JEE Mathematics Trigonometrical Functions and Identities MCQs Set 09

Answer: (b) \( 12x^2-11x-15=0 \)

Explanation:
1. \( \cos A=-\tfrac45 \), and since A is an angle of a convex quadrilateral, \( 0\lt A\lt \pi \), so \( \sin A=\tfrac35 \).
2. Then \( \tan A=-\tfrac34 \) and \( \operatorname{cosec}A=\tfrac53 \).
3. Sum of roots \( =\tfrac{11}{12} \), product \( =-\tfrac54 \), giving \( 12x^2-11x-15=0 \).

Teacher's Note:
1. Angles of a convex quadrilateral lie between 0 and 180 degrees, so the sine is positive.

Answer: (b) 0

Explanation:
1. By AM-GM, \( 2^x+2^{-x}\ge2 \) for every real x.
2. But \( \sin(e^x)\le1 \).
3. So the two sides can never be equal and there is no real solution.

Teacher's Note:
1. Compare the largest value of the left side with the smallest value of the right side.

Answer: (a) 0

Explanation:
1. Let \( t=e^{\sin x} \). Then \( t-\tfrac1t=4 \), so \( t^2-4t-1=0 \) and \( t=2+\sqrt5 \) (the other root is negative).
2. Then \( \sin x=\ln(2+\sqrt5)\approx1.44 \).
3. This exceeds 1, so no real x exists.

Teacher's Note:
1. Put e to the power sin x equal to t and solve the quadratic; then check whether sin x is allowed to take that value.

Answer: (c) \( px^2-2x+p=0 \)

Explanation:
1. Product of roots \( =\tan\tfrac\alpha2\cot\tfrac\alpha2=1 \).
2. Sum \( =\tan\tfrac\alpha2+\cot\tfrac\alpha2=\dfrac{2}{\sin\alpha}=\dfrac2p \).
3. The equation is \( x^2-\tfrac2px+1=0 \), i.e. \( px^2-2x+p=0 \).

Teacher's Note:
1. tan(t) + cot(t) = 2 / sin(2t).

Answer: (c) 6

Explanation:
1. \( (3\sin x-1)(\sin x-2)=0 \), and \( \sin x=2 \) is impossible, so \( \sin x=\tfrac13 \).
2. In each of the intervals \( (0,\pi) \), \( (2\pi,3\pi) \), \( (4\pi,5\pi) \) where sine is positive, there are 2 solutions.
3. Total \( =6 \).

Teacher's Note:
1. Count the solutions of sin x = 1/3 over each interval where sine is positive.

Answer: (c) \( 2(b+q)=a^2+p^2-2 \)

Explanation:
1. Then \( \cos\alpha+\cos\beta=-a \), \( \cos\alpha\cos\beta=b \), \( \sin\alpha+\sin\beta=-p \), \( \sin\alpha\sin\beta=q \).
2. Adding the squares of the sums, \( a^2+p^2=2+2\cos(\alpha-\beta) \).
3. Adding the products, \( b+q=\cos(\alpha-\beta) \).
4. Hence \( a^2+p^2=2+2(b+q) \).

Teacher's Note:
1. Use Vieta's formulas for both equations and combine with cos(A - B) = cos A cos B + sin A sin B.

Answer: (c) \( \pi \)

Explanation:
1. Rewrite as \( (x-1)^2+3+3\cos(ax+b)=0 \). Both \( (x-1)^2 \) and \( 3+3\cos(ax+b) \) are non-negative, so each must be 0.
2. Thus \( x=1 \) and \( \cos(a+b)=-1 \), i.e. \( a+b \) is an odd multiple of \( \pi \).
3. Since \( 0\lt a+b\le6\lt 3\pi \), \( a+b=\pi \).

Teacher's Note:
1. Complete the square in x and use the fact that a sum of non-negative terms is zero only if each term is zero.

Answer: (b) \( p\le\tfrac14 \)

Explanation:
1. A solution exists iff \( x+\tfrac px \) takes a value in \( [-1,1] \) for some \( x\ne0 \).
2. If \( p\le0 \), \( x+\tfrac px \) takes every real value, so this is possible.
3. If \( p>0 \), \( \left|x+\tfrac px\right|\ge2\sqrt p \), so we need \( 2\sqrt p\le1 \), i.e. \( p\le\tfrac14 \).
4. Together, \( p\le\tfrac14 \).

Teacher's Note:
1. Look at the range of x + p/x separately for p negative and p positive.

Answer: (d) \( [-1,+\infty) \)

Explanation:
1. With \( s=\sec\theta \) and \( |s|\ge1 \), \( \lambda=s^2+s-1 \).
2. For \( s\ge1 \) this takes all values in \( [1,\infty) \); for \( s\le-1 \) it takes all values in \( [-1,\infty) \) (it equals \( -1 \) at \( s=-1 \) and grows without bound).
3. The union is \( [-1,\infty) \).

Teacher's Note:
1. Replace tan squared by sec squared minus 1 and remember sec theta cannot lie between -1 and 1.

