JEE Mathematics Solutions of Equations and Inequations MCQs Set 03

Answer: (d) None of these

Explanation:
1. With \( c=\cos\theta \): \( c^2-3c+\sqrt3-1=0 \), so \( c=\dfrac{3-\sqrt{13-4\sqrt3}}{2}\approx0.268 \) (the other root exceeds 1).
2. In \( [0,3\pi] \) the value \( \cos\theta=0.268 \) occurs at \( \theta_1,\ 2\pi-\theta_1,\ 2\pi+\theta_1 \): 3 solutions, which is not among 4, 2, 0.

Teacher's Note:
1. Convert to a quadratic in cos θ.
2. Count the solutions in each period.

Answer: (a) 2

Explanation:
1. Multiply by cos θ: \( 13\cos\theta+12\sin\theta=5 \).
2. The amplitude is \( \sqrt{313}\gt5 \), so there are 2 solutions in one period.

Teacher's Note:
1. Rewrite in sine and cosine.
2. Use the amplitude test.

Answer: (d) infinite

Explanation:
1. Use \( \sin^2x=\tfrac{1-\cos2x}{2} \).
2. Matching coefficients: \( a_1+\tfrac{a_3}2=1 \) and \( a_2=\tfrac{a_3}2 \), two equations in three unknowns, so infinitely many triplets.

Teacher's Note:
1. Write in terms of cos 2x.
2. Count equations against unknowns.

Answer: (c) 2

Explanation:
1. Multiply by cos x: \( \sin x+1=2\cos^2x=2-2\sin^2x \), so \( 2\sin^2x+\sin x-1=0 \).
2. \( \sin x=\tfrac12 \) or \( -1 \). sin x = -1 makes sec x undefined, so only sin x = 1/2: \( x=\pi/6,\ 5\pi/6 \).

Teacher's Note:
1. Convert to sine and cosine.
2. Reject values where cos x = 0.

Answer: (c) 4

Explanation:
1. \( \sin\theta=1-\sqrt2\approx-0.414 \), which is negative, so solutions lie in \( (\pi,2\pi) \) mod 2π: \( \pi+a,\ 2\pi-a,\ 3\pi+a,\ 4\pi-a \).
2. The 4th solution is \( 4\pi-a\lt4\pi \) so the least n is 4.

Teacher's Note:
1. Solve the quadratic for sin θ.
2. List solutions in increasing order.

Answer: (d) 15

Explanation:
1. \( 2\sec^2x-5\sec x-3=0 \) gives \( \sec x=3 \) (sec x = -1/2 is impossible), so \( \cos x=\tfrac13 \).
2. With \( a=\cos^{-1}\tfrac13\approx1.23 \) the solutions are \( a,\ 2\pi-a,\ 2\pi+a,\dots,6\pi+a \) (7th).
3. Need \( 6\pi+a\le n\pi/2\lt8\pi-a \), i.e. n between 12.8 and 15.2, so the greatest n is 15.

Teacher's Note:
1. Convert to a quadratic in sec x.
2. List the solutions in order.

Answer: (b) \( n\pi,\ n\pi+\tfrac\pi4 \)

Explanation:
1. AP condition: \( \cos2x+\sin2x=1 \), i.e. \( \sin(2x+\tfrac\pi4)=\tfrac1{\sqrt2} \).
2. So \( 2x+\tfrac\pi4=2n\pi+\tfrac\pi4 \) or \( 2n\pi+\tfrac{3\pi}4 \), giving \( x=n\pi \) or \( n\pi+\tfrac\pi4 \).

Teacher's Note:
1. The middle term is the average of the others.
2. Use the R-formula.

Answer: (a) \( n\pi\pm\tfrac\pi3,\ n\in Z \)

Explanation:
1. \( \tan\left(\tfrac{3\pi}4+\theta\right)=\dfrac{t-1}{1+t} \) with \( t=\tan\theta \).
2. \( t+\dfrac{t-1}{t+1}=2 \) gives \( t^2=3 \), so \( \tan\theta=\pm\sqrt3 \) and \( \theta=n\pi\pm\tfrac\pi3 \).

Teacher's Note:
1. Expand tan(3π/4 + θ).
2. Solve for tan θ.

Answer: (b) 30\(\pi\)

Explanation:
1. The product equals \( \tfrac14\cos3x \), so \( \cos3x=1 \) and \( x=\tfrac{2k\pi}3 \).
2. In [0, 6π], k = 0, 1, ..., 9. Sum = \( \tfrac{2\pi}3\cdot45=30\pi \).

Teacher's Note:
1. Use \( \cos x\cos(\pi/3+x)\cos(\pi/3-x)=\tfrac14\cos3x \).
2. Add the arithmetic progression.

Answer: (a) \( n\pi-\tfrac\pi4 \)

Explanation:
1. \( \dfrac{(\sin x+i\cos x)(1-i)}2 \) has real part \( \dfrac{\sin x+\cos x}2 \).
2. Purely imaginary needs this to vanish: \( \tan x=-1 \), so \( x=n\pi-\tfrac\pi4 \).

Teacher's Note:
1. Multiply numerator and denominator by 1 - i.
2. Set the real part to zero.

