Jamia Millia Islamia Function Previous Year Question Paper and Solution with PDF

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Jamia Millia Islamia Function PYQ

Qus : 1 Jamia Millia Islamia PYQ

If $[x^2] - 5[x] + 6 = 0$, where $[\,]$ denotes the greatest integer function, then

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

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 $x \in [2,4)$

Qus : 2 Jamia Millia Islamia PYQ

The minimum value of $4^x + 4^{1-x},\ x \in \mathbb{R}$ is:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

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., $4^x = 2 \Rightarrow x = \dfrac{1}{2}$. $\boxed{\text{Answer: (B) 4}}$

Qus : 3 Jamia Millia Islamia PYQ

Let $f(x) = x - [x]$, where $[\,]$ denotes the greatest integer function. Then $f\left(\dfrac{5}{2}\right)$ is:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2021 PYQ

Solution

We have $x = \dfrac{5}{2} = 2.5$ $[x] = 2$ So, $f(x) = x - [x] = 2.5 - 2 = 0.5$ $\boxed{\text{Answer: (A) }\dfrac{3}{2}}$

Qus : 4 Jamia Millia Islamia PYQ

The floor function $[x]$ is

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2023 PYQ

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.

Qus : 5 Jamia Millia Islamia PYQ

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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2023 PYQ

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 $[4, \infty)$.

Qus : 6 Jamia Millia Islamia PYQ

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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

$(f\circ g)(x)=f(x/2)=x-3$. Its inverse solves $y=x-3\Rightarrow x=y+3$, so $(f\circ g)^{-1}(x)=x+3$.

Qus : 7 Jamia Millia Islamia PYQ

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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

Range of $x^{2}+5$ is $[5,\infty)$. Since $4$ is not in the range, $f^{-1}(4)$ does not exist.

Qus : 8 Jamia Millia Islamia PYQ

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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MCA 2017 PYQ

Solution

Solve $x^{2}-2=23 \Rightarrow x^{2}=25 \Rightarrow x=\pm5$.

Qus : 9 Jamia Millia Islamia PYQ

The value of $[1/2][5/2]$ is.....

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2020 PYQ

Solution

Here, $[x]$ denotes the greatest integer function (GIF). $[1/2] = 0$, and $[5/2] = 2$. Therefore, $[1/2][5/2] = 0 \times 2 = 0.$

Qus : 10 Jamia Millia Islamia PYQ

If $f(x)=ax^7+bx^3+cx-5$, where $a,b,c$ are real constants, and $f(-7)=7$, then the range of $f(7)+17\cos x$ is:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

$f(7)=-(f(-7))= -7 + 10 = 27$ (since odd-powered terms change sign). Actually, $f(7)=-f(-7)-10=...$ simplifying gives $f(7)=17$. Then $f(7)+17\cos x = 17 + 17\cos x$. Range = $[17-17,17+17]=[0,34]$.

Qus : 11 Jamia Millia Islamia PYQ

The domain of $\sqrt{|x-2|-1}+\sqrt{\,3-|x-2|\,}$ is:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2022 PYQ

Solution

$\sqrt{|x-2|-1}$ needs $|x-2|\ge1\Rightarrow x\le1$ or $x\ge3$. $\sqrt{3-|x-2|}$ needs $|x-2|\le3\Rightarrow x\in[-1,5]$. Intersection $\Rightarrow [-1,1]\cup[3,5]$.

Qus : 12 Jamia Millia Islamia PYQ

If $f : \mathbb{R} \to \mathbb{R}$ such that $f(x) = 3x$, then what type of function is $f$?

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2024 PYQ

Solution

Number of equivalence relations = Number of partitions of the set = Bell number $B_3 = 5$.

Qus : 13 Jamia Millia Islamia PYQ

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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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2024 PYQ

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.

Qus : 14 Jamia Millia Islamia PYQ

The function $f:[0,3] \to [1,29]$ defined by $f(x) = 2x^3 - 15x^2 + 36x + 1$ is:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

$f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3).$ Sign chart: • $f'$ > 0 for $x<2$; $f'$ < 0 for $2

Qus : 15 Jamia Millia Islamia PYQ

If $f:\mathbb{R}\to\mathbb{R}$ is given by $f(x) = (3 - x^3)^{1/3}$, then $f(f(f(f(x))))$ is:

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

Let $y=f(x)=(3 - x^3)^{1/3}$. Then $f(f(x)) = (3 - y^3)^{1/3} = (3 - (3 - x^3))^{1/3} = x$. Hence $f(f(f(f(x)))) = f(f(x)) = x$.

Qus : 16 Jamia Millia Islamia PYQ

Consider two functions $f(x)$ and $g(x)$ such that $f(x) = |x| + [x]$ and $g(x) = |x|[x]$, where $[x]$ denotes the greatest integer function.

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2019 PYQ

Solution

At $x=1^-$: $f(x)=|x|+[x]=1+0=1$; at $x=1$: $f(1)=|1|+[1]=2$. ⇒ jump ⇒ discontinuous. $g(x)=|x|[x]$: at $x=1^-$ → $1×0=0$; at $x=1$ → $1×1=1$. ⇒ discontinuous.

Qus : 17 Jamia Millia Islamia PYQ

Two men on a 3-D surface want to meet each other. The surface is given by $f(x,y) = \dfrac{x - 6y}{x + y}$ They move horizontally/vertically; one starts at $(200,400)$, other at $(100,100)$; meeting point $(0,0)$.

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

Both follow straight lines through origin → they meet.

Qus : 18 Jamia Millia Islamia PYQ

The graph of $y=f(x)$ is symmetrical about line $x=2$.

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Jamia Millia Islamia Previous Year PYQ Jamia Millia Islamia JAMIA MILLIA ISLAMIA MCA 2018 PYQ

Solution

For symmetry about $x=2$, $f(2+x)=f(2-x)$.

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