Answer: (a) 1 and 4 only

Explanation:
1. Using \( \sin P\sin Q=\tfrac12[\cos(P-Q)-\cos(P+Q)] \) on each product and adding, the terms cancel in pairs, so \( x+y+z=0 \) (statement 1 is true).
2. Then \( x+y-z=-2z \) and \( y+z-x=-2x \) are not identically zero, so 2 and 3 are false.
3. Since \( x+y+z=0 \), \( x^3+y^3+z^3=3xyz \) (statement 4 is true).

Teacher's Note:
1. Write each product as a difference of cosines and add the three results; use a^3 + b^3 + c^3 - 3abc = (a + b + c)(...).

Answer: (d) \( \dfrac1{8\sqrt2}\operatorname{cosec}\dfrac\pi{32} \)

Explanation:
1. The product equals \( \cos\tfrac\pi{32}\cos\tfrac\pi{16}\cos\tfrac\pi8 \).
2. Multiplying and dividing by \( \sin\tfrac\pi{32} \) and using \( \sin2x=2\sin x\cos x \) three times gives \( \dfrac{\sin\frac\pi4}{8\sin\frac\pi{32}} \).
3. With \( \sin\tfrac\pi4=\tfrac1{\sqrt2} \) this is \( \dfrac1{8\sqrt2\sin\frac\pi{32}} \).

Teacher's Note:
1. Convert to cos of the angles pi/32, pi/16, pi/8 and apply the doubling formula repeatedly.

Answer: (a) 1 and 2 only

Explanation:
1. \( 5\cos x+12\sin x \) takes every value in \( [-13,13] \), so the denominator \( D\ne0 \) lies in \( [-13,0)\cup(0,13] \).
2. Hence \( a\ge\tfrac1{13} \) or \( a\le-\tfrac1{13} \): the least positive value is \( \tfrac1{13} \) and the greatest negative value is \( -\tfrac1{13} \) (1 and 2 true).
3. Statement 3 is false (a = 1 is allowed, for example) and statement 4 is false (no value of a with absolute value below 1/13 is possible).

Teacher's Note:
1. The expression 5 cos x + 12 sin x ranges from -13 to 13, and its reciprocal skips the gap around zero.

Answer: (c) 2 and 4 only

Explanation:
1. \( y=\sin^2x+(1-\sin^2x)^2=1-\sin^2x+\sin^4x=1-s(1-s) \) with \( s=\sin^2x\in[0,1] \).
2. Hence \( y=\tfrac34+\left(s-\tfrac12\right)^2\in\left[\tfrac34,1\right] \).
3. So the maximum is 1 (statement 1 is false), the minimum is 3/4 (statement 2 is true), statement 3 is false and statement 4 is true.

Teacher's Note:
1. Write everything in terms of sin squared x and complete the square.

Answer: (a) 1, 3 and 4 only

Explanation:
1. \( \sin(60^\circ+x)\sin(60^\circ-x)=\sin^260^\circ-\sin^2x=\tfrac34-\sin^2x \).
2. So \( y=\sin x\left(\tfrac34-\sin^2x\right)=\tfrac14(3\sin x-4\sin^3x)=\tfrac14\sin3x \).
3. Hence \( -\tfrac14\le y\le\tfrac14 \): statements 1, 3 and 4 are true and 2 is false.

Teacher's Note:
1. Use sin(A + B) sin(A - B) = sin squared A - sin squared B, then the triple-angle formula.

Answer: (a) 2 and 3 only

Explanation:
1. \( x^2+y^2=X^2+Y^2 \) and \( xy=\tfrac12(X^2-Y^2)\sin2\theta-XY\cos2\theta \).
2. So \( x^2+4xy+y^2=X^2+Y^2+2(X^2-Y^2)\sin2\theta-4XY\cos2\theta \).
3. No XY term requires \( \cos2\theta=0 \), so \( \theta=\tfrac\pi4 \) and the expression is \( 3X^2-Y^2 \), i.e. \( a=3,\ b=-1 \).

Teacher's Note:
1. The cross term XY must vanish, which fixes the angle.

Answer: (a) 2 and 3 only

Explanation:
1. \( \cos A+\cos B=2\cos\tfrac{A+B}2\cos\tfrac{A-B}2=\sqrt3\cos\tfrac{A-B}2=1 \), so \( \cos\tfrac{A-B}2=\tfrac1{\sqrt3} \).
2. Then \( \cos(A-B)=2\cdot\tfrac13-1=-\tfrac13 \).
3. \( |\cos A-\cos B|=2\sin\tfrac{A+B}2\left|\sin\tfrac{A-B}2\right|=\left|\sin\tfrac{A-B}2\right|=\sqrt{1-\tfrac13}=\sqrt{\tfrac23} \).
4. So statements 2 and 3 are true.

Teacher's Note:
1. Use the sum-to-product formulas for cos A + cos B and cos A - cos B.

Answer: (a) 2 and 3 only

Explanation:
1. \( \cos(x-y)=-1 \) gives \( x-y=(2k+1)\pi \), so \( y=x-\pi \) (mod \( 2\pi \)).
2. Then \( \cos y=-\cos x \) and \( \sin y=-\sin x \).
3. So \( \cos x+\cos y=0 \) and \( \sin x+\sin y=0 \) are true, while \( \cos x-\cos y=2\cos x \) and \( \cos x+\sin y=\cos x-\sin x \) are not identically 0 or 1.

Teacher's Note:
1. Cosine equal to -1 means the two angles differ by an odd multiple of pi.