JEE Mathematics Miscellaneous Questions MCQs Set 02

Answer: (a) 1

Explanation:
1. By the triangle inequality \( |z - 1| + |z| \geq |1| = 1 \), attained for \( z \) on the segment from 0 to 1.

Teacher's Note:
1. Triangle inequality.

Answer: (b) 3

Explanation:
1. Each inverse cosine is at most \( \pi \), so all equal \( \pi \) and \( x = y = z = -1 \).
2. \( xy + yz + zx = 3 \).

Teacher's Note:
1. The maximum of each term is π.

Answer: (c) \( \lambda = \mu - 1 \)

Explanation:
1. The exterior angle gives \( A + B = \frac{3\pi}4 \), so \( \tan(A + B) = -1 = \frac\lambda{1 - \mu} \).
2. So \( \lambda = \mu - 1 \).

Teacher's Note:
1. Exterior angle equals A + B.

Answer: (c) onto and one-one

Explanation:
1. Odd \( n = 1, 3, 5, \dots \) map to \( 0, 1, 2, \dots \) and even \( n = 2, 4, 6, \dots \) map to \( -1, -2, -3, \dots \), covering every integer exactly once.

Teacher's Note:
1. List the first few images.

Answer: (b) \( \frac{5}{8} \)

Explanation:
1. \( P(X) = \frac13[P(X|A) + P(X|B) + P(X|C)] \), so \( \frac38 + \frac14 + p = \frac54 \).
2. \( p = \frac58 \).

Teacher's Note:
1. Total probability with equal weights.

Answer: (c) \( \frac{1}{n} \)

Explanation:
1. \( u_2\cdots u_n = \frac12\cdot\frac23\cdots\frac{n-1}n = \frac1n \).

Teacher's Note:
1. Telescoping product.

Answer: (b) \( 2\tan\alpha \)

Explanation:
1. By AM-GM, \( f(x) \geq 2\sqrt{\tan^2\alpha} = 2\tan\alpha \).

Teacher's Note:
1. AM-GM.

Answer: (c) \( 4(n+1) \)

Explanation:
1. The general term is \( (-1)^r(n + 2 - r){}^nC_r\,2^{n+1-r} \). Testing \( n = 1 \) gives 8, \( n = 2 \) gives 12, \( n = 3 \) gives 16 (verified by direct summation).
2. The pattern is \( 4(n+1) \).

Teacher's Note:
1. Test small values of n.

Answer: (a) \( \frac12\{(1 + i)z_1 + (1 - i)z_2\} \)

Explanation:
1. The ratio has modulus 1 and argument \( \frac\pi2 \), so \( z_2 - z_0 = i(z_0 - z_1) \).
2. \( z_0(1 + i) = z_2 + iz_1 \), so \( z_0 = \frac{(z_2 + iz_1)(1 - i)}2 = \frac12\{(1 + i)z_1 + (1 - i)z_2\} \).

Teacher's Note:
1. Ratio of modulus 1 and argument π/2 equals i.

Answer: (a) \( b^2 = ac \)

Explanation:
1. The terms are \( AR^5, AR^{10}, AR^{15} \), and \( b^2 = A^2R^{20} = ac \).

Teacher's Note:
1. Write the terms using the first term and ratio.

Answer: (d) 6

Explanation:
1. The altitude from \( A \) is perpendicular to \( BC \) (slope \( -\frac12 \)), so its slope is 2: \( y - 2 = 2(x - 1) \).
2. The orthocentre lies on it; at \( x = 3 \): \( y = 6 \).

Teacher's Note:
1. The orthocentre lies on the altitude from A.

Answer: (b) \( 2(-\hat{i} + \sqrt3\hat{j}) \)

Explanation:
1. Each successive side turns by \( 60^\circ \): \( \vec{AB} \) at \( 0^\circ \), \( \vec{BC} \) at \( 60^\circ \), \( \vec{CD} \) at \( 120^\circ \).
2. \( \vec{CD} = 4(\cos120^\circ\hat{i} + \sin120^\circ\hat{j}) = 2(-\hat{i} + \sqrt3\hat{j}) \).

Teacher's Note:
1. Rotate by 120°.

