JEE Mathematics Previous Year Questions MCQs Set 08

Answer: (b) \( (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0 \)

Explanation:
1. \( \alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta) \) gives \( q=-p^3+3p\alpha\beta \), so \( \alpha\beta=\frac{p^3+q}{3p} \).
2. Sum of the roots \( =\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{p^2}{\alpha\beta}-2=\frac{3p^3}{p^3+q}-2=\frac{p^3-2q}{p^3+q} \); product = 1.
3. So the equation is \( (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0 \).

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Answer: (d) 41

Explanation:
1. Each element goes to the first subset, the second subset or neither: 3⁴ = 81 ordered pairs.
2. The pair with both subsets empty is counted once, and every other unordered pair twice.
3. Unordered pairs = (81 − 1)/2 + 1 = 41.

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Answer: (d) \( C_{10}-B_{10} \)

Explanation:
1. Split the sum: \( B_{10}\sum_{r=1}^{10}\binom{10}r\binom{20}r-C_{10}\sum_{r=1}^{10}\binom{10}r^2 \).
2. By Vandermonde, \( \sum_{r=0}^{10}\binom{10}r\binom{20}{r}=\binom{30}{10}=C_{10} \) and \( \sum_{r=0}^{10}\binom{10}r^2=\binom{20}{10}=B_{10} \).
3. Each sum starts at r = 1, so the r = 0 term (equal to 1) is removed from each: the result is \( B_{10}(C_{10}-1)-C_{10}(B_{10}-1)=C_{10}-B_{10} \).

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Answer: (a) 0

Explanation:
1. A system of linear equations has either no solution, exactly one solution, or infinitely many.
2. If it had two distinct solutions u and v, then every point u + t(v − u) would also be a solution.
3. So exactly two solutions is impossible: the number of such matrices is 0.

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Answer: (d) \( \sqrt3 \)

Explanation:
1. A, B, C in AP gives B = 60°.
2. By the sine rule a/c = sin A/sin C, so the expression is \( \frac{\sin A}{\sin C}2\sin C\cos C+\frac{\sin C}{\sin A}2\sin A\cos A=2\sin(A+C) \).
3. Since A + C = 120°, the value is 2 sin 120° = √3.

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Answer: (a) parallelogram, which is neither a rhombus nor a rectangle

Explanation:
1. PQ = (6, 1) and SR = (6, 1), so PQ ∥ SR with equal length: a parallelogram.
2. PQ² = 37 but PS = (−1, 3) gives PS² = 10, so adjacent sides are unequal: not a rhombus.
3. PQ · PS = −6 + 3 = −3 ≠ 0, so not a rectangle.

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Answer: (b) \( \frac{\sqrt{17}}9 \)

Explanation:
1. |AB| = 15, |AD| = 3 and \( \vec{AB}\cdot\vec{AD}=-2+20+22=40 \), so the angle θ between them has cos θ = 40/45 = 8/9.
2. To make AD′ perpendicular to AB it must turn through α = 90° − θ.
3. So \( \cos\alpha=\sin\theta=\sqrt{1-\frac{64}{81}}=\frac{\sqrt{17}}9 \).

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Answer: (c) x − 2y + z = 0

Explanation:
1. The normal of the second plane is \( (3,4,2)\times(4,2,3)=(8,-1,-10) \).
2. The required plane contains direction (2, 3, 4) and the vector (8, −1, −10), so its normal is \( (2,3,4)\times(8,-1,-10)=(-26,52,-26)\propto(1,-2,1) \).
3. It passes through the origin: x − 2y + z = 0.

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Answer: (a) \( \left(\frac83,\frac43,-\frac73\right) \)

Explanation:
1. Distance: \( \frac{|1-4-2-\alpha|}3=5 \) gives α = 10 (α > 0).
2. The foot is P + t(1, 2, −2) with t = (10 − (−5))/9 = 5/3.
3. Foot = (1 + 5/3, −2 + 10/3, 1 − 10/3) = (8/3, 4/3, −7/3).

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Answer: (b) \( \frac1{12} \)

Explanation:
1. Near 0, \( \ln(1+t)\approx t \) and \( t^4+4\approx4 \), so the integrand is about \( \frac{t^2}4 \).
2. The integral is about \( \frac{x^3}{12} \), so the limit is 1/12 (also by L'Hôpital: \( \frac{x\ln(1+x)}{3x^2(x^4+4)}\to\frac1{12} \)).

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Answer: (b) \( \frac13 \)

Explanation:
1. Put x = 0: f(0) = 2, so f⁻¹(2) = 0.
2. Differentiating: \( e^{-x}(f'(x)-f(x))=\sqrt{x^4+1} \); at x = 0, f′(0) − 2 = 1, so f′(0) = 3.
3. (f⁻¹)′(2) = 1/f′(0) = 1/3.

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Answer: (c) \( \frac29 \)

Explanation:
1. The sum of three cube roots of unity is zero only when they are 1, ω and ω² in some order, i.e. r₁, r₂, r₃ have distinct remainders on division by 3.
2. Each remainder class has 2 numbers on a die. Favourable outcomes: 3! × 2 × 2 × 2 = 48.
3. Probability = 48/216 = 2/9.

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Answer: (c) \( \frac{20}{23} \)

Explanation:
1. P(B receives G | original G) = (3/4)(3/4) + (1/4)(1/4) = 5/8. P(B receives G | original R) = 2 × (3/4)(1/4) = 3/8.
2. Bayes: \( P=\frac{\frac45\cdot\frac58}{\frac45\cdot\frac58+\frac15\cdot\frac38}=\frac{20}{23} \).

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Answer: (b) \( \sqrt3+1 \)

Explanation:
1. Cosine rule with C = 30°: \( c^2=a^2+b^2-\sqrt3\,ab \).
2. Expanding and factorising gives \( (x-1)(x+1)\left(x-1-\sqrt3\right)\left(x+2+\sqrt3\right)=0 \).
3. The side b = x² − 1 must be positive, so x > 1, which leaves x = 1 + √3.

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