CBSE Class 12 Mathematics Integrals MCQs Set 10

Answer: (b) 2

Explanation:
1. \( \frac{1}{2}(1 - \cos 2x) = \sin^2 x \), so the integrand is |sin x|, an even function.
2. \( \int_{-\pi/2}^{\pi/2} |\sin x|\,dx = 2\int_{0}^{\pi/2} \sin x\,dx = 2 \).

Teacher's Note:
1. √(sin²x) = |sin x|, not sin x; dropping the modulus here would wrongly give 0.
2. 1 − cos 2x = 2sin²x is the identity that removes the square root.

Answer: (c) 0

Explanation:
1. x¹⁷ is odd and cos⁴x is even, so x¹⁷cos⁴x is odd.
2. Hence \( \int_{-1}^{1} x^{17}\cos^4 x\,dx = 0 \).

Teacher's Note:
1. Odd × even = odd, and odd functions integrate to zero over symmetric limits.
2. Integration by parts would be extremely long; symmetry gives the answer at once.

Answer: (d) 2

Explanation:
1. \( f(-a) = \frac{e^{-a}}{1 + e^{-a}} = \frac{1}{1 + e^a} = 1 - f(a) \), so the limits add up to 1.
2. Use x → 1 − x (allowed since the limits sum to 1); x(1 − x) is unchanged, so \( I_1 = \int (1 - x)g\{x(1 - x)\}\,dx = I_2 - I_1 \).
3. Hence \( 2I_1 = I_2 \) and \( \frac{I_2}{I_1} = 2 \).

Teacher's Note:
1. ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx; find a + b first and use it.
2. x(1 − x) is symmetric under x → 1 − x, which is why the factor x can be removed.

Answer: (c) \( \frac{4}{3} \)

Explanation:
1. \( \cos x - \cos^3 x = \cos x\sin^2 x \), so the integrand is \( \sqrt{\cos x}\,|\sin x| \), an even function.
2. \( 2\int_{0}^{\pi/2} \sqrt{\cos x}\,\sin x\,dx = 2\left[-\frac{2}{3}(\cos x)^{3/2}\right]_{0}^{\pi/2} = 2\cdot\frac{2}{3} = \frac{4}{3} \).

Teacher's Note:
1. √(sin²x) = |sin x|, which makes the integrand even rather than odd; ignoring the modulus would give 0.
2. With t = cos x, the integral of √t is (2/3)t^(3/2).

Answer: (a) 0

Explanation:
1. Replace x by π − x: sin 2m(π − x) = −sin 2mx, while sin(π − x) = sin x.
2. So the integrand changes sign and I = −I.
3. Hence I = 0.

Teacher's Note:
1. Use ∫₀^(2a) f = 0 when f(2a − x) = −f(x).
2. For odd multiples, sin(2m + 1)x/sin x behaves differently (the integral is π); here the multiple 2m is even.

Answer: (c) \( \int_{0}^{a} f(x)\,dx + \int_{0}^{a} f(2a - x)\,dx \)

Explanation:
1. \( \int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{a}^{2a} f(x)\,dx \).
2. In the second, put x = 2a − t: \( \int_{a}^{2a} f(x)\,dx = \int_{0}^{a} f(2a - t)\,dt \).
3. So \( \int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{0}^{a} f(2a - x)\,dx \).

Teacher's Note:
1. This general identity leads to the special cases 2∫₀^a f (if f(2a − x) = f(x)) and 0 (if f(2a − x) = −f(x)).
2. Options (a) and (b) hold only under extra conditions; (c) always holds.

Answer: (b) f(2a − x) = f(x)

Explanation:
1. Always \( \int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{0}^{a} f(2a - x)\,dx \).
2. This equals 2∫₀^a f(x) dx exactly when the second integral equals the first, which holds when f(2a − x) = f(x).

Teacher's Note:
1. f(2a − x) = f(x) means the graph is symmetric about the line x = a.
2. If instead f(2a − x) = −f(x), the integral over [0, 2a] would be 0.