Answer: (b) 1

Explanation:
1. The left side is decreasing and the right side is increasing, so at most one solution; \( x = 1 \) gives \( \frac35 + \frac75 = 2 \), so exactly one.

Teacher's Note:
1. Compare monotonicity.

Answer: (a) \( \frac{7!\times8!}{3!\times12!} \)

Explanation:
1. Seat the 8 girls in \( 8! \) ways, then put the 4 boys into 4 of the 7 inner gaps: \( {}^7C_4\cdot4! \) ways.
2. Probability \( = \frac{8!\cdot{}^7C_4\cdot4!}{12!} = \frac{7!\,8!}{3!\,12!} \) (since \( {}^7C_4\cdot4! = \frac{7!}{3!} \)).

Teacher's Note:
1. Girls first, boys in the gaps.

Answer: (c) \( \frac{2}{3} \)

Explanation:
1. Roots \( \alpha, -\alpha, \beta \). Sum of products: \( -\alpha^2 = -3 \). Product: \( -\alpha^2\beta = -2 \) gives \( \beta = \frac23 \).
2. \( m = \alpha - \alpha + \beta = \frac23 \).

Teacher's Note:
1. Use Vieta with roots α, -α, β.

Answer: (c) \( \dfrac1{x^{1/2^{n}} - 1} \)

Explanation:
1. Multiply both sides by \( x^{1/2^n} - 1 \). Repeated use of \( (u - 1)(u + 1) = u^2 - 1 \) collapses the product to \( x^2 - 1 \).
2. So \( f(x) = \frac1{x^{1/2^n} - 1} \).

Teacher's Note:
1. Telescope with difference of squares.

Answer: (d) \( (\sqrt7 + 1)x + (\sqrt7 + 2)y + (\sqrt7 - 4)z = 5 + \sqrt7 \)

Explanation:
1. Normals \( (1,1,1) \) and \( (1,2,-4) \) with lengths \( \sqrt3 \) and \( \sqrt{21} = \sqrt3\sqrt7 \). The region containing the origin has the acute angle (outward normals make an obtuse angle), so the obtuse bisector is the other one.
2. \( \frac{x + y + z - 1}{\sqrt3} = -\frac{x + 2y - 4z - 5}{\sqrt{21}} \) gives \( (\sqrt7 + 1)x + (\sqrt7 + 2)y + (\sqrt7 - 4)z = 5 + \sqrt7 \).

Teacher's Note:
1. Decide which bisector is obtuse using the sign of the normals' dot product.

Answer: (d) 1 : 3

Explanation:
1. For side \( a \): \( r = \frac{a}{2\sqrt3} \) and \( r_0 = \frac\Delta{s - a} = \frac{\frac{\sqrt3}4a^2}{a/2} = \frac{\sqrt3a}2 \).
2. \( r : r_0 = 1 : 3 \).

Teacher's Note:
1. Use r = Δ/s and r₀ = Δ/(s - a).

Answer: (a) -1

Explanation:
1. The determinant \( \begin{vmatrix}1&a&0\\0&1&a\\a&0&1\end{vmatrix} = 1 + a^3 = 0 \).
2. So \( a = -1 \).

Teacher's Note:
1. Set the determinant to zero.

Answer: (a) 4

Explanation:
1. \( \tan81^\circ = \cot9^\circ \), \( \tan63^\circ = \cot27^\circ \). So the expression is \( (\tan9^\circ + \cot9^\circ) - (\tan27^\circ + \cot27^\circ) = \frac2{\sin18^\circ} - \frac2{\sin54^\circ} \).
2. \( = 2(3.236 - 1.236) = 4 \) (exactly \( 2\cdot\frac{\sin54^\circ - \sin18^\circ}{\sin18^\circ\sin54^\circ} = 4 \)).

Teacher's Note:
1. Use tan x + cot x = 2/sin 2x.

Answer: (b) a^2 = 3b

Explanation:
1. For an equilateral triangle with a vertex at 0: \( z_1^2 + z_2^2 = z_1z_2 \), i.e. \( (z_1 + z_2)^2 = 3z_1z_2 \).
2. So \( a^2 = 3b \).

Teacher's Note:
1. Use the condition z₁² + z₂² = z₁z₂.

