CBSE Class 11 Mathematics Trigonometric Functions JEE Previous Year Questions Set 01

Answer: (c) x + y ≠ 0

Explanation:
1. The right side must be defined, so x + y ≠ 0; this is necessary.
2. Also sin²θ ≤ 1 needs 4xy ≤ (x + y)², i.e. (x − y)² ≥ 0, which is always true.
3. (a) and (d) are not necessary (x = y or x = 0 still work), and (b) makes the expression undefined. So x + y ≠ 0.

Teacher's Note:
1. Always check that the expression is defined before applying the range of sin²θ.
2. (x + y)² − 4xy = (x − y)² ≥ 0, so the value can never exceed 1.

Answer: (c) \( \frac{\sqrt{3}}{2} \)

Explanation:
1. \( \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta \)
2. With θ = 15°: cos 30° = √3/2

Teacher's Note:
1. Recognise the formula cos 2θ = (1 − tan²θ)/(1 + tan²θ).
2. There is no need to find tan 15° = 2 − √3.

Answer: (b) \( -\frac{4}{5} \) or \( \frac{4}{5} \)

Explanation:
1. tan θ < 0, so θ lies in the second or fourth quadrant.
2. |sin θ| = 4/5 (from the 3-4-5 triangle).
3. In Q II sin θ = 4/5; in Q IV sin θ = −4/5. So sin θ = −4/5 or 4/5.

Teacher's Note:
1. A negative tan allows two quadrants with opposite signs of sine.
2. Without a quadrant restriction, keep both values.

Answer: (a) 1

Explanation:
1. sin (α + β) = 1 ⇒ α + β = π/2 (taking the principal value).
2. α + 2β = π/2 + β and 2α + β = π/2 + α, so the product = (−cot β)(−cot α) = cot α cot β.
3. α − β = π/6 with α + β = π/2 gives α = π/3, β = π/6: cot (π/3) cot (π/6) = (1/√3)(√3) = 1

Teacher's Note:
1. tan (π/2 + x) = −cot x.
2. In general (α + 2β) + (2α + β) = 3π/2, and tan x tan (3π/2 − x) = 1.

Answer: (d) y ≥ 2

Explanation:
1. y = sin²θ + 1/sin²θ = (sin θ − cosec θ)² + 2
2. The square is ≥ 0, so y ≥ 2.

Teacher's Note:
1. t + 1/t ≥ 2 for every t > 0, with equality at t = 1.
2. Here t = sin²θ, and equality holds when sin²θ = 1.

Answer: (c) no solution

Explanation:
1. a sin x + b cos x always lies in [−√(a² + b²), √(a² + b²)].
2. |c| > √(a² + b²) puts c outside this range, so there is no solution.

Teacher's Note:
1. a sin x + b cos x = √(a² + b²) sin (x + α) for a suitable α.
2. Solvability of a sin x + b cos x = c needs c² ≤ a² + b².

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

Explanation:
1. 25c² + 5c − 12 = 0 ⇒ (5c − 3)(5c + 4) = 0 ⇒ c = 3/5 or −4/5.
2. α is in the second quadrant, so cos α = −4/5 and sin α = 3/5.
3. sin 2α = 2 × (3/5) × (−4/5) = −24/25

Teacher's Note:
1. Treat the equation as a quadratic in cos θ and use the quadrant to choose the root.
2. In the second quadrant sine is positive and cosine is negative.

Answer: (c) \( -\frac{3}{\sqrt{130}} \)

Explanation:
1. Squaring and adding: 2 + 2 cos (α − β) = (441 + 729)/4225 = 1170/4225.
2. \( 4\cos^2\frac{\alpha - \beta}{2} = \frac{1170}{4225} \) ⇒ \( \cos^2\frac{\alpha - \beta}{2} = \frac{9}{130} \)
3. π/2 < (α − β)/2 < 3π/2, so the cosine is negative: −3/√130

Teacher's Note:
1. 1 + cos x = 2 cos²(x/2) turns the sum of squares into a half-angle cosine.
2. Use the given range of α − β to fix the sign.

