Mathematics - MCQ Practice Questions

Understanding:

We must identify the measure of an angle inscribed in a semicircle.

Formula:

The inscribed angle theorem states:

Step 1: Identify the central angle for a semicircle.

A semicircle subtends an arc of $180^{\circ }$ at the centre.

Step 2: Apply the inscribed angle theorem.

Answer:

The angle inscribed in a semicircle is always $90^{\circ }$.

Quick Tip:

This is Thales' theorem — any angle inscribed in a semicircle (i.e., the angle at the circumference subtended by the diameter) is a right angle. It is one of the most frequently tested circle theorems.

Understanding:

We must find the area of a rectangle given its perimeter and diagonal.

Formula:

Let the length and width be $l$ and $w$. Then:

Step 1: Find $l+w$.

Step 2: Find $l^2+w^2$.

Step 3: Compute the area.

Verification:

$l$ and $w$ satisfy $t^2-35t+300=0$.

So $l=20\text{ cm}$ and $w=15\text{ cm}$. Check: $20+15=35$ ✓, $\sqrt{20^2+15^2}=\sqrt{625}=25$ ✓, Area $=20\times 15=300$ ✓.

Answer:

The area of the rectangle is 300 square centimetres.

Quick Tip:

The identity $\text{Area}=\frac{(l+w{)}^2-(l^2+w^2)}{2}$ converts the perimeter and diagonal directly into the area without solving for $l$ and $w$ individually.

Understanding:

We must compute the dot product of two vectors.

Formula:

Step 1: Multiply corresponding components

Answer:

The dot product $\vec{a}\cdot \vec{b}$ equals $-8$.

Quick Tip:

A common error is to forget the sign of the third component; always carry the minus sign through each multiplication carefully.

Understanding:

We must find $|\vec{a}\times \vec{b}|$.

Formula:

Step 1: Evaluate the determinant

Step 2: Find the magnitude

Answer:

The magnitude of $\vec{a}\times \vec{b}$ is $\sqrt{3}$.

Quick Tip:

$|\vec{a}\times \vec{b}|$ gives the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$; here that area is $\sqrt{3}$.

Understanding:

We must find the angle $\theta$ between two vectors using the dot product formula.

Formula:

Step 1: Compute the dot product

Step 2: Compute the magnitudes

Step 3: Find $\cos \theta$

Step 4: Find $\theta$

Answer:

The angle between the vectors is ${\cos }^{-1}\left(\frac{1}{3}\right)$.

Understanding:

We must find the unit vector along $\vec{v}=3\hat{i}-4\hat{j}$.

Formula:

Step 1: Compute the magnitude

Step 2: Divide by the magnitude

Answer:

The unit vector in the direction of $\vec{v}$ is $\frac{3}{5}\hat{i}-\frac{4}{5}\hat{j}$.

Quick Tip:

A common trap is preserving the sign: the $j$-component is negative, matching the original vector's direction.

Understanding:

We must find the distance between points $A$ and $B$ given their position vectors.

Formula:

Step 1: Compute $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

Step 2: Find the magnitude

Answer:

The length of $AB$ is $2\sqrt{6}$.

Understanding:

We must evaluate the scalar triple product (box product) $[\vec{a}\vec{b}\vec{c}]$.

Formula:

Step 1: Set up the determinant

Step 2: Evaluate

Answer:

The scalar triple product equals $1$.

Quick Tip:

For the standard basis vectors $\hat{i},\hat{j},\hat{k}$ in order, the scalar triple product is always $1$; reversing any two gives $-1$.

Understanding:

We must find the direction cosines of line $OP$.

Formula:

Step 1: Find $r=|OP|$

Step 2: Compute direction cosines

Verification:

Answer:

The direction cosines are $\frac{1}{3},\frac{2}{3},\frac{2}{3}$.

Quick Tip:

Always verify: the sum of squares of direction cosines must equal $1$.

Understanding:

We must find $\theta$ between $\vec{a}$ and $\vec{b}$.

Formula:

Step 1: Substitute values

Step 2: Find $\theta$

Answer:

The angle between the vectors is $\frac{\pi }{3}$.

Understanding:

We must find the area of triangle $ABC$ using the cross product formula.

Formula:

Step 1: Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$

Step 2: Compute $\overrightarrow{AB}\times \overrightarrow{AC}$

Step 3: Find the magnitude

Step 4: Compute area

Answer:

The area of the triangle is $\frac{\sqrt{21}}{2}$.

Quick Tip:

Remember to halve $|\overrightarrow{AB}\times \overrightarrow{AC}|$; omitting the factor of $\frac{1}{2}$ is the most common error here.

Understanding:

We must find $\det (A^2-5A)$ where $A =

$.

Formula:

The Cayley–Hamilton theorem states that every square matrix satisfies its own characteristic equation. For a $2\times 2$ matrix:

so $A^2-(\text{tr}A)A+\det (A)I=0$.

Step 1: Compute the trace and determinant of $A$.

