Mathematics - MCQ Practice Questions

Understanding:

We must evaluate ${\cos }^{2}48°-{\sin }^{2}12°$.

Formula:

Step 1: Identify $A=48°$ and $B=12°$, then apply the identity.

Step 2: Substitute exact values.

Step 3: Multiply.

Verification:

Numerically, ${\cos }^{2}48°\approx 0.4472$ and ${\sin }^{2}12°\approx 0.0432$, so the difference $\approx 0.404$. Also $\frac{\sqrt{5}+1}{8}\approx \frac{3.236}{8}\approx 0.405$ ✓

Answer:

The value of ${\cos }^{2}48°-{\sin }^{2}12°$ is $\frac{\sqrt{5}+1}{8}$.

Quick Tip:

The identity ${\cos }^{2}A-{\sin }^{2}B=\cos (A+B)\cos (A-B)$ is the key tool for this class of problems.

Understanding:

Given $\sin \theta +{\sin }^2\theta =1$, we must find the value of ${\cos }^2\theta +{\cos }^4\theta$.

Formula:

The Pythagorean identity:

Step 1: From the given equation, express $\sin \theta$ in terms of ${\cos }^2\theta$.

Step 2: Now compute ${\cos }^2\theta +{\cos }^4\theta$.

Step 3: Substitute ${\cos }^2\theta =\sin \theta$.

Answer:

The value of ${\cos }^2\theta +{\cos }^4\theta$ is $1$.

Quick Tip:

The key insight is recognising that $\sin \theta =1-{\sin }^2\theta ={\cos }^2\theta$ — this substitution converts the target expression back to the given condition.

Understanding:

We must find the maximum value of the expression $3\sin \theta +4\cos \theta$.

Formula:

The maximum value of $a\sin \theta +b\cos \theta$ is:

Step 1: Identify $a$ and $b$.

Step 2: Apply the formula.

Verification:

Write $3\sin \theta +4\cos \theta =5\left(\frac{3}{5}\sin \theta +\frac{4}{5}\cos \theta \right)=5\sin (\theta +\varphi )$ where $\cos \varphi =\frac{3}{5}$. The maximum of $5\sin (\theta +\varphi )$ is indeed $5$ ✓

Answer:

The maximum value of $3\sin \theta +4\cos \theta$ is $5$.

Quick Tip:

The $3$-$4$-$5$ right triangle appears here: the coefficients form a Pythagorean triple, making $\sqrt{3^2+4^2}=5$ immediate.

Understanding:

Given $\cos \alpha +\cos \beta =0$ and $\sin \alpha +\sin \beta =0$, we must find $\cos (\alpha -\beta )$.

Formula:

Step 1: From the given equations, write:

Step 2: This means $\beta =\pi +\alpha$ (or equivalently $\alpha$ and $\beta$ differ by $\pi$).

Step 3: Substitute into the formula.

Verification:

If $\alpha =0°$ and $\beta =180°$, then $\cos 0°+\cos 180°=1-1=0$ ✓ and $\sin 0°+\sin 180°=0$ ✓. Also, $\cos (0°-180°)=\cos (-180°)=-1$ ✓

Answer:

The value of $\cos (\alpha -\beta )$ is $-1$.

Quick Tip:

The conditions $\cos \alpha =-\cos \beta$ and $\sin \alpha =-\sin \beta$ together force $\alpha$ and $\beta$ to be supplementary in the sense $\beta =\alpha \pm \pi$, which always gives $\cos (\alpha -\beta )=-1$.

Understanding:

We must find the exact value of $\tan 75°$.

Formula:

Step 1: Write $75°=45°+30°$.

Step 2: Apply the addition formula with $\tan 45°=1$ and $\tan 30°=\frac{1}{\sqrt{3}}$.

Step 3: Rationalise.

Verification:

Numerically, $\tan 75°\approx 3.732$ and $2+\sqrt{3}\approx 2+1.732=3.732$ ✓

Answer:

The value of $\tan 75°$ is $2+\sqrt{3}$.

Quick Tip:

$\tan 75°$ and $\tan 15°=2-\sqrt{3}$ are reciprocals of each other — a quick way to cross-check since $\tan 75°\cdot \tan 15°=1$.