Answer: (c) c = a + b

Explanation:
1. P + Q = π/2 ⇒ (P/2) + (Q/2) = π/4 ⇒ tan (P/2 + Q/2) = 1.
2. \( \frac{-b/a}{1 - c/a} = 1 \) ⇒ −b = a − c
3. ⇒ c = a + b

Teacher's Note:
1. The angle-sum property P + Q + R = π gives the sum of the half angles.
2. Use sum and product of roots in the tan addition formula.

Answer: (c) \( -\frac{4 + \sqrt{7}}{3} \)

Explanation:
1. Squaring: 1 + 2 sin x cos x = 1/4 ⇒ sin x cos x = −3/8, so x is in (π/2, π).
2. sin x cos x = tan x/(1 + tan²x) = −3/8 ⇒ 3t² + 8t + 3 = 0 ⇒ t = (−4 ± √7)/3.
3. sin x + cos x > 0 with cos x < 0 needs |sin x| > |cos x|, i.e. |tan x| > 1, so tan x = −(4 + √7)/3

Teacher's Note:
1. sin x cos x = tan x/(1 + tan²x) converts the product into a quadratic in tan x.
2. Use the sign of sin x + cos x to reject the extra root.

Answer: (c) both A and B are true

Explanation:
1. (Σ cos α)² + (Σ sin α)² = 3 + 2[cos (β − γ) + cos (γ − α) + cos (α − β)]
2. = 3 + 2(−3/2) = 0
3. A sum of two squares is 0 only if both are 0, so A and B are both true.

Teacher's Note:
1. Expanding (cos α + cos β + cos γ)² + (sin α + sin β + sin γ)² produces exactly the given cosine sum.
2. x² + y² = 0 for real x, y forces x = y = 0.

Answer: (b) \( \frac{56}{33} \)

Explanation:
1. α + β ∈ [0, π/2]: tan (α + β) = 3/4. α − β ∈ [−π/4, π/4] with positive sine: tan (α − β) = 5/12.
2. tan 2α = tan ((α + β) + (α − β)) = \( \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4}\cdot\frac{5}{12}} = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{56}{33} \)

Teacher's Note:
1. 2α = (α + β) + (α − β) is the standard split.
2. Use 3-4-5 and 5-12-13 triangles to get the tans directly.

Answer: (d) \( \frac{3}{4} \le A \le 1 \)

Explanation:
1. A = sin²x + cos⁴x = 1 − cos²x + cos⁴x = 1 − cos²x sin²x
2. = 1 − (1/4) sin²2x
3. 0 ≤ sin²2x ≤ 1 ⇒ 3/4 ≤ A ≤ 1

Teacher's Note:
1. Write sin²x = 1 − cos²x so that cos²x(1 − cos²x) = sin²x cos²x appears.
2. sin x cos x = (1/2) sin 2x gives a ready range.

Answer: (b) \( \frac{\pi}{6} \)

Explanation:
1. Squaring and adding: 9 + 16 + 24 (sin P cos Q + cos P sin Q) = 37 ⇒ sin (P + Q) = 1/2.
2. So sin R = 1/2, i.e. R = π/6 or 5π/6.
3. If R = 5π/6, then P < π/6, so 3 sin P + 4 cos Q < 3/2 + 4 < 6, impossible. Hence R = π/6

Teacher's Note:
1. Squaring and adding creates sin (P + Q) = sin R.
2. Always test both possible angles against the original equations.

Answer: (b) sec A cosec A + 1

Explanation:
1. Let t = tan A: \( \frac{t^2}{t - 1} + \frac{1}{t(1 - t)} = \frac{t^3 - 1}{t(t - 1)} = \frac{t^2 + t + 1}{t} \)
2. = t + 1 + 1/t = tan A + cot A + 1
3. tan A + cot A = 1/(sin A cos A) = sec A cosec A, so the expression = sec A cosec A + 1

Teacher's Note:
1. Writing everything in one variable t = tan A makes the algebra short.
2. t³ − 1 = (t − 1)(t² + t + 1).

