamia MCA Function Previous Year Questions (PYQs) – Page 1 of 3
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If $[x^2] - 5[x] + 6 = 0$, where $[\,]$ denotes the greatest integer function, then
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Solution
Let $[x] = n$.
Then $[x^2] = n^2$ (since $x^2$ lies between $n^2$ and $(n+1)^2$).
Equation: $n^2 - 5n + 6 = 0$
$\Rightarrow (n - 2)(n - 3) = 0$
$\Rightarrow n = 2$ or $n = 3$
For $n = 2 \Rightarrow x \in [2,3)$
For $n = 3 \Rightarrow x \in [3,4)$
Hence $x \in [2,4)$
The minimum value of $4^x + 4^{1-x},\ x \in \mathbb{R}$ is:
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Solution
Let $4^x = t,\ t > 0$.
Then expression $= t + \dfrac{4}{t}$.
By AM ≥ GM,
$t + \dfrac{4}{t} \ge 2\sqrt{t \cdot \dfrac{4}{t}} = 4.$
Equality when $t = 2$, i.e., $4^x = 2 \Rightarrow x = \dfrac{1}{2}$.
$\boxed{\text{Answer: (B) 4}}$
Let $f(x) = x - [x]$, where $[\,]$ denotes the greatest integer function.
Then $f\left(\dfrac{5}{2}\right)$ is:
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Solution
We have $x = \dfrac{5}{2} = 2.5$
$[x] = 2$
So, $f(x) = x - [x] = 2.5 - 2 = 0.5$
$\boxed{\text{Answer: (A) }\dfrac{3}{2}}$
The floor function $[x]$ is
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Solution
The floor function $f(x) = [x]$ maps every real number to the greatest integer less than or equal to $x$.
Different real numbers can have the same floor value, so it is not one-to-one,
but for every integer $n \in \mathbb{Z}$, there exists an $x \in \mathbb{R}$ such that $[x]=n$.
Hence, it is onto but not one-to-one.
The domain of the real-valued function
$f(x) = \sqrt{x - 3} + \sqrt{x - 4}$
is the set of all values of $x$ satisfying
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Solution
For $f(x)$ to be real, both radicals must be defined:
$x - 3 \ge 0 \quad \text{and} \quad x - 4 \ge 0$
$\Rightarrow x \ge 4$
So the domain is $[4, \infty)$.
Let $f:\mathbb{R}\to\mathbb{R},\; g:\mathbb{R}\to\mathbb{R}$ be given by $f(x)=2x-3$ and $g(x)=x/2$.
Then $(f\circ g)^{-1}(x)$ is equal to
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Solution
$(f\circ g)(x)=f(x/2)=x-3$. Its inverse solves $y=x-3\Rightarrow x=y+3$,
so $(f\circ g)^{-1}(x)=x+3$.
Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x)=x^{2}+5$. Then the value of $f^{-1}(4)$ is
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Solution
Range of $x^{2}+5$ is $[5,\infty)$. Since $4$ is not in the range,
$f^{-1}(4)$ does not exist.
If $g:\mathbb{R}\to\mathbb{R}$ is defined by $g(x)=x^{2}-2$, then the value of $g^{-1}(23)$ is
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Solution
Solve $x^{2}-2=23 \Rightarrow x^{2}=25 \Rightarrow x=\pm5$.
The value of $[1/2][5/2]$ is.....
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Solution
Here, $[x]$ denotes the greatest integer function (GIF).
$[1/2] = 0$, and $[5/2] = 2$.
Therefore, $[1/2][5/2] = 0 \times 2 = 0.$
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