Answer: (c) 0

Explanation:
1. Replace x by π − x: cos x → −cos x, while cos(x/2) ↔ sin(x/2) swap, so the denominator is unchanged.
2. Thus the integrand changes sign and I = −I.
3. Hence I = 0.

Teacher's Note:
1. cos((π − x)/2) = sin(x/2) and sin((π − x)/2) = cos(x/2); a symmetric denominator survives the substitution.
2. Look for the a − x substitution whenever the answer options include 0.

Answer: (d) 0

Explanation:
1. cos(2π − x) = cos x, so \( \int_{0}^{2\pi} \cos^{99} x\,dx = 2\int_{0}^{\pi} \cos^{99} x\,dx \).
2. cos(π − x) = −cos x and 99 is odd, so \( \int_{0}^{\pi} \cos^{99} x\,dx = 0 \).
3. Hence the value is 0.

Teacher's Note:
1. An odd power of cos x integrates to 0 over [0, π] and over [0, 2π].
2. Even powers behave differently: ∫₀^(2π) cos²ⁿx dx is positive.

Answer: (c) 8/3

Explanation:
1. |sin³x| has period π, so \( \int_{0}^{2\pi} |\sin^3 x|\,dx = 2\int_{0}^{\pi} \sin^3 x\,dx \).
2. \( \int_{0}^{\pi} \sin^3 x\,dx = \left[-\cos x + \frac{\cos^3 x}{3}\right]_{0}^{\pi} = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{4}{3} \).
3. Value \( = 2\cdot\frac{4}{3} = \frac{8}{3} \).

Teacher's Note:
1. sin³x is negative on (π, 2π), so the modulus matters; without it the answer would be 0.
2. Use sin³x = (1 − cos²x) sin x and t = cos x.

Answer: (a) \( 2\pi \log_e\left(\frac{1}{2}\right) \)

Explanation:
1. \( \int_{0}^{\pi} \log\sin^2 x\,dx = 2\int_{0}^{\pi} \log\sin x\,dx = 2\cdot 2\int_{0}^{\pi/2} \log\sin x\,dx \).
2. \( = 4\left(-\frac{\pi}{2}\log 2\right) = -2\pi\log 2 = 2\pi\log_e\frac{1}{2} \).

Teacher's Note:
1. Use log sin²x = 2 log sin x (sin x > 0 on (0, π)) and the symmetry sin(π − x) = sin x.
2. A definite integral has no constant of integration, so the options with + c are wrong.

Answer: (a) 0

Explanation:
1. sin(2π − x) = −sin x, so sin⁵(2π − x) = −sin⁵x.
2. By \( \int_{0}^{2a} f = 0 \) when f(2a − x) = −f(x), the value is 0.

Teacher's Note:
1. Odd powers of sin x integrate to 0 over a full period [0, 2π].
2. 16/15 is ∫₀^π sin⁵x dx; over [0, 2π] the positive and negative halves cancel.

Answer: (b) 0

Explanation:
1. \( \log\left(\frac{1}{x} - 1\right) = \log\frac{1 - x}{x} \).
2. Replace x by 1 − x: the integrand becomes \( \log\frac{x}{1 - x} = -\log\frac{1 - x}{x} \), so I = −I.
3. Hence I = 0.

Teacher's Note:
1. Rewrite 1/x − 1 as (1 − x)/x to see the symmetry under x → 1 − x.
2. The graph is antisymmetric about x = ½, so the positive and negative areas cancel.

Answer: (a) 0

Explanation:
1. Replace x by π/2 − x: sin(4π − 8x) = −sin 8x, cos(π − 2x) = −cos 2x, and log cot x becomes log tan x = −log cot x.
2. The three sign changes give an overall minus, so I = −I.
3. Hence I = 0.

Teacher's Note:
1. Count the sign changes factor by factor; an odd number of changes gives I = −I.
2. Note \( \frac{\sin 8x}{\cos 2x} \) is finite at x = π/4, so the integrand has no real singularity there.

Answer: (a) 0

Explanation:
1. Replace x by π − x: sin 2(π − x) = −sin 2x and sin³(π − x) = sin³x.
2. So the integrand changes sign and I = −I, i.e. I = 0.

