JEE Mathematics Trigonometric Functions MCQs Set 10

Answer: (d) -1

Explanation:
1. Square and add: \( 9+4+12(\cos x\cos3x+\sin x\sin3x)=1 \).
2. \( 13+12\cos2x=1 \Rightarrow \cos2x=-1 \).

Teacher's Note:
1. Squaring and adding removes y.
2. cos x cos 3x + sin x sin 3x = cos 2x.

Answer: (c) \( -1-\cot\alpha \)

Explanation:
1. \( \frac1{\sin^2\alpha}=1+\cot^2\alpha \), so the expression is \( \sqrt{(1+\cot\alpha)^2}=|1+\cot\alpha| \).
2. For \( \frac{3\pi}4 \lt \alpha \lt \pi \), \( \cot\alpha \lt -1 \), so \( 1+\cot\alpha \lt 0 \) and the value is \( -1-\cot\alpha \).

Teacher's Note:
1. Complete the square using cosec^2 = 1 + cot^2.
2. The sign is decided by the interval of alpha.

Answer: (b) \( (-\infty,\infty) \)

Explanation:
1. \( \frac{\sin\theta(1+2\cos\theta)}{\cos\theta(1+2\cos\theta)}=\tan\theta \), valid because \( 1+2\cos\theta \gt 0 \) on this range.
2. \( \tan\theta \) takes every real value on \( (-\frac\pi2,\frac\pi2) \).

Teacher's Note:
1. Use 1 + cos 2t = 2cos^2 t and sin 2t = 2 sin t cos t.
2. Check that the factor 1 + 2cos t that was cancelled is never zero.

Answer: (c) infinite

Explanation:
1. \( \sin^2x=\frac{1-\cos2x}2 \), so the identity becomes \( \left(a_1+\frac{a_3}2\right)+\left(a_2-\frac{a_3}2\right)\cos2x=0 \).
2. Both coefficients must vanish: \( a_1=-\frac{a_3}2,\ a_2=\frac{a_3}2 \). Every choice of \( a_3 \) gives a triplet.

Teacher's Note:
1. Rewrite everything in terms of cos 2x.
2. There are three unknowns but only two independent conditions.

Answer: (b) 2

Explanation:
1. \( \cos A=-\cos(B+C)=\sin B\sin C-\cos B\cos C \).
2. Given \( \cos A=\cos B\cos C \): \( \sin B\sin C=2\cos B\cos C \), so \( \tan B\tan C=2 \).

Teacher's Note:
1. Replace cos A by -cos(B + C).
2. Divide by cos B cos C.

Answer: (a) \( \alpha=(6n+1)\frac\pi{12},\ n\in I \)

Explanation:
1. \( 1-\tan^2\alpha=2\sqrt3\tan\alpha \), so \( \tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac1{\sqrt3} \).
2. \( 2\alpha=n\pi+\frac\pi6 \Rightarrow \alpha=\frac{n\pi}2+\frac\pi{12}=(6n+1)\frac\pi{12} \).

Teacher's Note:
1. Rearrange so that the double-angle tangent formula appears.
2. Write the answer with a common denominator to match the options.

Answer: (d) \( \frac{n\pi}3+\frac\pi{12} \)

Explanation:
1. The identity \( \tan x+\tan(x+\frac\pi3)+\tan(x+\frac{2\pi}3)=3\tan3x \) holds.
2. So \( \tan3x=1 \Rightarrow 3x=n\pi+\frac\pi4 \Rightarrow x=\frac{n\pi}3+\frac\pi{12} \).

Teacher's Note:
1. The three angles are in arithmetic progression with common difference pi/3.
2. The sum equals 3 tan 3x.

