JEE Mathematics - MCQ Practice Questions

Step 1: Apply Vieta's Formulas

For the equation $x^2-5x+6=0$, Vieta's formulas give:

Step 2: Apply the Sum of Cubes Identity

Using the algebraic identity:

substitute the known values:

Step 1: Count the letters

The word $\text{MISSISSIPPI}$ has $11$ letters with the following frequencies:

Step 2: Apply the multinomial formula

The number of distinct arrangements of $n$ objects where groups of identical objects have sizes $n_1,n_2,\dots ,n_k$ is

Step 3: Substitute the values

Step 4: Evaluate

Final Answer:

Step 1: Find the intersection points.

Setting $y=x^2$ equal to $y=x$ gives $x^2=x$, so $x(x-1)=0$, yielding intersections at $x=0$ and $x=1$.

Step 2: Determine which curve is on top.

For $0\le x\le 1$, we have $x\ge x^2$, so the curve $y=x$ lies above $y=x^2$ on this interval.

Step 3: Set up and evaluate the integral.

The dot product of two vectors $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ is given by

Step 1: Identify the components.

Step 2: Substitute and simplify.

Using the standard limits $\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$ and $\underset{\theta \to 0}{\lim }\frac{\tan \theta }{\theta }=1$, we rewrite the expression by multiplying and dividing by the appropriate factors:

Step 1: Identify the series parameters.

The given series $1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots$ is an infinite geometric series with first term $a=1$ and common ratio $r=\frac{1}{3}$.

Step 2: Verify convergence.

Since $|r|=\frac{1}{3}<1$, the infinite geometric series converges and its sum is given by

Step 3: Compute the sum.

Final Answer:

Step 1: Factor the function.

Notice that

Step 2: Differentiate.

**Step 3: Solve $f^{\prime }(x)=0$.**

Conclusion.

The equation $f^{\prime }(x)=0$ has exactly one real root, $x=1$, which is a repeated (double) root.

Step 1: Find the radius.

The radius is the distance from the centre $(3,-4)$ to the origin $(0,0)$:

Step 2: Write the standard circle equation.

Using centre $(3,-4)$ and $r=5$:

Step 3: Expand and simplify.

Final Answer:

Step 1: Rewrite the equation

Starting from $\tan x=\sin x$, we write

Multiplying both sides by $\cos x$ (valid where $\cos x\ne 0$) gives

Step 2: Solve each factor

The equation $\sin x(1-\cos x)=0$ holds when either

**Step 3: Find solutions in $[0,2\pi ]$

Step 4: Collect distinct solutions

Taking the union of the two solution sets, the distinct values are

Conclusion

The total number of solutions is $\boxed{3}$.

Step 1: Apply the half-angle identity.

Using the identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$, we rewrite the integrand:

Step 2: Integrate term by term.

Step 3: Evaluate at the bounds.

Final Answer:

Understanding:

We need to find the inverse of the $2\times 2$ matrix $A =

$.

Formula:

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

$,

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 matrix — swap the diagonal entries and negate the off-diagonal entries.

Understanding:

We must evaluate the limit $\underset{x\to 0}{\lim }\frac{e^{3x}-1}{\sin 2x}$, which is a $\frac{0}{0}$ indeterminate form.

Formula:

We use the standard limits:

Step 1: Rewrite by multiplying and dividing by convenient factors.

Step 2: Take the limit of each factor separately.

Step 3: Combine.

Answer:

The value of the limit is $\frac{3}{2}$.

Quick Tip:

For limits of the form $\frac{e^{ax}-1}{\sin bx}$ as $x\to 0$, the answer is always $\frac{a}{b}$ — a very handy result to remember for JEE.

Understanding:

We need the angle between the two straight lines:

Formula:

For two lines with slopes $m_1$ and $m_2$, the angle $\theta$ between them satisfies:

If $1+m_1{m}_2=0$, the lines are perpendicular.

Step 1: Find the slopes.

Step 2: Check the perpendicularity condition.

Since $m_1{m}_2=-1$, the lines are perpendicular.

Answer:

The angle between the two lines is $90°$.

Quick Tip:

Two lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ are perpendicular if and only if $a_1{a}_2+b_1{b}_2=0$. Here: $(2)(1)+(-1)(2)=2-2=0$ — confirmed in one step.

Understanding:

We must solve the equation ${\log }_{2}x+{\log }_{4}x+{\log }_{16}x=\frac{21}{4}$.

