JEE Mathematics Trigonometric Equations MCQs Set 10

Answer: (b) \(2\le a\le6\)

Explanation:
1. Using \(\cos2x=1-2\sin^2x\) with \(s=\sin x\): \(2s^2-as+2a-8=0\), i.e. \(a(2-s)=2(2-s)(2+s)\).
2. Since \(s\le1 \lt 2\), divide by \(2-s\): \(a=2(2+s)\). With \(s\in[-1,1]\), \(a\in[2,6]\).

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Answer: (b) \([-1,\infty)\)

Explanation:
1. Put \(t=\sec x\), \(|t|\ge1\): \(a=t^2+t-1\).
2. For \(t\ge1\): \(a\ge1\). For \(t\le-1\): \(a\) decreases from large values to \(-1\) at \(t=-1\). So \(a\) takes every value in \([-1,\infty)\).

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Answer: (a) -1

Explanation:
1. The exponents are \(1+u\) and \(2-u\) with \(u=\sin2x+\cos2x\). With \(y=4^u\): \(4y+\dfrac{16}y=65\Rightarrow4y^2-65y+16=0\Rightarrow y=16\) or \(\tfrac14\).
2. \(u=2\) is impossible since \(|u|\le\sqrt2\). So \(4^u=\tfrac14\), \(u=-1\).

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Answer: (d) None of these

Explanation:
1. \(4s^4+(1-s^2)^2=1\Rightarrow5s^4-2s^2=0\Rightarrow s=0\) or \(s^2=\tfrac25\).
2. So \(x=n\pi\) or \(\sin^2x=\tfrac25\). Since the second family is not covered by (a), (b) or (c), the answer is none of these.

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Answer: (d) \(\dfrac\pi6\)

Explanation:
1. Squaring and adding: \(9+16+24\sin(P+Q)=37\Rightarrow\sin(P+Q)=\tfrac12\), so \(P+Q=30^\circ\) or \(150^\circ\).
2. If \(P+Q=30^\circ\), then \(3\sin P\le1.5\) and \(4\cos Q\le4\), so \(3\sin P+4\cos Q \lt 6\), impossible. Hence \(P+Q=150^\circ\) and \(R=30^\circ=\dfrac\pi6\).

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Answer: (a) \(\left[-\pi,-\dfrac{3\pi}4\right]\cup\left[-\dfrac\pi4,\dfrac\pi4\right]\cup\left[\dfrac{3\pi}4,\pi\right]\)

Explanation:
1. \(|\tan x|\le1\) when \(x\) is within \(\dfrac\pi4\) of a multiple of \(\pi\).
2. In \([-\pi,\pi]\) the multiples of \(\pi\) are \(-\pi,0,\pi\), giving the three intervals in (a).

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Answer: (d) None of these

Explanation:
1. \(\tan x+\cot x=2\Rightarrow\tan x=1\), so \(\sin^2x=\cos^2x=\tfrac12\).
2. \(\sin^{2n}x+\cos^{2n}x=2\left(\tfrac12\right)^n=2^{1-n}\), which equals the given options only for special \(n\) (for example \(\tfrac12\) at \(n=2\)). So in general none of these.

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Answer: (a) 2

Explanation:
1. \(s+\dfrac1s=2\Rightarrow(s-1)^2=0\Rightarrow\sin\theta=1\), \(\operatorname{cosec}\theta=1\).
2. \(1^8+1^8=2\).

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Answer: (b) (2) and (4) only

Explanation:
1. \(-\pi \lt x-y \lt \pi\) gives \(x-y=\dfrac\pi6\) or \(\dfrac{5\pi}6\); \(0 \lt x+y \lt 2\pi\) gives \(x+y=\dfrac\pi3\) or \(\dfrac{5\pi}3\).
2. \(\dfrac\pi6,\dfrac\pi3\): \(x=\dfrac\pi4,y=\dfrac\pi{12}\). \(\dfrac\pi6,\dfrac{5\pi}3\): \(x=\dfrac{11\pi}{12},y=\dfrac{3\pi}4\). \(\dfrac{5\pi}6,\dfrac\pi3\): \(y \lt 0\), rejected. \(\dfrac{5\pi}6,\dfrac{5\pi}3\): \(x=\dfrac{5\pi}4>\pi\), rejected.
3. So (2) and (4) are correct.

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Answer: (c) (22, 8)

Explanation:
1. Let \(s=\sin x+\cos x\), \(p=\sin x\cos x=\dfrac{s^2-1}2\). The equation becomes \(s+\dfrac{1+s}{p}=7\), i.e. \(s+\dfrac2{s-1}=7\), so \(s^2-8s+9=0\) and \(s=4\pm\sqrt7\).
2. Only \(s=4-\sqrt7\) satisfies \(|s|\le\sqrt2\). Then \(\sin2x=s^2-1=22-8\sqrt7\), so \((a,b)=(22,8)\).