Answer: (d) \( \left\{\frac\pi8,\frac{13\pi}8\right\} \)

Explanation:
1. \( 2+2\sin2x=2(\sin x+\cos x)^2 \), so the equation is \( 2\cos x=\sqrt2\,|\sin x+\cos x| \) with \( \cos x\ge0 \).
2. Squaring: \( 4\cos^2x=2(1+\sin2x) \Rightarrow \cos2x=\sin2x \), i.e. \( \tan2x=1 \): \( x=\frac\pi8,\frac{5\pi}8,\frac{9\pi}8,\frac{13\pi}8 \).
3. Keep the ones with \( \cos x\ge0 \): \( x=\frac\pi8 \) and \( \frac{13\pi}8 \).

Teacher's Note:
1. Squaring can introduce extra roots, so check the sign of cos x.
2. Principal solutions are those in [0, 2 pi).

Answer: (b) \( x\in\left[n\pi,n\pi+\frac\pi4\right],\ n\in I \)

Explanation:
1. With \( u=2^{\tan x} \): \( u^2-3u+2\le0 \Rightarrow 1\le u\le2 \).
2. \( 1\le2^{\tan x}\le2 \Rightarrow 0\le\tan x\le1 \), so \( x\in\left[n\pi,n\pi+\frac\pi4\right] \).

Teacher's Note:
1. Substitute u = 2^(tan x) to get a quadratic.
2. Convert back using the monotonic exponential.

Answer: (b) \( \left[\pi\frac{12n-7}6,\pi\frac{12n+1}6\right],\ n\in Z \)

Explanation:
1. If \( 6\sin x-1 \lt 0 \) the inequality is automatic. Otherwise square: \( 5-2s\ge36s^2-12s+1 \Rightarrow 18s^2-5s-2\le0 \Rightarrow -\frac29\le s\le\frac12 \).
2. Combining both cases, the inequality holds exactly when \( \sin x\le\frac12 \).
3. \( \sin x\le\frac12 \) for \( x\in\left[2n\pi-\frac{7\pi}6,2n\pi+\frac\pi6\right]=\left[\pi\frac{12n-7}6,\pi\frac{12n+1}6\right] \).

Teacher's Note:
1. Treat the cases of a negative and a non-negative right-hand side separately.
2. sin x <= 1/2 on an arc of length 4 pi / 3 per period.

Answer: (b) 0.71

Explanation:
1. \( 5\cos\theta-11=12\sin\theta \). Squaring: \( 169\cos^2\theta-110\cos\theta-23=0 \), so \( \cos\alpha+\cos\beta=\frac{110}{169} \) and \( \cos\alpha\cos\beta=-\frac{23}{169} \).
2. \( \sin\alpha=\frac{5\cos\alpha-11}{12} \), so \( \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\frac{10\cos\alpha\cos\beta-11(\cos\alpha+\cos\beta)}{12} \).
3. \( =\frac{-230-1210}{12\cdot169}=-\frac{120}{169} \), so \( |\sin(\alpha+\beta)|=\frac{120}{169}\approx0.71 \).

Teacher's Note:
1. Treat the two values of theta as the two roots of a quadratic in cos theta.
2. Express sin alpha through cos alpha with the given equation.

Answer: (a) 2

Explanation:
1. \( 4\cos^2\left(\frac\pi4-\frac x2\right)=2\left(1+\cos\left(\frac\pi2-x\right)\right)=2+2\sin x \).
2. \( \sqrt{4\sin^4x+4\sin^2x\cos^2x}=2|\sin x|=-2\sin x \) because \( \sin x \lt 0 \).
3. Sum = 2.

Teacher's Note:
1. Use 2cos^2 t = 1 + cos 2t for the first term.
2. Factor the radicand as 4 sin^2 x (sin^2 x + cos^2 x).

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

Explanation:
1. If \( a+b+c=0 \) then \( a^3+b^3+c^3=3abc \). Put \( a=\cos A \), etc.
2. \( \cos3A+\cos3B+\cos3C=4\sum\cos^3A-3\sum\cos A=4\cdot3\cos A\cos B\cos C=12\cos A\cos B\cos C \).
3. So \( \lambda=\frac1{12} \).