Formula:

Change of base: ${\log }_{a}x=\frac{{\log }_{2}x}{{\log }_{2}a}$, so every term can be written in terms of ${\log }_{2}x$:

Step 1: Convert all logarithms to base 2.

Step 2: Substitute and simplify. Let $t={\log }_{2}x$.

Step 3: Solve for $x$.

Answer:

The value of $x$ is $8$.

Quick Tip:

Converting all logarithms to the same base is the fastest approach whenever the bases are powers of a common number.

Understanding:

We need the number of distinct real roots of $x^4-5x^2+4=0$.

Formula:

Substitute $t=x^2$ to reduce this to a quadratic:

Step 1: Solve the quadratic in $t$.

Step 2: Back-substitute $t=x^2$.

Both values of $t$ are positive, so all four solutions are real.

Step 3: Count distinct roots.

The four roots $-2,-1,1,2$ are all distinct, giving $4$ distinct real roots.

Answer:

The equation has $4$ distinct real roots.

Quick Tip:

A biquadratic $x^4+b{x}^2+c=0$ has $4$ distinct real roots when both roots of the auxiliary quadratic are distinct and positive.

Understanding:

We need the coefficient of $x^3$ in the binomial expansion of $(1+2x{)}^6$.

Formula:

The general term in the expansion of $(1+ax{)}^n$ is:

Step 1: Identify the term with $x^3$ by setting $r=3$.

Step 2: Compute the coefficient.

Answer:

The coefficient of $x^3$ in $(1+2x{)}^6$ is $160$.

Quick Tip:

A common mistake is to forget to raise the coefficient of $x$ (here $2$) to the required power. Always compute $\left(\genfrac{}{}{0ex}{}{n}{r}\right)\cdot a^r$ in full.

Understanding:

We must find the distance of the point $P(3,-4,5)$ from the origin $O(0,0,0)$ in 3D space.

Formula:

Step 1: Substitute the coordinates.

Answer:

The distance from the origin to the point $(3,-4,5)$ is $5\sqrt{2}$.

Quick Tip:

$\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$. Always simplify surds fully — JEE options are usually given in simplified surd form.

Understanding:

Given $f(x)=\frac{x-1}{x+1}$, we need to find $f\left(f\left(\frac{1}{x}\right)\right)$.

Formula:

Apply the function definition $f(t)=\frac{t-1}{t+1}$ twice, first with $t=\frac{1}{x}$, then with the result.

Step 1: Compute $f\left(\frac{1}{x}\right)$.

Step 2: Compute $f\left(\frac{1-x}{1+x}\right)$.

Answer:

The value of $f\left(f\left(\frac{1}{x}\right)\right)$ is $-x$.

Quick Tip:

For functional iteration questions, intermediate simplification (multiplying numerator and denominator by the same factor) avoids messy compound fractions.

Understanding:

We need the sum of all integers in $\{1,2,\dots ,100\}$ that are divisible by $3$ or $5$. We use the inclusion-exclusion principle.

Formula:

where $S(k)$ denotes the sum of multiples of $k$ up to $100$, and $S(15)$ covers multiples of both (i.e., $\text{lcm}(3,5)=15$).

Step 1: Sum of multiples of $3$ up to $100$. The multiples are $3,6,\dots ,99$; there are $\lfloor 100/3\rfloor =33$ terms.

Step 2: Sum of multiples of $5$ up to $100$. Multiples: $5,10,\dots ,100$; there are $20$ terms.

Step 3: Sum of multiples of $15$ up to $100$. Multiples: $15,30,\dots ,90$; there are $6$ terms.

Step 4: Apply inclusion-exclusion.

Answer:

The required sum is $2418$.

Quick Tip:

Always subtract the sum of multiples of $\text{lcm}(3,5)=15$ to avoid double-counting integers divisible by both $3$ and $5$.

Understanding:

We need the area enclosed between the parabola $y=x^2-4$ and the $x$-axis ($y=0$).

Formula:

The area between a curve $y=f(x)$ and the $x$-axis from $x=a$ to $x=b$ where $f(x)\le 0$ is:

Step 1: Find the intersection points with $y=0$.

Step 2: Note that $y=x^2-4\le 0$ for $x\in [-2,2]$, so take the negative of the integral.

Step 3: Evaluate the integral using symmetry about the $y$-axis.

Answer:

The area of the bounded region is $\frac{32}{3}$ square units.

Quick Tip:

For a parabola $y=x^2-a^2$ between its $x$-intercepts, the area formula gives $\frac{4a^3}{3}$. Here $a=2$, so Area $=\frac{4\times 8}{3}=\frac{32}{3}$ — a useful pattern to remember.