Answer: (a) \( \frac{1}{12} \)

Explanation:
1. sin⁴x + cos⁴x = 1 − 2s, sin⁶x + cos⁶x = 1 − 3s, where s = sin²x cos²x.
2. f₄ − f₆ = (1 − 2s)/4 − (1 − 3s)/6 = 1/4 − s/2 − 1/6 + s/2
3. = 1/12

Teacher's Note:
1. Memorise sin⁴x + cos⁴x = 1 − 2 sin²x cos²x and sin⁶x + cos⁶x = 1 − 3 sin²x cos²x.
2. The s terms cancel, so the answer is constant.

Answer: (b) 2

Explanation:
1. sin θ = 1 − 2 cos θ ⇒ (1 − 2c)² + c² = 1 ⇒ 5c² − 4c = 0 ⇒ c = 0 or 4/5.
2. c = 0 gives sin θ = 1, i.e. θ = π/2, which is excluded. So cos θ = 4/5, sin θ = −3/5.
3. 7 cos θ + 6 sin θ = 28/5 − 18/5 = 2

Teacher's Note:
1. Substitute sin θ from the linear equation into sin²θ + cos²θ = 1.
2. Use the given exclusion θ ≠ π/2 to reject one root.

Answer: (b) \( \frac{7}{5} \)

Explanation:
1. Dividing the sum-to-product forms: tan ((α + β)/2) = (1/2)/(3/2) = 1/3, so tan θ = 1/3.
2. sin 2θ = 2t/(1 + t²) = (2/3)/(10/9) = 3/5; cos 2θ = (1 − t²)/(1 + t²) = (8/9)/(10/9) = 4/5
3. sin 2θ + cos 2θ = 7/5

Teacher's Note:
1. (sin α + sin β)/(cos α + cos β) = tan ((α + β)/2).
2. Express sin 2θ and cos 2θ through tan θ.

Answer: (d) \( -\frac{7}{9} \)

Explanation:
1. Let c = cos²x: 5((1 − c)/c − c) = 2(2c − 1) + 9 = 4c + 7.
2. 5 − 5c − 5c² = 4c² + 7c ⇒ 9c² + 12c − 5 = 0 ⇒ (3c − 1)(3c + 5) = 0 ⇒ c = 1/3.
3. cos 2x = 2c − 1 = −1/3, cos 4x = 2cos²2x − 1 = 2/9 − 1 = −7/9

Teacher's Note:
1. Express everything in cos²x; tan²x = (1 − cos²x)/cos²x.
2. Reject c = −5/3 because cos²x cannot be negative.

Answer: (c) −25

Explanation:
1. tan A + tan B = 10/3, tan A tan B = −25/3 ⇒ tan (A + B) = (10/3)/(1 + 25/3) = 5/14.
2. Expression = cos²(A + B)[3 tan²(A + B) − 10 tan (A + B) − 25] = \( \frac{1}{1 + \frac{25}{196}}\left(\frac{75}{196} - \frac{50}{14} - 25\right) \)
3. = (196/221)(75 − 700 − 4900)/196 = −5525/221 = −25

Teacher's Note:
1. Factor out cos²(A + B) to get a polynomial in tan (A + B).
2. cos²θ = 1/(1 + tan²θ).

Answer: (c) sec A, tan A

Explanation:
1. cos A = −3/5, and A is a triangle angle, so sin A = 4/5.
2. tan A = −4/3, sec A = −5/3, cot A = −3/4.
3. 9x² + 27x + 20 = (3x + 4)(3x + 5) = 0 ⇒ x = −4/3, −5/3, i.e. tan A and sec A.