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Answer: (c) no solution in any quadrant

Explanation:
1. \(\sin x-\sin3x=-2\cos2x\sin x\), so the equation is \(2\sin2x-2\sin x\cos2x=3\), i.e. \(\sin x\,(4\cos x-2\cos2x)=3\).
2. Evaluating \(g(x)=\sin x+2\sin2x-\sin3x\) over \([0,2\pi]\) (checked numerically), its maximum is about \(2.73 \lt 3\). So the equation has no solution.

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Answer: (b) \(-\dfrac{24}{25}\)

Explanation:
1. \(\cos\theta=\dfrac{-5\pm35}{50}=\dfrac35\) or \(-\dfrac45\). In the second quadrant \(\cos\alpha=-\dfrac45\), \(\sin\alpha=\dfrac35\).
2. \(\sin2\alpha=2\cdot\dfrac35\cdot\left(-\dfrac45\right)=-\dfrac{24}{25}\).

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Answer: (d) 8

Explanation:
1. Let \(A,B\) be the two roots. \(A^2-B^2=16\cos2x\). If \(A-B=k=\pm1\), then \(A+B=16k\cos2x\) and \(A=\dfrac{k(1+16u)}2\) with \(u=\cos2x\).
2. Also \(A^2=\dfrac{u^2+16u+19}2\). Equating gives \(254u^2=37\), \(u=\pm\sqrt{37/254}\); both values give \(A,B\ge0\) (for \(u>0\) take \(k=1\), for \(u \lt 0\) take \(k=-1\)).
3. Each value of \(\cos2x\) has \(4\) solutions in \([0,2\pi]\), so there are \(8\) solutions (confirmed numerically).

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Answer: (b) 2

Explanation:
1. With \(t=\tan\dfrac\theta2\): \(2(1-t^2)+2t=1+t^2\Rightarrow3t^2-2t-1=0\Rightarrow t=1\) or \(-\tfrac13\). \(t=1\) is \(\theta=\dfrac\pi2\), excluded.
2. For \(t=-\tfrac13\): \(\cos\theta=\dfrac45\), \(\sin\theta=-\dfrac35\). Then \(7\cdot\dfrac45-6\cdot\dfrac35=\dfrac{10}5=2\).

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Answer: (a) 3

Explanation:
1. \(2\sin x\cos x-2\cos x+4\sin x-4=0\Rightarrow(\sin x-1)(2\cos x+4)=0\).
2. \(\cos x=-2\) is impossible, so \(\sin x=1\): \(x=\dfrac\pi2,\dfrac{5\pi}2,\dfrac{9\pi}2\), i.e. \(3\) solutions.

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Answer: (b) Statement I is false and statement II is true

Explanation:
1. Statement II: the identity holds for \(x\in[-1,1]\). With \(s=\sin^{-1}x\in[-\tfrac\pi2,\tfrac\pi2]\), \(s-\tfrac\pi4\in[-\tfrac{3\pi}4,\tfrac\pi4]\), so the square lies in \([0,\tfrac{9\pi^2}{16}]\). Statement II is true (for the values of x in the domain).
2. Statement I: \(s^3+c^3=\dfrac\pi2\left(\dfrac{\pi^2}4-3s\left(\dfrac\pi2-s\right)\right)\), whose range is \(\left[\dfrac{\pi^3}{32},\dfrac{7\pi^3}8\right]\). So a solution exists only for \(\dfrac1{32}\le a\le\dfrac78\), not for all \(a\ge\dfrac1{32}\). Statement I is false.

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Answer: (a) \(-\dfrac12\)

Explanation:
1. \(\sin(\cot^{-1}(1+x))=\dfrac1{\sqrt{1+(1+x)^2}}\) and \(\cos(\tan^{-1}x)=\dfrac1{\sqrt{1+x^2}}\).
2. Equating: \((1+x)^2=x^2\Rightarrow x=-\dfrac12\).

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Answer: (d) 3

Explanation:
1. Take sine of both sides (\(x\in[-1,1]\)): \(x=\sin(2\tan^{-1}x)=\dfrac{2x}{1+x^2}\), so \(x=0\) or \(1+x^2=2\Rightarrow x=\pm1\).
2. Check: \(x=1\): \(\dfrac\pi2=2\cdot\dfrac\pi4\) true; \(x=-1\): \(-\dfrac\pi2=-\dfrac\pi2\) true; \(x=0\) true. So \(3\) solutions.

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Answer: (d) \(P=Q\)

Explanation:
1. \(P\): \(\sin\theta=(1+\sqrt2)\cos\theta\Rightarrow\tan\theta=1+\sqrt2\).
2. \(Q\): \(\cos\theta=(\sqrt2-1)\sin\theta\Rightarrow\tan\theta=\dfrac1{\sqrt2-1}=1+\sqrt2\). Both sets are the solutions of the same equation, so \(P=Q\).

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Answer: (c) 3

Explanation:
1. \(2s^3-7s^2+7s-2=(s-1)(2s-1)(s-2)\), so \(\sin\alpha=1,\tfrac12\) (\(2\) rejected).
2. \(\sin\alpha=1\): \(1\) value; \(\sin\alpha=\tfrac12\): \(2\) values in \([0,2\pi]\). Total \(3\).

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