Understanding:

$\alpha$ and $\beta$ are both solutions of $a\cos \theta +b\sin \theta =c$. We need $\cos (\alpha +\beta )$.

Formula:

For $\alpha$ and $\beta$ satisfying $a\cos \theta +b\sin \theta =c$, use the Weierstrass (half-angle) substitution $t=\tan \frac{\theta }{2}$:

Step 1: Substitute into the equation.

Step 2: The roots are $t_1=\tan \frac{\alpha }{2}$ and $t_2=\tan \frac{\beta }{2}$. By Vieta's formulas:

Step 3: Use $\tan \frac{\alpha +\beta }{2}=\frac{t_1+t_2}{1-t_1{t}_2}$.

Step 4: Compute $\cos (\alpha +\beta )$ using $\cos (\alpha +\beta )=\frac{1-{\tan }^2\frac{\alpha +\beta }{2}}{1+{\tan }^2\frac{\alpha +\beta }{2}}$.

Answer:

The value of $\cos (\alpha +\beta )$ is $\frac{a^2-b^2}{a^2+b^2}$.

Quick Tip:

The Weierstrass substitution converts a trigonometric equation into a quadratic, allowing Vieta's formulas to extract symmetric functions of the roots elegantly.

Understanding:

We must find the distance between two points in a coordinate plane.

Formula:

Step 1: Compute the differences.

Step 2: Substitute into the distance formula.

Answer:

The distance between $A$ and $B$ is $10$ units.

Quick Tip:

The pair $(-8,6)$ is a multiple of $(4,3)$ — a scaled $3$-$4$-$5$ right triangle — so $d=2\times 5=10$ can be spotted without a calculator.

Understanding:

We must find the slope of the line through two given points.

Formula:

Step 1: Substitute the coordinates.

Answer:

The slope of the line is $3$.

Quick Tip:

A common mistake is to reverse numerator and denominator, giving $\frac{1}{3}$. Always place the $y$-difference in the numerator.

Understanding:

We must write the equation of a line given its slope and $y$-intercept.

Formula:

Step 1: Substitute $m=-2$ and $c=7$.

Answer:

The equation of the line is $y=-2x+7$.

Understanding:

We must find the midpoint of the segment $PQ$.

Formula:

Step 1: Apply the formula.

Answer:

The midpoint of segment $PQ$ is $(1,1)$.

Quick Tip:

Add the coordinates first, then halve — never halve each coordinate separately before adding.

Understanding:

We must find the area of a triangle given its three vertices.

Formula:

Step 1: Substitute the coordinates.

Answer:

The area of the triangle is $12$ square units.

Quick Tip:

Note that $A$ and $C$ both have $y=2$, so $AC$ is a horizontal base of length $6$ and the height is the perpendicular distance from $B$ to that base, which is $6-2=4$. Area $=\frac{1}{2}\times 6\times 4=12$ — a faster route when two vertices share the same ordinate.

Understanding:

We must find the point $P$ that divides $AB$ internally in the ratio $m:n=1:2$.

Formula:

Step 1: Compute the $x$-coordinate.

Step 2: Compute the $y$-coordinate.

Answer:

The required point is $(4,1)$.

Quick Tip:

In the section formula, the ratio $m:n$ is applied as $m$ nearer to $B$ and $n$ nearer to $A$. A sign error on $y_1=-3$ is the most common slip here.

Understanding:

We must find the equation of a line through a given point and perpendicular to a given line.

Formula:

If the slope of a line is $m$, the slope of a perpendicular line is $-\frac{1}{m}$.

Step 1: Find the slope of the given line.

So $m=\frac{2}{3}$.

Step 2: The perpendicular slope is:

Step 3: Write the equation through $(1,-2)$.

Verification:

Check $(1,-2)$: $3(1)+2(-2)+1=3-4+1=0$. ✓

Answer:

The required equation is $3x+2y+1=0$.

Quick Tip:

For lines $ax+by+c=0$ and $bx-ay+k=0$ — swapping coefficients of $x$ and $y$ and flipping one sign always gives a perpendicular line.

Understanding:

We must find the centroid of a triangle given its three vertices.

Formula:

Step 1: Compute the $x$-coordinate.

Step 2: Compute the $y$-coordinate.