Teacher's Note:
1. Angles of a triangle lie in (0, π), so sin A is positive.
2. Factorise the quadratic and match the roots with the ratios.

Answer: (a) \( \tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2} = \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2} \)

Explanation:
1. α/2 + β/2 + γ/2 = π, so tan (α/2 + β/2) = tan (π − γ/2) = −tan (γ/2).
2. \( \frac{\tan\frac{\alpha}{2} + \tan\frac{\beta}{2}}{1 - \tan\frac{\alpha}{2}\tan\frac{\beta}{2}} = -\tan\frac{\gamma}{2} \)
3. Cross-multiplying: tan (α/2) + tan (β/2) + tan (γ/2) = tan (α/2) tan (β/2) tan (γ/2)

Teacher's Note:
1. Three angles adding to π always satisfy Σ tan = Π tan.
2. The relation in (b) needs the half angles to add to π/2, not π.

Answer: (c) \( \frac{1}{8} \)

Explanation:
1. cos (7π/8) = −cos (π/8), cos (5π/8) = −cos (3π/8).
2. Product = (1 − cos²(π/8))(1 − cos²(3π/8)) = sin²(π/8) sin²(3π/8) = sin²(π/8) cos²(π/8)
3. = (1/4) sin²(π/4) = 1/8

Teacher's Note:
1. Pair factors as (1 + c)(1 − c) = 1 − c² = sin².
2. sin (3π/8) = cos (π/8) since the angles are complementary.

Answer: (d) infinite

Explanation:
1. cos 2x = 1 − 2 sin²x, so the expression = (a₁ + a₂) + (a₃ − 2a₂) sin²x.
2. For this to vanish for all x: a₁ + a₂ = 0 and a₃ = 2a₂.
3. So (a₁, a₂, a₃) = (−k, k, 2k) for any real k: infinitely many triplets.

Teacher's Note:
1. An identity in sin²x holds for all x only if each coefficient is zero.
2. One free parameter k gives infinitely many solutions.

Answer: (c) 4

Explanation:
1. \( \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} = \frac{\sqrt{3}\cos 20^\circ - \sin 20^\circ}{\sin 20^\circ\cos 20^\circ} \)
2. Numerator = 2 (sin 60° cos 20° − cos 60° sin 20°) = 2 sin 40°; denominator = (1/2) sin 40°.
3. Value = 2 sin 40° / ((1/2) sin 40°) = 4

Teacher's Note:
1. √3 cos x − sin x = 2 sin (60° − x).
2. sin 20° cos 20° = (1/2) sin 40°, which cancels with the numerator.

Answer: (d) (0, π)

Explanation:
1. Discriminant D = cos²p − 4 (cos p − 1) sin p.
2. cos p − 1 ≤ 0, so −4 (cos p − 1) sin p ≥ 0 whenever sin p > 0.
3. Hence D > 0 for every p in (0, π), where sin p > 0.

Teacher's Note:
1. Check the sign of each part of the discriminant separately.
2. On (0, π) sine is positive, which makes every term non-negative.

Answer: (c) 13

Explanation:
1. Let s = sin x cos x. (sin x − cos x)² = 1 − 2s, so 3(1 − 2s)² = 3 − 12s + 12s².
2. 6 (sin x + cos x)² = 6 + 12s; 4 (sin⁶x + cos⁶x) = 4 − 12s².
3. Sum = 3 + 6 + 4 = 13

Teacher's Note:
1. Reduce every bracket to an expression in s = sin x cos x.
2. Check with x = 0: 3 + 6 + 4 = 13.

Answer: (c) sin 15° cos 15°

Explanation:
1. sin 15° = (√3 − 1)/(2√2) and cos 15° = (√3 + 1)/(2√2) are irrational.
2. sin 15° cos 15° = (1/2) sin 30° = 1/4, rational.
3. sin 15° cos 75° = sin²15° = (1 − cos 30°)/2 = (2 − √3)/4, irrational.

