JEE Mathematics — MCQ Practice Questions & Answers

Question 1

If \alpha and \beta are the roots of the equation x^2 - 5x + 6 = 0 , then the value of \alpha^3 + \beta^3 is:

Accepted Answer

Correct answer: A. 35 — **Step 1: Apply Vieta's Formulas** For the equation x^2 - 5x + 6 = 0 , Vieta's formulas give: \alpha + \beta = 5, \qquad \alpha\beta = 6. **Step 2: Apply the Sum of Cubes Identity** Using the algebraic identity: \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta), substitute the known values: \begin{aligned} \alpha^3 + \beta^3 &= (5)^3 - 3(6)(5) \ &= 125 - 90 \ &= \boxed{35}. \end{aligned}

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B. 45

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C. 125

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D. 215

Question 2

The number of ways to arrange the letters of the word 	ext{MISSISSIPPI} is:

Accepted Answer

Correct answer: A. \dfrac{11!}{4!\,4!\,2!} — **Step 1: Count the letters** The word 	ext{MISSISSIPPI} has 11 letters with the following frequencies: - M appears 1 time - I appears 4 times - S appears 4 times - P appears 2 times **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, \ldots, n_k is \frac{n!}{n_1!\, n_2!\, \cdots\, n_k!}. **Step 3: Substitute the values** \begin{aligned} 	ext{Number of arrangements} &= \frac{11!}{1!\,4!\,4!\,2!} \ 6pt] &= \frac{11!}{4!\,4!\,2!}. \end{aligned} **Step 4: Evaluate** \begin{aligned} \frac{11!}{4!\,4!\,2!} &= \frac{39{,}916{,}800}{24 	imes 24 	imes 2} \ 6pt] &= \frac{39{,}916{,}800}{1{,}152} \ 6pt] &= 34{,}650. \end{aligned} **Final Answer:** \boxed{34{,}650}

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B. \dfrac{11!}{4!\,4!}

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C. \dfrac{11!}{4!\,2!}

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D. \dfrac{11!}{4!\,4!\,2!\,1!}

Question 3

The area enclosed between the curves y = x^2 and y = x is:

Accepted Answer

Correct answer: A. \dfrac{1}{6} — **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.** \begin{aligned} 	ext{Area} &= \int_0^1 \left(x - x^2ight)dx \ 6pt] &= \left[\frac{x^2}{2} - \frac{x^3}{3}ight]_0^1 \ 6pt] &= \frac{1}{2} - \frac{1}{3} \ 6pt] &= \boxed{\dfrac{1}{6}} \end{aligned}

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B. \dfrac{1}{3}

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C. \dfrac{1}{2}

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D. \dfrac{1}{4}

Question 4

If \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} and \vec{b} = 2\hat{i} - \hat{j} + \hat{k} , then \vec{a} \cdot \vec{b} equals:

Accepted Answer

Correct answer: A. 3 — 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 \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3. **Step 1: Identify the components.** \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \implies a_1 = 1,\quad a_2 = 2,\quad a_3 = 3, \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \implies b_1 = 2,\quad b_2 = -1,\quad b_3 = 1. **Step 2: Substitute and simplify.** \begin{aligned} \vec{a} \cdot \vec{b} &= (1)(2) + (2)(-1) + (3)(1) \ &= 2 - 2 + 3 \ &= \boxed{3}. \end{aligned}

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B. 6

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C. 1

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D. 7

Question 5

The value of \displaystyle\lim_{x 	o 0} \frac{\sin 3x}{	an 5x} is:

Accepted Answer

Correct answer: A. \dfrac{3}{5} — Using the standard limits \displaystyle\lim_{	heta 	o 0}\frac{\sin	heta}{	heta} = 1 and \displaystyle\lim_{	heta 	o 0}\frac{	an	heta}{	heta} = 1 , we rewrite the expression by multiplying and dividing by the appropriate factors: \begin{aligned} \lim_{x	o 0}\frac{\sin 3x}{	an 5x} &= \lim_{x	o 0}\left(\frac{\sin 3x}{3x}\cdot\frac{5x}{	an 5x}\cdot\frac{3x}{5x}ight) \ 6pt] &= \lim_{x	o 0}\frac{\sin 3x}{3x}\cdot\lim_{x	o 0}\frac{5x}{	an 5x}\cdot\frac{3}{5} \ 6pt] &= 1 \cdot 1 \cdot \frac{3}{5} \ 6pt] &= \boxed{\dfrac{3}{5}}. \end{aligned}

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B. \dfrac{5}{3}

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C. 1

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D. 0

Question 6

The sum of the infinite geometric series 1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \cdots is:

Accepted Answer

Correct answer: A. \dfrac{3}{2} — **Step 1: Identify the series parameters.** The given series 1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \cdots is an infinite geometric series with first term a = 1 and common ratio r = \dfrac{1}{3} . **Step 2: Verify convergence.** Since |r| = \dfrac{1}{3} < 1 , the infinite geometric series converges and its sum is given by S = \frac{a}{1 - r}. **Step 3: Compute the sum.** \begin{aligned} S &= \frac{a}{1-r} \ &= \frac{1}{1 - \dfrac{1}{3}} \ &= \frac{1}{\,\dfrac{2}{3}\,} \ &= \frac{3}{2}. \end{aligned} **Final Answer:** \boxed{\dfrac{3}{2}}

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B. 2

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C. \dfrac{2}{3}

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D. 3

Question 7

If f(x) = x^3 - 3x^2 + 3x - 1 , then f'(x) = 0 has:

Accepted Answer

Correct answer: A. Exactly one real root — **Step 1: Factor the function.** Notice that f(x) = x^3 - 3x^2 + 3x - 1 = (x-1)^3. **Step 2: Differentiate.** f'(x) = 3(x-1)^2. **Step 3: Solve f'(x) = 0 .** \begin{aligned} 3(x-1)^2 &= 0 \ (x-1)^2 &= 0 \ x &= 1. \end{aligned} **Conclusion.** The equation f'(x) = 0 has exactly one real root, x = 1 , which is a repeated (double) root. \boxed{x = 1 	ext{ (one repeated real root)}}

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B. Two distinct real roots

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C. Three distinct real roots

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D. No real roots

Question 8

The equation of the circle with centre (3, -4) and passing through the origin is:

Accepted Answer

Correct answer: A. x^2 + y^2 - 6x + 8y = 0 — **Step 1: Find the radius.** The radius is the distance from the centre (3, -4) to the origin (0, 0) : r = \sqrt{(3-0)^2 + (-4-0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5. **Step 2: Write the standard circle equation.** Using centre (3, -4) and r = 5 : (x - 3)^2 + (y + 4)^2 = 25. **Step 3: Expand and simplify.** \begin{aligned} (x-3)^2 + (y+4)^2 &= 25 \ x^2 - 6x + 9 + y^2 + 8y + 16 &= 25 \ x^2 + y^2 - 6x + 8y + 25 &= 25 \ x^2 + y^2 - 6x + 8y &= 0. \end{aligned} **Final Answer:** \boxed{x^2 + y^2 - 6x + 8y = 0}

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B. x^2 + y^2 + 6x - 8y = 0

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C. x^2 + y^2 - 6x - 8y = 0

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D. x^2 + y^2 + 6x + 8y = 0

JEE Mathematics - MCQ Practice Questions