Teacher's Note:
1. Directly: 2∫₀^π sin⁴x cos x dx = [2sin⁵x/5]₀^π = 0, confirming the symmetry argument.
2. sin 2x = 2 sin x cos x makes the substitution t = sin x natural.

Answer: (b) \( \int_{0.5}^{1.5} f(x)\,dx \)

Explanation:
1. The limits 0.5 and 1.5 add up to 2, so replace x by 2 − x: \( I = \int_{0.5}^{1.5} (2 - x)f(2 - x)\,dx = \int_{0.5}^{1.5} (2 - x)f(x)\,dx \).
2. So \( I = 2\int_{0.5}^{1.5} f(x)\,dx - I \), giving \( I = \int_{0.5}^{1.5} f(x)\,dx \).

Teacher's Note:
1. ∫ₐᵇ x f(x) dx = ((a + b)/2)∫ₐᵇ f(x) dx when f(a + b − x) = f(x).
2. Here (a + b)/2 = 1, so the factor x is simply replaced by 1.

Answer: (c) \( \frac{na - 2}{2n} \)

Explanation:
1. The limits 1/n and (an − 1)/n add up to a. Replace x by a − x: \( I = \int \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}}dx \) over the same limits.
2. Adding: \( 2I = \int_{1/n}^{(an - 1)/n} 1\,dx = \frac{an - 1}{n} - \frac{1}{n} = \frac{an - 2}{n} \).
3. So \( I = \frac{na - 2}{2n} \).

Teacher's Note:
1. The answer is half the length of the interval, as for every g(x)/(g(x) + g(a + b − x)) integral.
2. Check that the limits sum to the constant a inside √(a − x); that is what makes the method work.

Answer: (c) \( \frac{2}{\pi} I_1 \)

Explanation:
1. The limits a and π − a add up to π. Replace x by π − x: \( I_1 = \int_{a}^{\pi - a} (\pi - x)f(\sin x)\,dx = \pi I_2 - I_1 \).
2. So \( 2I_1 = \pi I_2 \), i.e. \( I_2 = \frac{2}{\pi}I_1 \).

Teacher's Note:
1. sin(π − x) = sin x keeps f(sin x) unchanged under the substitution.
2. Read the question carefully: it asks for I₂ in terms of I₁, not the other way round.

Answer: (c) 1/2

Explanation:
1. The limits 1 − k and k add up to 1. Replace x by 1 − x; x(1 − x) is unchanged, so \( I_1 = \int_{1 - k}^{k} (1 - x)f\{x(1 - x)\}\,dx = I_2 - I_1 \).
2. So \( 2I_1 = I_2 \) and \( \frac{I_1}{I_2} = \frac{1}{2} \).

Teacher's Note:
1. The condition 2k − 1 > 0 just ensures k > 1 − k, so the interval is non-empty; positivity of f makes I₂ ≠ 0.
2. The factor x averages to the midpoint ½ of the interval.

Answer: (c) \( \pi \)

Explanation:
1. Put x = 2 + sin²θ: dx = 2 sin θ cos θ dθ, x − 2 = sin²θ, 3 − x = cos²θ; θ: 0 → π/2.
2. \( \int_{0}^{\pi/2} \frac{2\sin\theta\cos\theta\,d\theta}{\sin\theta\cos\theta} = \int_{0}^{\pi/2} 2\,d\theta = \pi \).

Teacher's Note:
1. In general \( \int_{a}^{b} \frac{dx}{\sqrt{(x - a)(b - x)}} = \pi \) for any a < b.
2. The substitution x = a cos²θ + b sin²θ always handles (x − a)(b − x).

Answer: (b) \( 200\sqrt{2} \)

Explanation:
1. \( \sqrt{1 - \cos 2x} = \sqrt{2\sin^2 x} = \sqrt{2}\,|\sin x| \), which has period π.
2. \( I = 100\sqrt{2}\int_{0}^{\pi} \sin x\,dx = 100\sqrt{2}\cdot 2 = 200\sqrt{2} \).

