Mathematics - MCQ Practice Questions

Understanding:

We must find the total number of positive divisors of $720$.

Formula:

If $n=p_1^{a_1}\cdot p_2^{a_2}\cdots p_k^{a_k}$, then the number of divisors is:

Step 1: Prime factorisation of $720$

So $720=2^4\cdot 3^2\cdot 5^1$.

Step 2: Apply the divisor formula

Answer:

The number of positive divisors of $720$ is $30$.

Quick Tip:

$720=6!$ — factorising factorials is a common source of divisor-count questions in Indian exams.

Understanding:

We must find $2^{100}(\mathrm{mod}7)$.

Formula:

By Fermat's Little Theorem, for a prime $p$ and $\gcd (a,p)=1$:

Step 1: Identify the period

Since $7$ is prime and $\gcd (2,7)=1$:

Step 2: Reduce the exponent modulo $6$

Step 3: Compute

since $16=2\times 7+2$.

Answer:

The remainder when $2^{100}$ is divided by $7$ is $2$.

Quick Tip:

Always reduce the exponent modulo $(p-1)$ when the divisor is prime — Fermat's Little Theorem is the fastest route.

Understanding:

We must find $\gcd (252,198)$.

Formula:

The Euclidean algorithm repeatedly applies:

Step 1: First application

Step 2: Second application

Step 3: Third application

Step 4: Final step

Verification:

$252=18\times 14$ and $198=18\times 11$, and $\gcd (14,11)=1$ ✓

Answer:

The greatest common divisor of $252$ and $198$ is $18$.

Understanding:

We must count integers in $\{1,2,\dots ,200\}$ not divisible by $3$ or $5$.

Formula:

By the Inclusion–Exclusion Principle:

where $A$ = multiples of $3$, $B$ = multiples of $5$, and $A\cap B$ = multiples of $15$.

Step 1: Count multiples of $3$, $5$, and $15$ up to $200$

Step 2: Apply inclusion–exclusion

Step 3: Count integers divisible by neither

Answer:

The number of integers from $1$ to $200$ divisible by neither $3$ nor $5$ is $107$.

Quick Tip:

A common error is to forget to subtract the double-counted multiples of $\text{lcm}(3,5)=15$.

Understanding:

We must find the second number given $\text{LCM}$, $\text{HCF}$, and one number.

Formula:

For any two positive integers $a$ and $b$:

Step 1: Solve for the second number $b$

Verification:

$\gcd (180,168)$: $180=1\times 168+12$, $168=14\times 12+0$, so $\gcd =12$ ✓

$\text{lcm}(180,168)=\frac{180\times 168}{12}=\frac{30240}{12}=2520$ ✓

Answer:

The other number is $168$.

Understanding:

We must determine for which positive integers $n$ the expression $n^2-n$ is divisible by $6$.

Formula:

Divisibility by $6$ requires divisibility by both $2$ and $3$:

Step 1: Divisibility by $2$

$n(n-1)$ is the product of two consecutive integers. Among any two consecutive integers, one is always even, so:

Step 2: Divisibility by $3$

Among any three consecutive integers, exactly one is divisible by $3$. Since $n-1$, $n$, $n+1$ are three consecutive integers, either $n-1$ or $n$ is divisible by $3$ (or the next one $n+1$ is, but that doesn't affect our pair unless $n\equiv 0$ or $n\equiv 1(\mathrm{mod}3)$).

Check all residues of $n(\mathrm{mod}3)$:

For $n\equiv 2(\mathrm{mod}3)$: $n(n-1)=2\times 1=2$ in residue, so $3\nmid n(n-1)$?

Actual check: $n=2$: $n(n-1)=2$, and $3\nmid 2$. But $6\nmid 2$.

So the claim must be re-examined. $n=2$: $4-2=2$, not divisible by $6$. $n=3$: $9-3=6$, divisible by $6$. $n=4$: $16-4=12$, divisible by $6$. $n=1$: $1-1=0$, divisible by $6$.

Thus $n=2$ fails. The correct characterisation: $6\mid n(n-1)$ if and only if $n\not\equiv 2(\mathrm{mod}3)$, i.e. $n\equiv 0$ or $1(\mathrm{mod}3)$.

Step 3: Conclude

Since $n=2$ gives $n(n-1)=2$ which is not divisible by $6$, the condition does NOT hold for all positive integers. The smallest $n$ for which $n(n-1)$ is divisible by $6$ (and $n\ge 1$) is $n=1$ (giving $0$, divisible by everything), and the next is $n=3$ (giving $6$).

Among the options, the correct statement is: divisibility holds for $n=1$ (trivially, as $1\times 0=0$) and the options ask us to identify which is smallest. $n=1$ gives $0$, which is divisible by $6$.

Answer:

The smallest positive integer $n$ such that $n^2-n$ is divisible by $6$ is $n=1$, since $1^2-1=0$ and $6\mid 0$.

Quick Tip:

$0$ is divisible by every positive integer by definition ($0=6\times 0$), so $n=1$ is always the trivial solution to such problems.

Understanding:

We are given $\gcd (a,b)=1$ (i.e., $a$ and $b$ are coprime) and must identify which statement must always be true.

Formula:

A standard result in number theory states:

Step 1: Verify option A

If a prime $p$ divided both $a^2$ and $b$, then since $p$ is prime and $p\mid a^2$, we get $p\mid a$. Combined with $p\mid b$, this means $p\mid \gcd (a,b)=1$, a contradiction. Therefore:

Step 2: Check the other options with counterexamples

Step 3: Conclude

Only $\gcd (a^2,b)=1$ is guaranteed whenever $\gcd (a,b)=1$.

Answer:

If $\gcd (a,b)=1$, then $\gcd (a^2,b)=1$ must be true.

Quick Tip:

More generally, if $\gcd (a,b)=1$ then $\gcd (a^m,b^n)=1$ for all positive integers $m,n$.

Understanding:

Trailing zeros in $n!$ come from factors of $10=2\times 5$. Since factors of $2$ are more abundant than factors of $5$, we count the number of times $5$ divides $50!$.

Formula:

The exponent of a prime $p$ in $n!$ is given by Legendre's formula:

Step 1: Apply the formula for $p=5$, $n=50$

Step 2: Confirm factors of $2$ exceed $12$

So the limiting factor is indeed $5$, and the number of trailing zeros equals $12$.

Answer:

The number of trailing zeros in $50!$ is $12$.

Quick Tip:

Multiples of $25$ contribute two factors of $5$ each — forgetting this is the most common error in trailing-zero questions.

Understanding:

We are given that the sum of all positive divisors of $496$ is $992$, and we must classify $496$ accordingly.

Formula:

A positive integer $n$ is classified as:

Step 1: Compare $\sigma (496)$ with $2\times 496$

Step 2: Classify

Since $\sigma (496)=2n$, the number $496$ is a perfect number.

Step 3: Distinguish from the other options

Answer:

$496$ is a perfect number.

Quick Tip:

The first four perfect numbers are $6,28,496,8128$ — all of the form $2^{p-1}(2^p-1)$ where $2^p-1$ is a Mersenne prime.

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 $g$.

Understanding:

We are given $\sin \theta +\cos \theta =\sqrt{2}\cos \theta$ and must find $\cot \theta$.

Formula:

Step 1: Rearrange the given equation.

Step 2: Divide both sides by $\sin \theta$.

Step 3: Rationalise the denominator.

Answer:

The value of $\cot \theta$ is $\sqrt{2}+1$.

Quick Tip:

Rationalising $\frac{1}{\sqrt{2}-1}$ is a very common step in trigonometry problems — always multiply by the conjugate.