Answer: (a) \( \frac{3}{4} \)

Explanation:
1. \( P(E) + P(E') = 1 \) gives \( 2\lambda^2 + 3\lambda + 1 = 0 \), so \( \lambda = -1 \) or \( -\frac12 \).
2. \( \lambda = -1 \) gives \( P(E) = 1 \) (certain, excluded). With \( \lambda = -\frac12 \), \( P(E) = \frac34 \).

Teacher's Note:
1. Probabilities of complementary events add to 1.

Answer: (b) \( \frac{2}{5} \)

Explanation:
1. \( f(x) = \cos^{-1}\frac{x^2 - 1}{x^2 + 1} \). For \( x \lt 0 \), with \( x = \tan\phi \) (\( \phi \in (-\frac\pi2, 0) \)), \( f = \pi + 2\tan^{-1}x \).
2. \( f'(x) = \frac2{1 + x^2} \), so \( f'(-2) = \frac25 \). (Numerically, \( f(-1.9) = 0.970 \) and \( f(-2.1) = 0.889 \), a rising slope of 0.4.)

Teacher's Note:
1. Be careful with the sign of x.

Answer: (c) \( {}^{12}C_6 + 2 \)

Explanation:
1. \( x^{24} \) arises as: \( x^{24} \) from the first factor (coefficient 1); \( x^{12}\cdot x^{12} \) (coefficient \( {}^{12}C_6 \)); and \( x^{24} \) from the last factor (coefficient 1).
2. Total \( {}^{12}C_6 + 2 \).

Teacher's Note:
1. List how 24 can be formed from the three factors.

Answer: (b) 3

Explanation:
1. \( n = 0 \): 2; \( n = 1 \): 0; \( n = 2 \): \( -2 \); \( n = 3 \): 0, and the pattern repeats.
2. The set is \( \{2, 0, -2\} \).

Teacher's Note:
1. Use the period 4.

Answer: (c) \( \alpha \lt \gamma \lt \beta \) or \( \beta \lt \gamma \lt \alpha \)

Explanation:
1. \( \gamma = -\frac b{2a} = \frac{\alpha + \beta}2 \) is the midpoint of the roots, so it lies between them.

Teacher's Note:
1. Midpoint of the roots.

Answer: (a) \( \pm1 \)

Explanation:
1. A line parallel to the \( y \)-axis has infinite slope, so the quadratic must drop in degree: \( a^2 - 1 = 0 \).
2. So \( a = \pm1 \).

Teacher's Note:
1. Infinite slope means the m² coefficient vanishes.

Answer: (c) \( 2\sqrt{b^2 - c^2\sin^2B} \)

Explanation:
1. \( a^2 - 2ac\cos B + c^2 - b^2 = 0 \) has roots \( a_1, a_2 \).
2. \( |a_1 - a_2| = \sqrt{4c^2\cos^2B - 4(c^2 - b^2)} = 2\sqrt{b^2 - c^2\sin^2B} \).

Teacher's Note:
1. Use the cosine rule as a quadratic in a.

Answer: (c) \( [\vec r\,\vec a\,\vec b]\vec b + (\vec r\cdot\vec b)\vec b\times\vec a \)

Explanation:
1. Resolve along \( \vec a, \vec b, \vec a\times\vec b \). \( \vec r\times\vec a \) is perpendicular to \( \vec a \); its \( \vec b \) component is \( [\vec r\,\vec a\,\vec b] \).
2. Its component along \( \vec a\times\vec b \) is \( -\vec r\cdot\vec b \), which gives the second term as \( (\vec r\cdot\vec b)\vec b\times\vec a \).

Teacher's Note:
1. Expand in the orthonormal basis a, b, a×b.

Answer: (b) \( \left[\frac13, \frac12\right] \)

Explanation:
1. \( P(C) \geq 0 \) gives \( x \leq \frac12 \). Total \( \leq 1 \): \( \frac{13}{12} - \frac x4 \leq 1 \) gives \( x \geq \frac13 \).
2. The other bounds are weaker, so \( x \in \left[\frac13, \frac12\right] \).

Teacher's Note:
1. Each probability is non-negative and the sum is at most 1.