JEE Mathematics Properties and Applications of Definite Integrals MCQs Set 07

Answer: (d) none of these

Explanation:
1. Replace x by π/2 − x: I = ∫ f(π/2 − x)/(f(x) + f(π/2 − x)) dx.
2. Adding, 2I = ∫₀^(π/2) dx = π/2, so I = π/4, which is not among the first three options.

Teacher's Note:
1. Use the property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a − x)dx.
2. Add the two forms of the integral.

Answer: (b) \( \int_{-a}^{a}f(x)dx \)

Explanation:
1. ∫₀ᵃ f(−x)dx = ∫_{−a}^0 f(t)dt by the substitution t = −x.
2. So the sum is ∫₀ᵃ f + ∫_{−a}^0 f = ∫_{−a}^{a} f(x)dx.

Teacher's Note:
1. Substitute t = −x in the second integral.

Answer: (b) λ and ν

Explanation:
1. The term μ sin x/(1 + cos x) is odd, so its integral over [−π/4, π/4] is 0.
2. The remaining terms give 2λ(1 − cos π/4) + νπ/2 = 0, a relation between λ and ν only.

Teacher's Note:
1. Odd integrands vanish over symmetric limits.

Answer: (b) 2

Explanation:
1. f is the fractional part, periodic with period 1 and ∫ over one period = 1/2.
2. The interval [−2, 2] covers 4 periods, so the integral is 4 × 1/2 = 2.

Teacher's Note:
1. Use the integral over one period times the number of periods.

Answer: (a) 4

Explanation:
1. The integrand is even, so I = 2∫₀²|1 − x²|dx.
2. ∫₀¹(1 − x²)dx = 2/3 and ∫₁²(x² − 1)dx = 4/3, giving 2, so I = 4.

Teacher's Note:
1. Split at x = 1 and use evenness.

Answer: (a) logₑ 2

Explanation:
1. 2e⁻ˣ lies in [1, 2) for 0 < x ≤ ln 2, so [2e⁻ˣ] = 1 there.
2. For x > ln 2, 0 < 2e⁻ˣ < 1 so the value is 0.
3. The integral = ln 2.

Teacher's Note:
1. Find where the greatest integer changes from 1 to 0.

Answer: (d) −π/2

Explanation:
1. For 0 < x < π/2, cos x ∈ (0, 1) so [cos x] = 0; for π/2 < x ≤ π, cos x ∈ [−1, 0) so [cos x] = −1.
2. The integral = −(π − π/2) = −π/2.

Teacher's Note:
1. Evaluate the greatest integer on each half of the interval.

Answer: (b) \( \frac92 \)

Explanation:
1. On [−1, 0): 2 − x ≥ 2 and 2 − x ≥ 1 + x, so the maximum is 2 − x.
2. On [0, 1]: 2 ≥ 2 − x and 2 ≥ 1 + x, so the maximum is 2.
3. ∫₋₁⁰(2 − x)dx = 5/2 and ∫₀¹2 dx = 2, giving 9/2.

Teacher's Note:
1. Find which of the three expressions is largest on each interval.

Answer: (c) π/2

Explanation:
1. Replace x by π − x: I = ∫ dx/(1 + 3^(−cos x)) = ∫ 3^(cos x)/(1 + 3^(cos x)) dx.
2. Adding, 2I = ∫₀^π dx = π, so I = π/2.

Teacher's Note:
1. Use the property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a − x)dx.
2. Add the two forms of the integral.

Answer: (a) π

Explanation:
1. Replace x by 2π − x: sin → −sin, so I = ∫ dx/(e^(−sin x) + 1) = ∫ e^(sin x)/(1 + e^(sin x)) dx.
2. Adding, 2I = 2π, so I = π.

Teacher's Note:
1. Use the property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a − x)dx.
2. Add the two forms of the integral.

Answer: (b) \( 1-\frac{1}{\sqrt2} \)

Explanation:
1. On (0, π/4), [x] = 0 so d(x − [x]) = dx.
2. ∫₀^(π/4) sin x dx = 1 − cos(π/4) = 1 − 1/√2.

Teacher's Note:
1. The greatest integer is constant on the interval, so its differential is 0.

Answer: (c) \( \frac13(\alpha^3+\beta^3) \)

Explanation:
1. ∫_α^0 (−x²)dx = α³/3 (since x|x| = −x² for x < 0).
2. ∫₀^β x² dx = β³/3.
3. The total is (α³ + β³)/3.

Teacher's Note:
1. Split at 0 to remove the modulus.

Answer: (a) \( \frac\pi8\log_e2 \)

Explanation:
1. Using x → π/4 − x: 1 + tan(π/4 − x) = 2/(1 + tan x).
2. So I = ∫(ln 2 − ln(1 + tan x))dx = (π/4)ln 2 − I, giving I = (π/8) ln 2.

Teacher's Note:
1. King property with x → π/4 − x.

Answer: (b) \( \frac{17}{6} \)

Explanation:
1. ∫₋₁⁰(1 − x)dx = 1 + 1/2 = 3/2.
2. ∫₀¹(1 + x²)dx = 4/3.
3. Total 3/2 + 4/3 = 17/6.

Teacher's Note:
1. Remove the modulus on each piece.

Answer: (c) \( \frac{11}{4} \)

Explanation:
1. x³ − 3x² + 2x = x(x − 1)(x − 2): positive on (0, 1), negative on (1, 2), positive on (2, 3).
2. With F = x⁴/4 − x³ + x²: ∫₀¹ = 1/4, ∫₁² = 1/4 in absolute value, ∫₂³ = 9/4.
3. Total = 11/4.

Teacher's Note:
1. Find the sign changes and integrate piece by piece.

Answer: (b) a/2

Explanation:
1. Replace x by a − x: I = ∫ dx/(1 + e^(−f(x))) = ∫ e^(f)/(1 + e^(f)) dx.
2. Adding, 2I = a, so I = a/2.

Teacher's Note:
1. Use the property ∫₀ᵃ f(x)dx = ∫₀ᵃ f(a − x)dx.
2. Add the two forms of the integral.

Answer: (b) 0

Explanation:
1. f(−x) = ln(√(1 + x²) − x) = −ln(√(1 + x²) + x) = −f(x), so f is odd.
2. The integral over a symmetric interval is 0.

Teacher's Note:
1. Rationalise to test the parity.

Answer: (c) q, a

Explanation:
1. tan³x and sin x are odd, so their integrals vanish over [−a, a].
2. Only q∫cos²x dx remains, which depends on q and a.

Teacher's Note:
1. Discard the odd terms.

Answer: (c) 200√2

Explanation:
1. √(1 + cos 2x) = √2|cos x|.
2. |cos x| has period π and ∫ over one period = 2, so over 100π it is 200.
3. The integral = 200√2.

Teacher's Note:
1. Use the double-angle formula and periodicity.