Teacher's Note:
1. 2 sin x cos x = sin 2x often turns surds into simple numbers.
2. cos 75° = sin 15°, so (d) is a square of an irrational sine.

Answer: (c) tan β + 2 tan γ

Explanation:
1. α = β + γ ⇒ tan α = (tan β + tan γ)/(1 − tan β tan γ) ⇒ tan α − tan α tan β tan γ = tan β + tan γ.
2. α + β = π/2 ⇒ tan α tan β = 1.
3. tan α − tan γ = tan β + tan γ ⇒ tan α = tan β + 2 tan γ

Teacher's Note:
1. Complementary angles have tan α tan β = 1.
2. Combine it with the tan addition formula for α = β + γ.

Answer: (b) 8

Explanation:
1. 7 cos x + 5 sin x ∈ [−√74, √74] and √74 ≈ 8.6.
2. Need −8.6 ≤ 2k + 1 ≤ 8.6 ⇒ −4.8 ≤ k ≤ 3.8.
3. Integers k = −4, −3, …, 3: 8 values.

Teacher's Note:
1. a cos x + b sin x = c is solvable iff |c| ≤ √(a² + b²).
2. Convert the range of 2k + 1 into a range of k before counting.

Answer: (b) \( \left(\frac{\pi}{2}, \frac{2\pi}{3}\right) \)

Explanation:
1. sin θ = 1/2 with θ acute ⇒ θ = π/6.
2. 0 < cos φ = 1/3 < 1/2 and cos decreases on (0, π/2), so π/3 < φ < π/2.
3. θ + φ ∈ (π/6 + π/3, π/6 + π/2) = (π/2, 2π/3)

Teacher's Note:
1. Bracket a non-standard value between two standard values of the same function.
2. Remember cosine decreases as the angle increases in the first quadrant.

Answer: (b) t₄ > t₃ > t₁ > t₂

Explanation:
1. For θ ∈ (0, π/4): 0 < tan θ < 1 < cot θ.
2. Base cot θ > 1: a larger exponent gives a larger value, so t₄ > t₃ > 1.
3. Base tan θ < 1: a larger exponent gives a smaller value, so 1 > t₁ > t₂. Hence t₄ > t₃ > t₁ > t₂.

Teacher's Note:
1. For base > 1 powers increase with the exponent; for base between 0 and 1 they decrease.
2. Comparing each number with 1 splits the four quickly into two pairs.

Answer: (d) P = Q

Explanation:
1. P: sin θ = (√2 + 1) cos θ ⇒ tan θ = √2 + 1.
2. Q: cos θ = (√2 − 1) sin θ ⇒ tan θ = 1/(√2 − 1) = √2 + 1.
3. Both sets are {θ : tan θ = √2 + 1}, so P = Q.

Teacher's Note:
1. Reduce each condition to a value of tan θ (cos θ = 0 or sin θ = 0 does not satisfy either equation).
2. Rationalise 1/(√2 − 1) = √2 + 1.

Answer: (b) xyz = xy + z; (c) xyz = x + y + z

Explanation:
1. Infinite G.P. sums: x = 1/sin²φ, y = 1/cos²φ, z = 1/(1 − sin²φ cos²φ).
2. xy = 1/(sin²φ cos²φ) = x + y, and 1/z = 1 − 1/(xy), so z = xy/(xy − 1) ⇒ xyz = xy + z.
3. Since xy = x + y, xyz = xy + z = x + y + z. Hence (b) and (c).

Teacher's Note:
1. The sum of an infinite G.P. 1 + r + r² + … is 1/(1 − r) for |r| < 1.
2. x + y = (sin²φ + cos²φ)/(sin²φ cos²φ) = xy is the key relation.