Teacher's Note:
1. Use cos 3A = 4cos^3 A - 3cos A and the sum condition.
2. Apply a^3 + b^3 + c^3 = 3abc when a + b + c = 0.

Answer: (b) \( 2a^{2/3} \)

Explanation:
1. \( m+n=a(\cos\alpha+\sin\alpha)^3 \) and \( m-n=a(\cos\alpha-\sin\alpha)^3 \).
2. \( (m+n)^{2/3}+(m-n)^{2/3}=a^{2/3}\left[(\cos\alpha+\sin\alpha)^2+(\cos\alpha-\sin\alpha)^2\right]=2a^{2/3} \).

Teacher's Note:
1. Recognise the expansions of (cos + sin)^3 and (cos - sin)^3.
2. The cross terms cancel when squared and added.

Answer: (d) 8

Explanation:
1. Put \( t=\tan\frac x2 \): \( 2(1-t^2)+2t=1+t^2 \Rightarrow 3t^2-2t-1=0 \Rightarrow t=1 \) or \( t=-\frac13 \). (x = pi is not a solution.)
2. \( t=1 \): \( x=\frac\pi2 \), value \( 7\cdot0+6\cdot1=6 \). \( t=-\frac13 \): \( \sin x=-\frac35,\ \cos x=\frac45 \), value \( \frac{28}5-\frac{18}5=2 \).
3. Sum = 8.

Teacher's Note:
1. Use the half-angle tangent substitution for linear equations in sin and cos.
2. Find sin x and cos x for each root before evaluating.

Answer: (b) 17

Explanation:
1. On each interval \( (k\pi,(k+1)\pi) \) cot decreases from \( +\infty \) to \( -\infty \) while \( \frac\pi2+x \) increases, so there is exactly one intersection.
2. Full intervals from \( -\pi \) to \( 15\pi \): 16 intervals, 16 roots.
3. On \( \left(15\pi,\frac{31\pi}2\right) \) cot falls from \( +\infty \) to 0 while the line is about 49, so there is one more root. Total 17.

Teacher's Note:
1. Sketch y = cot x and the line y = pi/2 + x.
2. Treat the last partial interval separately.

Answer: (c) 20

Explanation:
1. \( 1-2s^2+as=2a-7 \Rightarrow 2s^2-as+2a-8=0 \Rightarrow 2(s-2)(s+2)=a(s-2) \).
2. Since \( s=\sin x\ne2 \), \( a=2(s+2)\in[2,6] \).
3. Integers 2, 3, 4, 5, 6: sum 20.

Teacher's Note:
1. Factor out (s - 2), which cannot vanish.
2. a is then a linear function of sin x.

Answer: (d) 24

Explanation:
1. \( \sin x=\pm\cos3x \Rightarrow \cos3x=\cos\left(\frac\pi2\mp x\right) \).
2. The solutions are \( x=\frac{(2k+1)\pi}8 \) (16 in the interval) and \( x=\frac{(2k+1)\pi}4 \) (8 in the interval).
3. These two families do not overlap: total 24.

Teacher's Note:
1. Square both sides or use cos A = cos B, whichever is easier.
2. Count each family separately and check for overlap.

Answer: (d) 2

Explanation:
1. The sum is \( \frac{\sum\cos A\sin A}{\sin A\sin B\sin C}=\frac{\frac12\sum\sin2A}{\sin A\sin B\sin C} \).
2. In a triangle \( \sum\sin2A=4\sin A\sin B\sin C \), so the value is 2.

Teacher's Note:
1. Put everything over the common denominator sin A sin B sin C.
2. The standard identity for sin 2A + sin 2B + sin 2C is needed.

Answer: (c) 17

Explanation:
1. \( 1+\cos2x+\cos4x+\cos6x=4\cos x\cos2x\cos3x \).
2. Zeros: \( \cos x=0 \) at \( \frac\pi2+k\pi \) (3 values), \( \cos2x=0 \) at \( \frac\pi4+\frac{k\pi}2 \) (7 values), \( \cos3x=0 \) at \( \frac\pi6+\frac{k\pi}3 \) (10 values, of which 3 coincide with cos x = 0).
3. Distinct solutions: 3 + 7 + 7 = 17.