Answer:

The centroid of the triangle is $\left(\frac{10}{3},1\right)$.

Quick Tip:

The centroid divides each median in the ratio $2:1$ from the vertex. It always lies inside the triangle.

Understanding:

We must find the inclination of the given line with the positive $x$-axis.

Formula:

If the slope of the line is $m$, the inclination $\theta$ satisfies:

Step 1: Find the slope $m$ from the given equation.

So $m=\sqrt{3}$.

Step 2: Find $\theta$.

Answer:

The line makes an angle of $60^{\circ }$ with the positive $x$-axis.

Quick Tip:

Memorize: $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$, $\tan 45^{\circ }=1$, $\tan 60^{\circ }=\sqrt{3}$, $\tan 120^{\circ }=-\sqrt{3}$. A positive slope greater than $1$ immediately points to $60^{\circ }$.

Understanding:

We need to count divisors of $720$ that are divisible by $4$ but NOT by $8$.

Formula:

First find the prime factorisation of $720$, then count divisors with exactly the right power of $2$.

Step 1: Prime factorise $720$

Step 2: Identify the condition

A divisor $d$ of $720$ has the form $d=2^a\cdot 3^b\cdot 5^c$ where $0\le a\le 4$, $0\le b\le 2$, $0\le c\le 1$.

Divisible by $4=2^2$ but NOT by $8=2^3$ means exactly $a=2$.

Step 3: Count valid divisors

With $a=2$ fixed, count choices for $b$ and $c$:

Total:

Verification:

The $6$ divisors are: $4\cdot 1=4,4\cdot 3=12,4\cdot 9=36,4\cdot 5=20,4\cdot 15=60,4\cdot 45=180$. Each is divisible by $4$ but not $8$. ✓

Answer:

The number of required divisors is $6$.

Quick Tip:

When counting divisors with an exact power of a prime, fix that prime's exponent and freely choose exponents for all remaining primes.

Understanding:

We are given:

Formula:

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

Step 1: Apply the formula

Step 2: Verify

Answer:

The other number is $60$.

Understanding:

Count integers from $1$ to $300$ divisible by neither $3$ nor $5$ using the Inclusion–Exclusion Principle.

Formula:

Step 1: Count multiples of $3$

Step 2: Count multiples of $5$

Step 3: Count multiples of $15$ (both $3$ and $5$)

Step 4: Apply Inclusion–Exclusion

Answer:

The number of integers divisible by neither $3$ nor $5$ is $160$.

Quick Tip:

Alternatively, use Euler's product: $300\times \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)=300\times \frac{2}{3}\times \frac{4}{5}=160$.

Understanding:

Find prime numbers $p$ and $q$ such that $p^2-2q^2=1$.

Formula:

Rearrange and test small primes:

Step 1: Test small prime values of $q$

Step 2: Check parity for $q>2$ (odd primes)

If $q$ is odd, $q^2$ is odd, so $2q^2$ is even, $2q^2+1$ is odd. $p^2$ is odd, so $p$ is odd — consistent. But checking shows no further prime solutions for small $q$.

For large $q$, $p=\sqrt{2q^2+1}$ is between $q\sqrt{2}$ and $q\sqrt{2}+1$, making integer solutions rare (Pell equation). The only small prime solution is $p=3,q=2$.

Step 3: Compute the sum

Answer:

The value of $p+q$ is $5$.

Quick Tip:

Whenever a Diophantine equation involves primes, always test $2$ first since it is the only even prime — it often yields the only solution.

Understanding:

The number of trailing zeros in $100!$ equals the highest power of $10$ dividing $100!$, which equals the highest power of $5$ (since there are always more factors of $2$ than $5$).

Formula:

Legendre's formula for the largest power of a prime $p$ dividing $n!$:

Step 1: Apply with $p=5$, $n=100$

Step 2: Sum the terms

Step 3: Confirm factor of $2$ is not limiting

So the number of trailing zeros $=24$.

Answer:

The number of trailing zeros in $100!$ is $24$.

Quick Tip:

Multiples of $25$ contribute two factors of $5$ each, which is why the second term $\lfloor 100/25\rfloor =4$ adds $4$ extra zeros beyond the naive count of $20$.