Answer: (d) none of these

Explanation:
1. The right side is at least 11 while \( \cos x\le1 \).
2. So there is no solution.

Teacher's Note:
1. Compare the ranges of the two sides.
2. |sin x| is non-negative.

Answer: (c) \( \pi/4 \)

Explanation:
1. Let \( s=1+\sin2x=(\sin x+\cos x)^2 \). Then \( (\sin x+\cos x)^{s}=s^{s/2}=2 \) gives s = 2.
2. So \( \sin2x=1 \) and \( x=\pi/4 \), where sin x + cos x = √2 and \( (\sqrt2)^2=2 \).

Teacher's Note:
1. Express the exponent through the base.
2. Test the options.

Answer: (a) \( \pi/2 \)

Explanation:
1. First: \( 3\sin^2\theta=\cos2\phi \). Second: \( 3\sin\theta\cos\theta=\sin2\phi \).
2. Then \( \cos(\theta+2\phi)=\cos\theta\cos2\phi-\sin\theta\sin2\phi=3\sin^2\theta\cos\theta-3\sin^2\theta\cos\theta=0 \).
3. With both angles in range, \( \theta+2\phi=\pi/2 \).

Teacher's Note:
1. Expand cos(θ + 2φ).
2. Use both given relations.

Answer: (c) \( 2n\pi+\tfrac{7\pi}6 \)

Explanation:
1. Both squares vanish: \( \sin\theta=-\tfrac12 \) and \( \tan\theta=\tfrac1{\sqrt3} \).
2. These hold together only in the third quadrant at \( \theta=\tfrac{7\pi}6 \), repeating every 2π.

Teacher's Note:
1. A sum of squares is zero only if both are zero.
2. Find the common quadrant.

Answer: (b) one

Explanation:
1. x = 0 is a solution.
2. For x \gt 0, \( 2\sin x\le2x\lt4x\lt x^3+x^2+4x \), so the left side is positive. Hence x = 0 is the only solution

Teacher's Note:
1. Check x = 0.
2. Compare the polynomial part with 2 sin x.

Answer: (c) an irrational number

Explanation:
1. Both cosines must equal 1: \( x=2k\pi \) and \( 2\lambda x=2m\pi \).
2. x = 0 always works. A non-zero solution exists when \( \lambda\cdot2k\pi=m\pi \) with \( \lambda=m/(2k) \) rational; so a unique solution needs an irrational \( \lambda \).

Teacher's Note:
1. Both terms must reach their maximum together.
2. Rational λ gives extra solutions.

Answer: (c) 1

Explanation:
1. Both cosines must be 1, so x = y = 0.
2. cos(x - y) = cos 0 = 1.

Teacher's Note:
1. Each cosine is at most 1.
2. Both must be 1.

Answer: (a) 6

Explanation:
1. Either (cos x, sin y) = (1, 1) or (-1, -1).
2. cos x = 1: x = 0, 2π; sin y = 1: y = π/2, 5π/2 gives 4 pairs. cos x = -1: x = π, 3π; sin y = -1: y = 3π/2 gives 2 pairs. Total 6.

Teacher's Note:
1. The product of numbers of modulus at most 1 equals 1 only at the extremes.
2. Count each case.

Answer: (b) \( [0,\tfrac{7\pi}6]\cup[\tfrac{11\pi}6,2\pi] \)

Explanation:
1. \( \sin x\ge-\tfrac12 \).
2. On [0, 2π], sin x is below -1/2 only for \( \tfrac{7\pi}6\lt x\lt\tfrac{11\pi}6 \), so the rest is the solution.

Teacher's Note:
1. Draw the sine curve.
2. Mark where it equals -1/2.

Answer: (a) \( [-\pi,-\tfrac\pi6)\cup(\tfrac\pi6,\pi] \)

Explanation:
1. \( \cos x\lt\tfrac{\sqrt3}2 \) fails only for \( |x|\le\pi/6 \).
2. The inequality is strict, so the endpoints \( \pm\pi/6 \) are excluded.

Teacher's Note:
1. Locate cos x = √3/2.
2. Check endpoint inclusion.

Answer: (d) \( [\tfrac\pi6,\tfrac{5\pi}6] \)

Explanation:
1. \( (2s-1)(2s-3)\le0 \) gives \( \tfrac12\le\sin x\le\tfrac32 \).
2. Since sin x \le 1, we need \( \sin x\ge\tfrac12 \): \( x\in[\tfrac\pi6,\tfrac{5\pi}6] \).

Teacher's Note:
1. Factorise in sin x.
2. Use the bound sin x \le 1.

Answer: (c) 4

Explanation:
1. \( \sqrt3\cos x-\sin x=2\cos(x+\tfrac\pi6) \) has absolute value at most 2.
2. Equality holds when \( \cos(x+\tfrac\pi6)=\pm1 \), i.e. \( x=\tfrac{5\pi}6,\tfrac{11\pi}6,\tfrac{17\pi}6,\tfrac{23\pi}6 \): 4 values.

Teacher's Note:
1. Use the R-formula.
2. The inequality can only hold with equality.