JEE Mathematics Properties and Applications of Definite Integrals MCQs Set 08

Answer: (b) λ ∈ (0, 1)

Explanation:
1. ∫₀¹ x eˣ² dx = (e − 1)/2 ≈ 0.859.
2. ∫₀¹ eˣ² dx ≈ 1.463, so λ ≈ 0.587, which lies in (0, 1).

Teacher's Note:
1. Compute the first integral exactly and estimate the second.

Answer: (b) 0

Explanation:
1. (eˣ − 1)/(eˣ + 1) = tanh(x/2) is odd in x, and sin of an odd function is odd.
2. The limits are symmetric (−ln 2 to ln 2), so the integral is 0.

Teacher's Note:
1. Check that the integrand is odd.

Answer: (a) λ + μ

Explanation:
1. ∫ₐ^(2a) f(x)dx with x = 2a − t becomes ∫₀ᵃ f(2a − t)dt = μ.
2. So ∫₀^(2a) f = λ + μ.

Teacher's Note:
1. Substitute x = 2a − t in the second half.

Answer: (b) an even function

Explanation:
1. f is odd. Let F(x) = ∫₀ˣ f(t)dt.
2. F(−x) = ∫₀^(−x) f(t)dt = −∫₀ˣ f(−u)du = ∫₀ˣ f(u)du = F(x) (substituting t = −u).
3. So F is even.

Teacher's Note:
1. The integral of an odd function from 0 is even.

Answer: (a) \( \int_0^af(x)dx \)

Explanation:
1. I = ∫ f(a − x)g(a − x)dx = ∫ f(x)g(a − x)dx.
2. Adding, 2I = ∫ f(x)[g(x) + g(a − x)]dx = 2∫ f(x)dx, so I = ∫₀ᵃ f(x)dx.

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) k only

Explanation:
1. For a function of period k, ∫ₐ^(a+k) f is the same for every a.
2. It depends only on k.

Teacher's Note:
1. Integral over one period is independent of the starting point.

Answer: (a) \( (\sqrt2-1)\pi \)

Explanation:
1. With x → π − x (limits swap into each other), I = (π/2)∫dx/(1 + sin x).
2. ∫ dx/(1 + sin x) = tan x − sec x, which between π/4 and 3π/4 gives (−1 + √2) − (1 − √2) = 2√2 − 2.
3. I = (π/2)(2√2 − 2) = (√2 − 1)π.

Teacher's Note:
1. Use the king property to remove the factor x.

Answer: (b) π/4

Explanation:
1. Replace x by π/2 − x: tan → cot, giving ∫ tan³x/(1 + tan³x)dx.
2. Adding, 2I = π/2, so I = π/4.

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: (d) 0

Explanation:
1. The first factor is even and the second is odd, so the product is odd.
2. Over a symmetric interval an odd function integrates to 0.

Teacher's Note:
1. Check the parity of each factor.

Answer: (a) |p| ≤ q

Explanation:
1. |∫ f| ≤ ∫ |f| (triangle inequality for integrals), so |p| ≤ q.

Teacher's Note:
1. Triangle inequality for integrals.

Answer: (c) 1

Explanation:
1. f′(x) = ln(1 + x²).
2. f″(x) = 2x/(1 + x²), so f″(1) = 1.

Teacher's Note:
1. Apply the fundamental theorem of calculus twice.

Answer: (a) 2λ

Explanation:
1. f(2a − x) = f(x) means the integral over [a, 2a] equals that over [0, a].
2. So ∫₀^(2a) f = 2λ.

Teacher's Note:
1. Use the symmetry of the function about x = a.

Answer: (d) 3

Explanation:
1. f′(x) = x(eˣ − 1)(x − 2)³(x − 3)⁵.
2. x(eˣ − 1) ≥ 0 and does not change sign at 0.
3. (x − 2)³(x − 3)⁵ is positive for x < 2, negative on (2, 3) and positive for x > 3.
4. So f′ changes from − to + at x = 3: a local minimum at x = 3.

Teacher's Note:
1. Find where f′ changes sign from negative to positive.

Answer: (a) 1

Explanation:
1. ∫₂^(f(x)) 2t dt = f(x)² − 4, so the limit is lim (f(x)² − 4)/(x − 1).
2. By L'Hôpital this is 2f(1)f′(1) = 4f′(1) = 4, so f′(1) = 1.

Teacher's Note:
1. The limit is a derivative of f(x)² at x = 1.

Answer: (b) 4

Explanation:
1. φ′(x) = 2x(x² − 1) − (x − 1) = 2x³ − 3x + 1 = (x − 1)(2x² + 2x − 1) ≥ 0 on [1, 2].
2. φ is increasing, so the greatest value is φ(2) = ∫₂⁴(t − 1)dt = [t²/2 − t]₂⁴ = 4.

Teacher's Note:
1. Use the Leibniz rule to show φ is increasing.

Answer: (a) at least one root in (1, 2)

Explanation:
1. The equality gives ∫₁² (1 + sin⁴x)(ax² + bx + c)dx = 0.
2. Since 1 + sin⁴x > 0, the quadratic must change sign in (1, 2), so it has at least one root there.

Teacher's Note:
1. A continuous function with zero integral over an interval has a zero in it.

Answer: (c) \( \left[-\frac14,2\right] \)

Explanation:
1. f′(x) = x(x − 1)(x − 2): negative on (1, 2), positive on (2, 3).
2. f(1) = 0, f(2) = −1/4 (minimum), f(3) = 2.
3. The range is [−1/4, 2].

Teacher's Note:
1. Find the critical point at x = 2 and the end values.

Answer: (c) 0 if n is even

Explanation:
1. For even n, sin nx/sin x is antisymmetric about x = π/2, so the integral is 0.
2. For odd n it equals π.

Teacher's Note:
1. Compare the integrand at x and π − x.

Answer: (b) 1 and 2 only

Explanation:
1. f′ = 3a(x − 1)(x − 5), so b = −9a and c = 15a.
2. ∫₋₁¹ f = 2b/3 = 6 gives b = 9, a = −1 and c = −15.
3. Statements 1 and 2 are true; c = 15 and a = 1 are false.

Teacher's Note:
1. The odd terms ax³ and cx integrate to 0 over [−1, 1].

Answer: (a) 1, 2 and 4 only

Explanation:
1. f′(x) = |x − 1|.
2. It is continuous everywhere, differentiable at x = 2, but has a corner at x = 1.

Teacher's Note:
1. Differentiate the integral, then examine |x − 1|.