Teacher's Note:
1. Factor the left side into a product of three cosines.
2. Count carefully to avoid double counting the common zeros.

Answer: (c) 15

Explanation:
1. \( 4\sin\theta\sin2\theta\sin4\theta=2\sin4\theta(\cos\theta-\cos3\theta)=\sin5\theta+\sin3\theta-\sin7\theta-\sin\theta \).
2. The equation becomes \( \sin5\theta-\sin7\theta-\sin\theta=0 \Rightarrow -\sin\theta(2\cos6\theta+1)=0 \).
3. \( \sin\theta=0 \): 3 solutions (0, pi, 2 pi). \( \cos6\theta=-\frac12 \): \( 6\theta\in[0,12\pi] \) gives 12 solutions. Total 15.

Teacher's Note:
1. Convert the product of sines to a sum using 2 sin A sin B = cos(A - B) - cos(A + B).
2. Count the solutions of each factor.

Answer: (d) \( 7\pi \)

Explanation:
1. \( \cos6x(1+\tan^2x)=1-\tan^2x \Rightarrow \cos6x=\frac{1-\tan^2x}{1+\tan^2x}=\cos2x \).
2. \( 6x=2n\pi\pm2x \Rightarrow x=\frac{n\pi}2 \) or \( x=\frac{n\pi}4 \), i.e. \( x=\frac{k\pi}4 \).
3. In \( [0,2\pi] \) the values \( x=\frac\pi2,\frac{3\pi}2 \) make \( \tan x \) undefined, so they are rejected. The remaining values are \( 0,\frac\pi4,\frac{3\pi}4,\pi,\frac{5\pi}4,\frac{7\pi}4,2\pi \).
4. Sum \( =\frac{\pi}4(0+1+3+4+5+7+8)=7\pi \).

Teacher's Note:
1. Divide through to reach cos 6x = cos 2x.
2. Remember to discard points where tan x does not exist.

Answer: (b) 4

Explanation:
1. \( \cos(\alpha-\beta)=1 \Rightarrow \alpha=\beta \) (the other possibility \( \alpha-\beta=\pm2\pi \) needs \( \alpha=\pm\pi,\ \beta=\mp\pi \)).
2. Case \( \alpha=\beta \): \( \cos2\alpha=\frac1e \) with \( 2\alpha\in[-2\pi,2\pi] \) has 4 solutions.
3. Case \( \alpha-\beta=\pm2\pi \): then \( \alpha+\beta=0 \) and \( \cos(\alpha+\beta)=1\ne\frac1e \). So exactly 4 pairs.

Teacher's Note:
1. Handle the possibility alpha - beta = +/- 2 pi, which also gives cos = 1.
2. Count the solutions of cos 2 alpha = 1/e on [-2 pi, 2 pi].

Answer: (a) \( 1440^\circ \)

Explanation:
1. \( \frac1{\sin^2\theta\cos^2\theta}+\frac2{\sin^2\theta}=8 \Rightarrow 1+2\cos^2\theta=8\sin^2\theta\cos^2\theta \).
2. With \( c=\cos^2\theta \): \( 8c(1-c)=1+2c \Rightarrow 8c^2-6c+1=0 \Rightarrow c=\frac12 \) or \( \frac14 \).
3. \( c=\frac12 \): \( 45^\circ,135^\circ,225^\circ,315^\circ \) (sum 720). \( c=\frac14 \): \( 60^\circ,120^\circ,240^\circ,300^\circ \) (sum 720). Total \( 1440^\circ \).

Teacher's Note:
1. Multiply through by sin^2 cos^2 and let c = cos^2 theta.
2. Each value of cos^2 gives four angles in a full turn, symmetric about 180 degrees.