Teacher's Note:
1. For a periodic function with period T, ∫₀^(nT) f = n∫₀^T f.
2. The modulus is essential: without it the integral of sin x over 100 periods would be 0.

Answer: (d) 18

Explanation:
1. |sin x| has period π, and \( \int_{0}^{\pi} |\sin x|\,dx = 2 \).
2. The interval [π, 10π] has length 9π, i.e. 9 periods.
3. So the value is 9 × 2 = 18.

Teacher's Note:
1. For a periodic integrand, any interval of length one period gives the same integral, regardless of where it starts.
2. Count the periods carefully: 10π − π = 9π, not 10π.

Answer: (c) 1000(e − 1)

Explanation:
1. x − [x] has period 1, so e^(x − [x]) also has period 1.
2. \( \int_{0}^{1000} e^{x - [x]}\,dx = 1000\int_{0}^{1} e^x\,dx = 1000(e - 1) \).

Teacher's Note:
1. On [0, 1), [x] = 0, so the integrand is just eˣ there.
2. Splitting into 1000 unit intervals shows each one contributes e − 1.

Answer: (c) n/2

Explanation:
1. x − [x] has period 1 and \( \int_{0}^{1} x\,dx = \frac{1}{2} \).
2. So \( \int_{0}^{n} \{x - [x]\}\,dx = n\cdot\frac{1}{2} = \frac{n}{2} \).

Teacher's Note:
1. The fractional part {x} forms a sawtooth graph with n identical triangles of area ½ on [0, n].
2. Do not integrate x and [x] separately unless you can sum [x] correctly; the period method is faster.

Answer: (d) \( 200\sqrt{2} \)

Explanation:
1. \( \sqrt{1 + \cos 2x} = \sqrt{2}\,|\cos x| \), with period π and \( \int_{0}^{\pi} |\cos x|\,dx = 2 \).
2. The interval has length 199π/2 + π/2 = 100π, i.e. 100 periods.
3. Value \( = 100\cdot\sqrt{2}\cdot 2 = 200\sqrt{2} \).

Teacher's Note:
1. The length of the interval, not its starting point, decides the number of periods.
2. √(cos²x) = |cos x|; the modulus is what keeps the answer non-zero.

Answer: (b) 100

Explanation:
1. \( \sin^4 x + \cos^4 x = 1 - \frac{1}{2}\sin^2 2x = \frac{3}{4} + \frac{1}{4}\cos 4x \), which has period π/2.
2. [0, 50π] contains 50π ÷ (π/2) = 100 periods.
3. So \( \int_{0}^{50\pi} = 100\int_{0}^{\pi/2} \left(\frac{3}{4} + \frac{1}{4}\cos 4x\right)dx \), giving k = 100.

Teacher's Note:
1. sin⁴x + cos⁴x = (sin²x + cos²x)² − 2sin²x cos²x is the quickest route to the simplified form.
2. sin⁴x + cos⁴x actually has period π/2 (not π), which doubles the count of periods.

Answer: (b) \( \frac{\pi}{2}\int_{0}^{\pi} f(\sin x)\,dx \)

Explanation:
1. Replace x by π − x: \( I = \int_{0}^{\pi} (\pi - x)f(\sin x)\,dx = \pi\int_{0}^{\pi} f(\sin x)\,dx - I \).
2. So \( I = \frac{\pi}{2}\int_{0}^{\pi} f(\sin x)\,dx \).

Teacher's Note:
1. This identity is used constantly: it removes the factor x from ∫₀^π x f(sin x) dx.
2. It works because sin(π − x) = sin x.

Answer: (b) \( \pi \)

Explanation:
1. By x → π − x, \( \int_{0}^{\pi} x f(\sin x)\,dx = \frac{\pi}{2}\int_{0}^{\pi} f(\sin x)\,dx \).
2. Since sin(π − x) = sin x, \( \int_{0}^{\pi} f(\sin x)\,dx = 2\int_{0}^{\pi/2} f(\sin x)\,dx \).
3. So the integral \( = \pi\int_{0}^{\pi/2} f(\sin x)\,dx \), giving A = π.