Step 2: Apply Cayley–Hamilton.

Step 3: Compute $A-5I$.

Step 4: Find the determinant.

Answer:

The determinant of $A^2-5A$ is $0$.

Quick Tip:

Whenever you see a polynomial in a matrix, apply Cayley–Hamilton first — it often collapses the expression to something simple.

Understanding:

We need $\det (B)$ where $B=A^{T}A$ and $A =

$.

Formula:

because $\det (A^T)=\det (A)$ for any square matrix.

Step 1: Use the multiplicative property.

Verification: Compute $A^{T}A$ directly.

Answer:

The determinant of $B=A^{T}A$ is $4$.

Quick Tip:

$\det (A^{T}A)=[\det (A){]}^2$ is always non-negative — a useful sanity check.

Understanding:

We must find $\det (\text{adj}A)$ given that $A$ is a $3\times 3$ matrix with $\det (A)=4$.

Formula:

For an $n\times n$ matrix $A$:

Step 1: Apply the formula with $n=3$ and $\det (A)=4$.

Answer:

The determinant of the adjugate of $A$ is $16$.

Quick Tip:

The companion formula $A\cdot \text{adj}(A)=\det (A)\cdot I$ gives $\det (A)\cdot \det (\text{adj}A)=[\det (A){]}^n$, which immediately yields the result.

Understanding:

A homogeneous system has a non-trivial solution if and only if the coefficient matrix is singular (determinant $=0$).

$

Formula:

Step 1: Expand the determinant along the first row.

Step 2: Evaluate each $2\times 2$ determinant.

Step 3: Substitute and simplify.

Step 4: Set equal to zero.

Answer:

The system has a non-trivial solution when $k=5$.

Quick Tip:

For a homogeneous system, always set $\det =0$; for a non-homogeneous system, use Cramer's rule or rank conditions instead.

Understanding:

We must find $A^{T}A$ for the rotation matrix $A =

$.

Formula:

A matrix is orthogonal when its rows (equivalently, columns) form an orthonormal set.

Step 1: Write $A^T$.

Step 2: Multiply $A^{T}A$.

using ${\cos }^2\theta +{\sin }^2\theta =1$.

Answer:

$A^{T}A=I$, the identity matrix.

Quick Tip:

Every rotation matrix is orthogonal, so $A^T=A^{-1}$ and $\det (A)=1$ — both facts are worth memorising for MCQs.

Understanding:

We must classify the matrix $A =

$.

Formula:

A matrix $A$ is skew-symmetric if and only if:

equivalently, $a_{ij}=-a_{ji}$ for all $i,j$, and all diagonal entries are $0$.

Step 1: Compute $A^T$.

Step 2: Check whether $A^T=-A$.

Since $A^T=-A$, the matrix is skew-symmetric.

Step 3: Verify it is not symmetric — clearly $A\ne A^T$.

Step 4: Check it is not orthogonal by noting $\det (A)=0$ (for any odd-order real skew-symmetric matrix, $\det =0$), so $A$ cannot be orthogonal.

Answer:

$A$ is a skew-symmetric matrix.

Quick Tip:

For any odd-order real skew-symmetric matrix, $\det (A)=0$ — this is a standard result worth remembering.

Understanding:

We identify the correct formula for the inverse of a product of two invertible matrices $A$ and $B$.

Formula:

For invertible matrices $A$ and $B$ of the same order:

This is the reversal (socks-and-shoes) rule.

Step 1: Verify by direct multiplication.

Step 2: Similarly check the other side.

Both products give $I$, confirming $B^{-1}{A}^{-1}$ is the inverse of $AB$.

Answer:

The correct identity is $(AB{)}^{-1}=B^{-1}{A}^{-1}$.

Quick Tip:

Matrix multiplication is not commutative, so the order reverses when taking inverses — just like putting on socks before shoes means removing shoes before socks.

Understanding:

We analyse the solution set of the system:

Formula:

For the system $AX=B$, the solution exists (and is unique or infinite) based on the rank condition:

Step 1: Form the coefficient matrix and augmented matrix.

Step 2: Apply $R_2\to R_2-2R_1$.

Step 3: Determine ranks.

Since both ranks are equal and less than $n=2$ (the number of unknowns), the system is consistent with infinitely many solutions.

Step 4: Note that equation (2) is simply $2\times$ equation (1), so the two equations represent the same line.

Answer:

The system has infinitely many solutions.

Quick Tip:

When one equation is a scalar multiple of the other (both the coefficients AND the RHS scale by the same factor), the two equations are dependent and the system has infinitely many solutions.

Understanding:

We must find the inverse of $A =

$.

Formula:

For a $2\times 2$ matrix $A =

$with$\det(A) \ne 0$:

Step 1: Compute the determinant.

Step 2: Apply the inverse formula.

Verification:

Answer:

The inverse of $A$ is $

$.

Quick Tip:

When $\det (A)=1$, the inverse is simply the adjugate — swap the diagonal entries and negate the off-diagonal entries.