JEE Mathematics Miscellaneous Questions MCQs Set 03

Answer: (d) none of these

Explanation:
1. Let \( u = \sec x + \tan x \), so \( du = u\sec x\,dx \) and the integral becomes \( \int_1^{1+\sqrt2}u^{-3}du = \frac12\left(1 - \frac1{(1+\sqrt2)^2}\right) \).
2. \( \frac1{(1+\sqrt2)^2} = 3 - 2\sqrt2 \), so the value is \( \sqrt2 - 1 \), which is none of (a)-(c).

Teacher's Note:
1. Substitute u = sec x + tan x.

Answer: (c) \( \frac{4}{5} \)

Explanation:
1. The smaller is at least 4 only if both numbers come from \( \{4, 5, 6\} \): \( \frac{\binom32}{\binom62} = \frac3{15} \).
2. Required probability \( = 1 - \frac15 = \frac45 \).

Teacher's Note:
1. Use the complement.

Answer: (c) \( \frac{-5}{4} \)

Explanation:
1. \( 4a + 2b + c = 0 \) with \( 2b = a + c \) gives \( 5a + 2c = 0 \), so \( \frac ca = -\frac52 \).
2. The product of roots is \( -\frac52 \), so the other root is \( -\frac54 \).

Teacher's Note:
1. Use the AP condition and the product of roots.

Answer: (c) a = 2, b = 2

Explanation:
1. \( b = 2 \). The slope is \( 2ax - 6 = 0 \) at \( x = \frac32 \), so \( 3a = 6 \) and \( a = 2 \).

Teacher's Note:
1. Use the point and the derivative condition.

Answer: (b) (4, -4)

Explanation:
1. Vertex \( (0, -\frac\mu\lambda) = (0, 1) \) gives \( \mu = -\lambda \), so \( x^2 = -\lambda(y - 1) \).
2. Here \( 4a = -\lambda \) and the directrix is \( a \) units from the vertex: \( a = 1 \) and the parabola opens downward, so \( \lambda = 4 \) and \( \mu = -4 \).

Teacher's Note:
1. Vertex gives μ = -λ.
2. Distance vertex to directrix is a = 1.

Answer: (b) 3

Explanation:
1. If \( [\cos x] = 0 \) then \( |\sin x| = 1 \), so \( x = \frac{3\pi}2 \) or \( \frac{5\pi}2 \). If \( [\cos x] = 1 \) then \( \sin x = 0 \) and \( x = 2\pi \).
2. If \( [\cos x] = -1 \), \( |\sin x| = 2 \) is impossible. So there are 3 solutions.

Teacher's Note:
1. Consider each integer value of [cos x].

Answer: (a) no solution

Explanation:
1. The first gives \( y \geq x + 1 \) and the second \( y \leq x \); these contradict each other.

Teacher's Note:
1. Combine the first two inequalities.

Answer: (b) \( \frac{52}{3} \)

Explanation:
1. From the two conditions: \( a = \frac{b + 4}4 \) and \( c = \frac{b^2}a \), giving \( 9b^2 - 88b - 240 = 0 \), so \( b = 12 \).
2. Then \( a = 4 \), \( c = 36 \) and the AM is \( \frac{52}3 \).

Teacher's Note:
1. Use b² = ac and the AP/GP relations.

Answer: (c) \( f(-3) \lt f(5) \)

Explanation:
1. \( f(-3) = f(3) \) and \( 3 \lt 5 \) within \( [2, 6] \), so \( f(3) \lt f(5) \).

Teacher's Note:
1. Use evenness to move to the positive side.

Answer: (a) \( \frac{2ab}{a + b} \)

Explanation:
1. For one-way distance \( d \), total time \( = \frac da + \frac db \), total distance \( 2d \).
2. Constant speed \( = \frac{2d}{d(\frac1a + \frac1b)} = \frac{2ab}{a + b} \).

Teacher's Note:
1. Average speed over equal distances is the harmonic mean.

Answer: (d) \( \left(0, \frac34\right] \)

Explanation:
1. \( A + C = \frac{2\pi}3 \) and \( x = \frac12[\cos(A - C) - \cos(A + C)] = \frac12\left[\cos(A - C) + \frac12\right] \).
2. \( \cos(A - C) \in \left(-\frac12, 1\right] \), so \( x \in \left(0, \frac34\right] \).

Teacher's Note:
1. Use product to sum.

Answer: (c) \( (-\infty, 0) \)

Explanation:
1. \( f'(x) = 2x\left[e^{-(x^2+1)^2} - e^{-x^4}\right] \) and the bracket is negative.
2. So \( f'(x) > 0 \) exactly when \( x \lt 0 \).

Teacher's Note:
1. Differentiate under the integral limits.

Answer: (a) m = 12, n = 9

Explanation:
1. Coefficient of \( x \): \( m - n = 3 \). Coefficient of \( x^2 \): \( \frac{(m-n)^2 - (m+n)}2 = -6 \), so \( m + n = 21 \).
2. Hence \( m = 12 \), \( n = 9 \).

Teacher's Note:
1. Expand to second order.

Answer: (b) HP

Explanation:
1. \( -\frac ba = \frac{b^2 - 2ac}{c^2} \) gives \( 2a^2c = ab^2 + bc^2 \).
2. Dividing by \( abc \): \( \frac{2a}b = \frac bc + \frac ca \), which is \( \frac2{b/a} = \frac1{a/c} + \frac1{c/b} \): an HP.

Teacher's Note:
1. Express the condition in the three ratios.