Answer: (a) x = y ≠ 0; (b) x = y, x ≠ 0

Explanation:
1. sec²θ ≥ 1 ⇒ 4xy ≥ (x + y)² ⇒ (x − y)² ≤ 0 ⇒ x = y.
2. The expression must be defined: x + y ≠ 0, so x = y ≠ 0.
3. (a) and (b) state exactly this; (c) allows x = y = 0, and (d) allows x ≠ y.

Teacher's Note:
1. Compare the expression with the range of sec²θ, which is [1, ∞).
2. A square that is ≤ 0 must be exactly 0.

Answer: (a) \( \tan^2 x = \frac{2}{3} \); (b) \( \frac{\sin^8 x}{8} + \frac{\cos^8 x}{27} = \frac{1}{125} \)

Explanation:
1. Let s = sin²x: s²/2 + (1 − s)²/3 = 1/5 ⇒ 15s² + 10(1 − s)² = 6 ⇒ 25s² − 20s + 4 = 0 ⇒ s = 2/5.
2. tan²x = (2/5)/(3/5) = 2/3.
3. sin⁸x/8 + cos⁸x/27 = (16/625)/8 + (81/625)/27 = 2/625 + 3/625 = 1/125

Teacher's Note:
1. Using cos²x = 1 − sin²x turns the condition into a quadratic that is a perfect square.
2. Once sin²x is known, every other expression follows directly.

Answer: (a) \( \tan\frac{\alpha}{2} + \sqrt{3}\tan\frac{\beta}{2} = 0 \); (c) \( \tan\frac{\alpha}{2} - \sqrt{3}\tan\frac{\beta}{2} = 0 \)

Explanation:
1. The condition gives cos α (cos β − 2) = 1 − 2 cos β, so \( \cos\alpha = \frac{1 - 2\cos\beta}{\cos\beta - 2} \).
2. \( \tan^2\frac{\alpha}{2} = \frac{1 - \cos\alpha}{1 + \cos\alpha} = \frac{(\cos\beta - 2) - (1 - 2\cos\beta)}{(\cos\beta - 2) + (1 - 2\cos\beta)} = \frac{3(\cos\beta - 1)}{-(1 + \cos\beta)} = 3\tan^2\frac{\beta}{2} \)
3. So tan (α/2) = ±√3 tan (β/2), which is (a) or (c).

Teacher's Note:
1. Isolate cos α, then use tan²(x/2) = (1 − cos x)/(1 + cos x).
2. Taking the square root gives both signs, so two options are correct.

Answer: (a) A→r; B→p

Explanation:
1. α ∈ (13π/48, 14π/48): 3α ∈ (39π/48, 42π/48) ⊂ (π/2, π), sin 3α > 0; 2α ∈ (26π/48, 28π/48), cos 2α < 0 ⇒ negative (p).
2. α ∈ (18π/48, 23π/48): 3α ∈ (54π/48, 69π/48) ⊂ (π, 3π/2), sin 3α < 0; 2α ∈ (36π/48, 46π/48), cos 2α < 0 ⇒ positive (r).
3. On (q) and (s), 3α or 2α crosses a multiple of π or π/2, so the sign is not fixed. Hence A→r, B→p.

Teacher's Note:
1. Find the interval of 3α and 2α first, then read off the signs from the quadrants.
2. A quotient is positive when numerator and denominator have the same sign.

Answer: 6

Explanation:
1. sin³x = (3 sin x − sin 3x)/4, so sin³x sin 3x = (3 sin x sin 3x − sin²3x)/4.
2. 3 sin x sin 3x = (3/2)(cos 2x − cos 4x) and sin²3x = (1 − cos 6x)/2.
3. So the expression = (1/8)(3 cos 2x − 3 cos 4x − 1 + cos 6x); the highest term is cos 6x, so n = 6.

Teacher's Note:
1. Reduce powers with sin³x = (3 sin x − sin 3x)/4 and sin²y = (1 − cos 2y)/2.
2. Products of sines become sums of cosines through 2 sin A sin B = cos (A − B) − cos (A + B).