JEE Mathematics Trigonometric Functions MCQs Set 09

Answer: (c) \( \left[\frac34,1\right] \)

Explanation:
1. \( f=\sin^4\theta-\sin^2\theta+1=\left(\sin^2\theta-\frac12\right)^2+\frac34 \).
2. As \( \sin^2\theta\in[0,1] \), the square ranges over \( [0,\frac14] \), so \( f\in\left[\frac34,1\right] \).

Teacher's Note:
1. Complete the square in sin^2 theta.
2. Evaluate at the endpoints 0, 1/2, 1 of sin^2 theta.

Answer: (b) \( \frac{16}3 \)

Explanation:
1. \( y=4+2s-3s^2 \) with \( s=\sin x\in[-1,1] \).
2. Maximum at \( s=\frac13 \): \( \frac{13}3 \). Minimum at \( s=-1 \): \( 4-2-3=-1 \).
3. Difference \( =\frac{13}3+1=\frac{16}3 \).

Teacher's Note:
1. The quadratic opens downward, so the maximum is at the vertex and the minimum at an endpoint.
2. Compare the values at s = -1 and s = 1.

Answer: (b) 4

Explanation:
1. With \( s=\sin\theta\in[-1,1] \), \( 12s-9s^2=4-(3s-2)^2 \).
2. \( s=\frac23 \) is allowed, so the maximum is 4.

Teacher's Note:
1. Complete the square in s.
2. Check that the vertex lies inside [-1, 1].

Answer: (a) 11, 1

Explanation:
1. \( y=6+4\cos2x-3\sin2x \).
2. \( 4\cos2x-3\sin2x \) ranges over \( [-5,5] \).
3. Greatest 11 and least 1.

Teacher's Note:
1. Express in terms of 2x using \( \cos^2x=\frac{1+\cos2x}2 \).
2. The range of \( a\cos t+b\sin t \) is \( \pm\sqrt{a^2+b^2} \).

Answer: (d) \( n\pi \) or \( 2m\pi\pm\frac{2\pi}3 \)

Explanation:
1. \( \sin\theta\left(2+\frac1{\cos\theta}\right)=0 \).
2. \( \sin\theta=0 \Rightarrow \theta=n\pi \); \( \cos\theta=-\frac12 \Rightarrow \theta=2m\pi\pm\frac{2\pi}3 \).

Teacher's Note:
1. Factor out sin theta after writing tan as sin over cos.
2. cos theta = -1/2 has the principal value 2 pi / 3.

Answer: (b) 15

Explanation:
1. \( \sin x\sin4x=\cos x\cos4x \Rightarrow \cos5x=0 \), provided \( \cos4x\ne0 \).
2. \( x=\frac{(2k+1)\pi}{10} \). In \( (-\pi,2\pi) \) these are \( (2k+1)\in\{-9,-7,\ldots,19\} \), i.e. 15 values.
3. None of them makes cos 4x = 0 (that needs 4x to be an odd multiple of pi/2, i.e. x an odd multiple of pi/8), so all 15 are valid.

Teacher's Note:
1. Reduce to cos 5x = 0 and then check the excluded points.
2. List the odd multiples of pi/10 inside the interval.

Answer: (d) 5

Explanation:
1. \( \sin7x+\sin x=2\sin4x\cos3x \), so the equation is \( \sin4x(2\cos3x+1)=0 \).
2. \( \sin4x=0 \) gives \( x=0,\frac\pi4,\frac\pi2 \).
3. \( \cos3x=-\frac12 \) with \( 3x\in[0,\frac{3\pi}2] \) gives \( 3x=\frac{2\pi}3,\frac{4\pi}3 \), i.e. \( x=\frac{2\pi}9,\frac{4\pi}9 \).
4. Total: 5 solutions.

Teacher's Note:
1. Combine the outer terms with the sum-to-product formula.
2. Count the roots of each factor inside [0, pi/2]; none coincide.

Answer: (a) \( 2n\pi-\frac\pi3 \lt x \lt 2n\pi+\frac\pi3 \)

Explanation:
1. \( (2\cos x-1)(\cos x-3) \lt 0 \). Since \( \cos x-3 \lt 0 \) always, we need \( 2\cos x-1 \gt 0 \).
2. \( \cos x \gt \frac12 \Rightarrow 2n\pi-\frac\pi3 \lt x \lt 2n\pi+\frac\pi3 \).

Teacher's Note:
1. Factorise the quadratic in cos x.
2. The factor cos x - 3 is never zero or positive.

Answer: (c) \( x\in\left[2n\pi-\frac{2\pi}3,2n\pi+\frac{2\pi}3\right] \)

Explanation:
1. \( 2\cos^2x-\cos x-1\le0 \Rightarrow (2\cos x+1)(\cos x-1)\le0 \).
2. Since \( \cos x-1\le0 \), either \( \cos x=1 \) or \( 2\cos x+1\ge0 \), i.e. \( \cos x\ge-\frac12 \).
3. So \( x\in\left[2n\pi-\frac{2\pi}3,2n\pi+\frac{2\pi}3\right] \).

