Mathematics - MCQ Practice Questions

Understanding:

We need to check whether $f(x)=\frac{x^2-1}{x^2+1}$ is one-one (injective) and onto (surjective) from $\mathbb{R}$ to $\mathbb{R}$.

Formula:

A function is one-one if $f(a)=f(b)\Rightarrow a=b$.

A function is onto if for every $y\in \mathbb{R}$, there exists $x\in \mathbb{R}$ with $f(x)=y$.

Step 1: Check one-one

Note that $f(x)$ depends on $x^2$, so $f(1)=f(-1)=0$ but $1\ne -1$.

Hence $f$ is not one-one.

Step 2: Find the range of $f$

Rewrite:

For $x^2\ge 0$ we need $\frac{1+y}{1-y}\ge 0$, which holds when $-1\le y<1$.

Also as $x\to \infty$, $f(x)\to 1$ but never reaches $1$.

So the range of $f$ is $[-1,1)$, which is a proper subset of $\mathbb{R}$.

Hence $f$ is not onto.

Answer:

$f$ is neither one-one nor onto.

Quick Tip:

Whenever $f(x)$ involves only $x^2$ (an even function), it automatically fails to be one-one on $\mathbb{R}$.

Understanding:

We need to find the number of elements in $(A\cup B)-(A\cap B)$, the symmetric difference of $A$ and $B$.

Formula:

This is the symmetric difference $A△B$.

Step 1: Find $A\cap B$

Step 2: Find $A\cup B$

Step 3: Compute $(A\cup B)-(A\cap B)$

Number of elements $=5$.

Answer:

The number of elements in $(A\cup B)-(A\cap B)$ is $5$.

Quick Tip:

$(A\cup B)-(A\cap B)$ is always the symmetric difference $A△B$; its size equals $|A|+|B|-2|A\cap B|$.

Understanding:

We need the domain of $f(x)=\sqrt{4-x^2}+\sqrt{x^2-1}$, i.e., the set of all real $x$ for which both square roots are defined.

Formula:

For $\sqrt{g(x)}$ to be real, we need $g(x)\ge 0$.

Step 1: Condition from first term

Step 2: Condition from second term

So $x\in (-\infty ,-1]\cup [1,\infty )$.

Step 3: Intersect both conditions

Answer:

The domain of $f$ is $[-2,-1]\cup [1,2]$.

Understanding:

We check the three properties of the divisibility relation on $\mathbb{N}$: $(a,b)\in R⟺a\mid b$.

Formula:

A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

Step 1: Reflexive

For every $a\in \mathbb{N}$, $a\mid a$, so $(a,a)\in R$. Hence $R$ is reflexive.

Step 2: Symmetric

Take $a=2$, $b=6$: $2\mid 6$ so $(2,6)\in R$.

But $6\nmid 2$, so $(6,2)\notin R$.

Hence $R$ is not symmetric.

Step 3: Transitive

Suppose $a\mid b$ and $b\mid c$. Then $b=ka$ and $c=lb$ for some $k,l\in \mathbb{N}$, giving $c=kla$, so $a\mid c$.

Hence $R$ is transitive.

Answer:

$R$ is reflexive and transitive but not symmetric.

Quick Tip:

Divisibility on $\mathbb{N}$ is a classic example of a partial order (reflexive, antisymmetric, transitive) — note antisymmetric $\ne$ symmetric.

Understanding:

We need the inverse of the composition $g\circ f$, where both $f$ and $g$ are bijections.

Formula:

For bijections, $(g\circ f{)}^{-1}=f^{-1}\circ g^{-1}$.

This is the socks-and-shoes rule: reverse the order and invert each.

Step 1: Verify symbolically

Let $y=(g\circ f)(x)=g(f(x))$.

Applying $g^{-1}$: $g^{-1}(y)=f(x)$.

Applying $f^{-1}$: $f^{-1}(g^{-1}(y))=x$.

So $(g\circ f{)}^{-1}(y)=(f^{-1}\circ g^{-1})(y)$.

Step 2: Confirm domain matching

$g:B\to C$, so $g^{-1}:C\to B$, and $f^{-1}:B\to A$.

Thus $f^{-1}\circ g^{-1}:C\to A$, which correctly maps back from $C$ to $A$.

Answer:

$(g\circ f{)}^{-1}=f^{-1}\circ g^{-1}$.

Quick Tip:

Remember: "put on socks then shoes; to undo, remove shoes first, then socks." The order reverses when taking inverses of compositions.

Understanding:

We must find the fourth iterate of $f(x)=\frac{x+1}{x-1}$.

Formula:

Compute $f(f(x))$ first, then apply again twice.

Step 1: Find $f(f(x))$

Step 2: Conclude the fourth iterate

Since $f(f(x))=x$, applying $f$ twice returns to $x$.

Answer:

$f(f(f(f(x))))=x$.

Quick Tip:

$f$ is its own inverse (an involution): $f\circ f=\text{id}$. Any even number of applications returns the original input.

Understanding:

We need $n(A^{\prime }\cap B^{\prime })$, the number of elements in $U$ that belong to neither $A$ nor $B$.

Formula:

By De Morgan's law:

Step 1: Find $A\cup B$

Step 2: Apply the formula

The elements are $\{8,9,10\}$.

Answer:

$n(A^{\prime }\cap B^{\prime })=3$.

Understanding:

We check whether $f(x)=3x+5$ is injective and surjective as a map from $\mathbb{R}$ to $\mathbb{R}$.

Formula:

A linear function $f(x)=ax+b$ with $a\ne 0$ is a bijection on $\mathbb{R}$.

Step 1: Check one-one

Suppose $f(a)=f(b)$:

So $f$ is one-one.

Step 2: Check onto

For any $y\in \mathbb{R}$, we need $x$ such that $f(x)=y$:

Such an $x$ always exists, so $f$ is onto.

Answer:

$f$ is both one-one and onto, i.e., a bijection.

Understanding:

We count the number of one-one (injective) functions from $A=\{1,2,3,4\}$ to itself.

Formula:

The number of one-one functions from a set of $n$ elements to itself equals $n!$ (the number of permutations):

Step 1: Apply the formula

Step 2: Reasoning

The first element can map to $4$ choices, the second to $3$, the third to $2$, and the fourth to $1$ (no repetition allowed for injectivity), giving $4\times 3\times 2\times 1=24$.

Answer:

The number of one-one functions is $24$.

Quick Tip:

For a finite set, a one-one function from $A$ to $A$ is automatically onto as well — such functions are exactly the permutations of $A$.

Understanding:

We need $(f\circ g)(2)=f(g(2))$.

Formula:

Step 1: Compute $g(2)$

Step 2: Compute $f(g(2))=f(5)$

Answer:

$(f\circ g)(2)=27$.

Quick Tip:

A common error is computing $g(f(2))$ instead of $f(g(2))$. Always evaluate the inner function first: $f\circ g$ means $f$ applied after $g$.