Teacher's Note:
1. Two properties combine: ∫₀^π x f(sin x) = (π/2)∫₀^π f(sin x) and ∫₀^π f(sin x) = 2∫₀^(π/2) f(sin x).
2. Check with f = 1: ∫₀^π x dx = π²/2 and A∫₀^(π/2) dx = Aπ/2, so A = π.

Answer: (a) \( \frac{2n}{n^2 - m^2} \)

Explanation:
1. \( \cos mx\sin nx = \frac{1}{2}[\sin(n + m)x + \sin(n - m)x] \).
2. For a non-zero integer k, \( \int_{0}^{\pi} \sin kx\,dx = \frac{1 - \cos k\pi}{k} = \frac{2}{k} \) when k is odd. Since n − m is odd, n + m is odd too.
3. So the integral \( = \frac{1}{2}\left[\frac{2}{n + m} + \frac{2}{n - m}\right] = \frac{1}{n + m} + \frac{1}{n - m} = \frac{2n}{n^2 - m^2} \).

Teacher's Note:
1. Convert products of sines and cosines into sums before integrating.
2. n + m and n − m differ by the even number 2m, so they are both odd or both even.

Answer: (d) \( \frac{\alpha}{2}(\sin \alpha)^{-1} \)

Explanation:
1. \( x^2 + 2x\cos\alpha + 1 = (x + \cos\alpha)^2 + \sin^2\alpha \).
2. \( \int_{0}^{1} = \frac{1}{\sin\alpha}\left[\tan^{-1}\frac{x + \cos\alpha}{\sin\alpha}\right]_{0}^{1} \).
3. At 1: \( \frac{1 + \cos\alpha}{\sin\alpha} = \cot\frac{\alpha}{2} \), giving \( \frac{\pi}{2} - \frac{\alpha}{2} \); at 0: cot α, giving \( \frac{\pi}{2} - \alpha \).
4. Value \( = \frac{1}{\sin\alpha}\cdot\frac{\alpha}{2} = \frac{\alpha}{2}(\sin\alpha)^{-1} \).

Teacher's Note:
1. Complete the square, then use half-angle forms like (1 + cos α)/sin α = cot(α/2).
2. tan⁻¹(cot θ) = π/2 − θ for 0 < θ < π; the condition 0 < α < π keeps this valid.

Answer: (b) \( F(t) = e^t - (1 + t) \)

Explanation:
1. \( F(t) = \int_{0}^{t} e^{t - y}\,y\,dy = e^t\int_{0}^{t} y e^{-y}\,dy \).
2. By parts, \( \int_{0}^{t} y e^{-y}\,dy = [-ye^{-y} - e^{-y}]_{0}^{t} = 1 - e^{-t}(1 + t) \).
3. So \( F(t) = e^t - (1 + t) \).

Teacher's Note:
1. e^(t − y) = eᵗ·e^(−y); the factor eᵗ is constant with respect to y and comes outside.
2. Check: F(0) = 0, and F(t) = eᵗ − 1 − t also gives F(0) = 0.

Answer: (d) None of these

Explanation:
1. f′ = f with f(0) = 1 gives f(x) = eˣ, so g(x) = x² − eˣ.
2. \( \int_{0}^{1} e^x(x^2 - e^x)\,dx = \int_{0}^{1} x^2 e^x\,dx - \int_{0}^{1} e^{2x}\,dx \).
3. \( \int_{0}^{1} x^2 e^x\,dx = [e^x(x^2 - 2x + 2)]_{0}^{1} = e - 2 \) and \( \int_{0}^{1} e^{2x}\,dx = \frac{e^2 - 1}{2} \).
4. Value \( = e - 2 - \frac{e^2 - 1}{2} = \frac{2e - e^2 - 3}{2} \approx -2.48 \), which matches none of (a), (b), (c).

Teacher's Note:
1. The differential equation f′ = f with f(0) = 1 has the unique solution eˣ.
2. For ∫xⁿeˣ, the tabular method gives eˣ(xⁿ − nxⁿ⁻¹ + …) quickly.