Teacher's Note:
1. cos 2x = 2cos^2 x - 1 turns this into a quadratic inequality in cos x.
2. The point cos x = 1 already lies inside the interval.

Answer: (d) \( \left(\frac{(4n+1)\pi}4,\frac{(3n+1)\pi}3\right),\ n\in Z \)

Explanation:
1. \( (\tan x-1)(\tan x-\sqrt3) \lt 0 \Rightarrow 1 \lt \tan x \lt \sqrt3 \).
2. So \( n\pi+\frac\pi4 \lt x \lt n\pi+\frac\pi3 \), which is \( \frac{(4n+1)\pi}4 \lt x \lt \frac{(3n+1)\pi}3 \).

Teacher's Note:
1. Factorise using the sum and product of the roots 1 and sqrt3.
2. Tangent has period pi.

Answer: (a) \( \tan^{-1}\frac29 \)

Explanation:
1. Take \( AB=1,\ AP=2 \): \( \tan\angle APC=\frac{1/2}2=\frac14 \) and \( \tan\angle APB=\frac12 \).
2. \( \tan\beta=\frac{\frac12-\frac14}{1+\frac18}=\frac{1/4}{9/8}=\frac29 \).

Teacher's Note:
1. beta is the difference of the elevations of B and C.
2. Use the tangent subtraction formula.

Answer: (c) \( r\operatorname{cosec}\frac\alpha2\sin\beta \)

Explanation:
1. The tangents from the eye to the balloon make an angle \( \alpha \), so the distance \( d \) from the eye to the centre satisfies \( \sin\frac\alpha2=\frac rd \).
2. Height of the centre \( =d\sin\beta=r\operatorname{cosec}\frac\alpha2\sin\beta \).

Teacher's Note:
1. Draw the right triangle formed by the eye, the centre and a point of tangency.
2. The line of sight to the centre bisects the angle between the tangents.

Answer: (b) 400 m

Explanation:
1. Let the cloud be \( h \) above the lake and the horizontal distance be \( d \).
2. \( \frac{h-200}d=\tan30^\circ \) and \( \frac{h+200}d=\tan60^\circ \). Dividing: \( \frac{h+200}{h-200}=3 \).
3. \( h+200=3h-600 \Rightarrow h=400 \) m.

Teacher's Note:
1. The reflection is as far below the lake surface as the cloud is above it.
2. Divide the two equations to remove d.

Answer: (c) 4

Explanation:
1. Group as \( (\tan9^\circ+\tan81^\circ)-(\tan27^\circ+\tan63^\circ) \).
2. \( \tan\theta+\cot\theta=\frac2{\sin2\theta} \), so the value is \( \frac2{\sin18^\circ}-\frac2{\sin54^\circ} \).
3. With \( \sin18^\circ=\frac{\sqrt5-1}4 \) and \( \sin54^\circ=\cos36^\circ=\frac{\sqrt5+1}4 \): \( 2\left(\frac4{\sqrt5-1}-\frac4{\sqrt5+1}\right)=8\cdot\frac{(\sqrt5+1)-(\sqrt5-1)}{4}=4 \).

Teacher's Note:
1. Pair tan t with tan(90 - t) = cot t.
2. The exact values of sin 18 and cos 36 finish the calculation.

Answer: (a) 4

Explanation:
1. \( \frac1{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}=\frac{\cos10^\circ-\sqrt3\sin10^\circ}{\sin10^\circ\cos10^\circ} \).
2. Numerator \( =2\cos(10^\circ+60^\circ)=2\sin20^\circ \); denominator \( =\frac12\sin20^\circ \).
3. Ratio 4.

Teacher's Note:
1. Use the auxiliary angle form cos t - sqrt3 sin t = 2 cos(t + 60).
2. sin 10 cos 10 = (1/2) sin 20.

Answer: (c) 3

Explanation:
1. \( \cot70^\circ+4\cos70^\circ=\frac{\cos70^\circ+4\sin70^\circ\cos70^\circ}{\sin70^\circ}=\frac{\cos70^\circ+2\sin140^\circ}{\sin70^\circ} \).
2. \( \cos70^\circ=\sin20^\circ,\ \sin140^\circ=\sin40^\circ \): the numerator is \( (\sin20^\circ+\sin40^\circ)+\sin40^\circ=\sin80^\circ+\sin40^\circ=2\sin60^\circ\cos20^\circ=\sqrt3\sin70^\circ \).
3. So the bracket is \( \sqrt3 \) and the value is \( \sqrt3\cdot\sqrt3=3 \).