Answer: (d) none of these

Explanation:
1. Place the 5 white flowers on the garland; the 4 coloured ones go into 4 of the 5 gaps, so one gap (with two whites together) stays empty. Fix that gap to break the rotation.
2. The 4 coloured flowers can then be arranged in \( 4! = 24 \) ways, and turning the garland over halves this: \( 12 \) ways.

Teacher's Note:
1. Fix the empty gap, then count the order and reflections.

Answer: (a) \( \frac{1}{12} \)

Explanation:
1. \( P(B\cap\bar C) = P(A\cap B\cap\bar C) + P(\bar A\cap B\cap\bar C) = \frac23 \).
2. \( P(B\cap C) = P(B) - P(B\cap\bar C) = \frac34 - \frac23 = \frac1{12} \).

Teacher's Note:
1. Split B into C and not-C.

Answer: (a) a + b = c

Explanation:
1. \( \frac P2 + \frac Q2 = \frac\pi4 \), so \( \tan\frac{P+Q}2 = 1 = \frac{-b/a}{1 - c/a} = \frac{-b}{a - c} \).
2. So \( -b = a - c \), i.e. \( a + b = c \).

Teacher's Note:
1. Use the tan addition formula on the roots.

Answer: (a) \( \frac{3}{\sqrt{38}} \)

Explanation:
1. Direction cross product \( (1,1,1)\times(1,-2,3) = (5,-2,-3) \), magnitude \( \sqrt{38} \).
2. The vector between the points is \( (0,0,1) \). Distance \( = \frac{|{-3}|}{\sqrt{38}} = \frac3{\sqrt{38}} \).

Teacher's Note:
1. Use
2. (a₂ - a₁)·(d₁ × d₂)
3. /
4. d₁ × d₂
5. .

Answer: (c) -a

Explanation:
1. The limit is \( f'(2) - g'(2) = a - 2a = -a \).

Teacher's Note:
1. It is the derivative of f - g at 2.

Answer: (a) \( \frac12\{\sin1 + \sin2 + (\pi - 3)\sin3\} \)

Explanation:
1. For \( x \in [0, \frac12) \): 0; \( [\frac12, 1) \): 1; \( [1, \frac32) \): 2; \( [\frac32, \frac\pi2) \): 3.
2. Integral \( = \frac12\sin1 + \frac12\sin2 + \left(\frac\pi2 - \frac32\right)\sin3 \).

Teacher's Note:
1. Split the range where [2x] is constant.

Answer: (b) \( (0, +\infty) \)

Explanation:
1. \( a^xa^y = a^x + a^y \) gives \( a^y(a^x - 1) = a^x \), so \( a^y = \frac{a^x}{a^x - 1} \).
2. This is positive only if \( a^x > 1 \), i.e. \( x > 0 \).

Teacher's Note:
1. Solve for a^y.
2. It must be positive.

Answer: (a) 20

Explanation:
1. All arrangements give 5-digit numbers. The odd digits 1, 3, 5 must appear in a fixed order: \( \frac{5!}{3!} = 20 \).

Teacher's Note:
1. Fix the relative order of the odd digits.

Answer: (d) \( (1, 1) \)

Explanation:
1. The chord has length \( 2\sqrt2 \), so the radius is \( 2 \) (a right angle at the centre).
2. Candidate centres are \( (1, 1) \) and \( (3, 3) \): from \( (1, 1) \) the distances to both ends are 2 and the radii are perpendicular, so \( (1, 1) \) is a centre.

Teacher's Note:
1. The radius is the chord divided by √2.
2. Check the distances from each candidate.

Answer: (a) 2

Explanation:
1. If the limit \( L \) exists, \( L = \frac12\left(L + \frac4L\right) \), so \( L^2 = 4 \).
2. Since \( f > 0 \), \( L = 2 \).

Teacher's Note:
1. Take the limit on both sides.

Answer: (c) 3

Explanation:
1. No two lines are parallel, giving at most \( {}^4C_3 = 4 \) triangles.
2. The lines \( y = x + 3 \), \( y = 2x + 3 \) and \( y + x = 3 \) all pass through \( (0, 3) \), so that triple gives no triangle. Answer: 3.

Teacher's Note:
1. Check for concurrent triples.

Answer: (a) {0, 2q}

Explanation:
1. The circle has centre \( (q\cos\alpha, q\sin\alpha) \) and radius \( |q| \).
2. Distance from the centre to the line is \( |q - p| = |q| \), so \( p = 0 \) or \( p = 2q \).

Teacher's Note:
1. Distance from centre equals radius.

Answer: (b) \( \left(0, -\frac23\right) \)

Explanation:
1. The algebraic sum of distances being zero means the line passes through the centroid, which is \( \left(0, -\frac23\right) \).

Teacher's Note:
1. Think of the centroid.

Answer: (d) \( \frac{-2}{n} \)

Explanation:
1. \( \ln(1 + x + x^2) = \ln(1 - x^3) - \ln(1 - x) \).
2. For \( n = 3m \): coefficient \( = -\frac1m + \frac1n = -\frac3n + \frac1n = -\frac2n \).

Teacher's Note:
1. Use 1 + x + x² = (1 - x³)/(1 - x).

Answer: (b) zero

Explanation:
1. The first equation forces \( \cos(\theta - \phi) = 1 \) with \( \tan\phi = \frac34 \); the second needs \( \tan\phi' = \frac43 \).
2. The solutions differ, so no value of \( \theta \) satisfies both.

Teacher's Note:
1. Write each as R cos(θ - φ).

Answer: (d) AP

Explanation:
1. \( s_t = u + \frac f2(2t - 1) \) is linear in \( t \), so \( s_2, s_4, s_6 \) form an AP.

Teacher's Note:
1. The distance in the nth second is linear in n.