Teacher's Note:
1. Put everything over sin 70 and use 2 sin t cos t = sin 2t.
2. Sum-to-product on sin 20 + sin 40 and then sin 80 + sin 40.

Answer: (a) 1

Explanation:
1. \( \sin^4\theta+\cos^4\theta=1-\frac12\sin^22\theta \).
2. This is largest when \( \sin2\theta=0 \), giving 1 (the least value is \( \frac12 \)).

Teacher's Note:
1. Use a^2 + b^2 = (a + b)^2 - 2ab twice.
2. The maximum is at theta = 0, pi/2 and so on.

Answer: (a) 2

Explanation:
1. \( 2\sin^2\theta+3\cos^2\theta=2+\cos^2\theta \).
2. The minimum of \( \cos^2\theta \) is 0, so the least value is 2.

Teacher's Note:
1. Rewrite using sin^2 + cos^2 = 1.
2. Then use the range of cos^2.

Answer: (d) 1

Explanation:
1. \( 1-\cos^2\theta+3\cos\theta=3 \Rightarrow \cos^2\theta-3\cos\theta+2=0 \Rightarrow (\cos\theta-1)(\cos\theta-2)=0 \).
2. \( \cos\theta=2 \) is impossible, so \( \cos\theta=1 \), giving \( \theta=0 \) only in \( [-\pi,\pi] \).

Teacher's Note:
1. Replace sin^2 by 1 - cos^2.
2. A cosine can never equal 2.

Answer: (d) 5

Explanation:
1. \( \sin x\sin4x=\cos x\cos4x \Rightarrow \cos5x=0 \), so \( x=\frac{(2k+1)\pi}{10} \).
2. In \( (0,\pi) \): \( \frac\pi{10},\frac{3\pi}{10},\frac{5\pi}{10},\frac{7\pi}{10},\frac{9\pi}{10} \).
3. None of these makes cos 4x zero, so all 5 are valid.

Teacher's Note:
1. The equation reduces to cos 5x = 0 as long as tan 4x is defined.
2. Check that tan 4x is defined at each root.

Answer: (a) 0

Explanation:
1. \( [\sin x]\ge-1 \) and \( [\sqrt2\cos x]\ge-2 \), so the sum \( -3 \) needs \( [\sin x]=-1 \) and \( [\sqrt2\cos x]=-2 \).
2. \( \sin x \lt 0 \Rightarrow x\in(\pi,2\pi) \) and \( \sqrt2\cos x \lt -1 \Rightarrow x\in\left(\frac{3\pi}4,\frac{5\pi}4\right) \). So \( x\in\left(\pi,\frac{5\pi}4\right) \).
3. There \( 2x\in(2\pi,\frac{5\pi}2) \) and \( \sin2x\in(0,1) \), so \( [\sin2x]=0 \).

Teacher's Note:
1. Both brackets must take their minimum possible values.
2. Find the interval of x that satisfies both and evaluate sin 2x on it.

Answer: (a) \( \tan B\tan C \lt 1 \)

Explanation:
1. \( \tan A \lt 0 \) means A is obtuse, so B and C are acute with \( B+C \lt \frac\pi2 \).
2. Then \( \tan(B+C)=\frac{\tan B+\tan C}{1-\tan B\tan C} \gt 0 \) with a positive numerator, so \( 1-\tan B\tan C \gt 0 \).

Teacher's Note:
1. An obtuse angle forces the other two angles to be acute with a small sum.
2. The sign of the denominator decides the answer.

Answer: (c) \( (\sqrt5+\sqrt3)/\sqrt5 \)

Explanation:
1. \( \sin^2A+\cos^2A=1 \Rightarrow \frac34\sin^2B+\frac54\cos^2B=1 \Rightarrow 3+2\cos^2B=4 \), so \( \cos^2B=\frac12 \) and \( B=\frac\pi4 \).
2. \( \tan A=\frac{\sin A}{\cos A}=\frac{\sqrt3\sin B}{\sqrt5\cos B}=\sqrt{\frac35} \) and \( \tan B=1 \).
3. \( \tan A+\tan B=\frac{\sqrt3+\sqrt5}{\sqrt5} \).

Teacher's Note:
1. Eliminate A by squaring and adding.
2. Then find tan A from the ratio of the two given ratios.

Answer: (c) \( \frac\pi8,\frac{3\pi}8 \)

Explanation:
1. The altitude on the hypotenuse is \( h=c\sin A\cos A \), so \( c=2\sqrt2\,c\sin A\cos A \).
2. \( \sin2A=\frac1{\sqrt2} \Rightarrow 2A=\frac\pi4 \) or \( \frac{3\pi}4 \), so \( A=\frac\pi8 \) or \( \frac{3\pi}8 \).

Teacher's Note:
1. Altitude to the hypotenuse = c sin A cos A.
2. The two acute angles are complementary.