Answer: (a) π log 2

Explanation:
1. By parts with \( \int \mathrm{cosec}^2\theta\,d\theta = -\cot\theta \): \( \int_{0}^{\pi/2} \theta^2\mathrm{cosec}^2\theta\,d\theta = [-\theta^2\cot\theta]_{0}^{\pi/2} + 2\int_{0}^{\pi/2} \theta\cot\theta\,d\theta \).
2. The bracket is 0 at both ends (θ² cot θ → 0 as θ → 0).
3. Again by parts, \( \int_{0}^{\pi/2} \theta\cot\theta\,d\theta = [\theta\log\sin\theta]_{0}^{\pi/2} - \int_{0}^{\pi/2} \log\sin\theta\,d\theta = \frac{\pi}{2}\log 2 \).
4. So the value is \( 2\cdot\frac{\pi}{2}\log 2 = \pi\log 2 \).

Teacher's Note:
1. Two rounds of integration by parts reduce this to the standard result ∫₀^(π/2) log sin θ dθ = −(π/2) log 2.
2. Check the boundary terms carefully: θ log sin θ → 0 as θ → 0.

Answer: (c) \( \frac{4}{3} \)

Explanation:
1. On [0, π], sin θ ≥ 0, so |sin³θ| = sin³θ.
2. \( \int_{0}^{\pi} \sin^3\theta\,d\theta = \int_{0}^{\pi} (1 - \cos^2\theta)\sin\theta\,d\theta = \left[-\cos\theta + \frac{\cos^3\theta}{3}\right]_{0}^{\pi} \).
3. \( = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \frac{4}{3} \).

Teacher's Note:
1. Check the sign of the expression on the interval before dropping the modulus.
2. sin³θ = (3 sin θ − sin 3θ)/4 is another route: (3·2 − 2/3)/4 = 4/3.

Answer: (b) \( \frac{\pi}{2} \log 2 \)

Explanation:
1. By parts: \( \int_{0}^{\pi/2} x\cot x\,dx = [x\log\sin x]_{0}^{\pi/2} - \int_{0}^{\pi/2} \log\sin x\,dx \).
2. The bracket is 0 at π/2 (log 1 = 0) and tends to 0 as x → 0.
3. So the value is \( -\left(-\frac{\pi}{2}\log 2\right) = \frac{\pi}{2}\log 2 \).

Teacher's Note:
1. ∫ cot x dx = log sin x, so take x as the first function.
2. The integrand is positive on (0, π/2), so the negative options are impossible.

Answer: (b) 2

Explanation:
1. Try n = 2: \( \int e^x(x - 1)^2\,dx = e^x[(x - 1)^2 - 2(x - 1) + 2] \).
2. \( \left[e^x\{(x - 1)^2 - 2(x - 1) + 2\}\right]_{0}^{1} = 2e - (1 + 2 + 2) = 2e - 5 \).
3. This matches, so n = 2 (n = 1 gives 2 − e and n = 3 gives 16 − 6e, which do not).

Teacher's Note:
1. Use \( \int e^x p(x)\,dx = e^x[p - p' + p'' - \dots] \) for a polynomial p.
2. When the answer depends on n, testing the options one by one is often the fastest route.

Answer: (a) \( \frac{1}{n - 1} \)

Explanation:
1. For 0 ≤ x ≤ π/4, [x] = 0, so d(x − [x]) = dx.
2. \( \tan^n x + \tan^{n-2} x = \tan^{n-2} x\sec^2 x \).
3. With t = tan x: \( \int_{0}^{1} t^{n-2}\,dt = \frac{1}{n - 1} \).

Teacher's Note:
1. Simplify the differential first: on an interval with no integer, x − [x] differs from x by a constant.
2. tanⁿ⁻²x(1 + tan²x) = tanⁿ⁻²x sec²x is the key factorisation.

Answer: (a) \( \left(\frac{6}{5}, \frac{36}{25}\right) \)

Explanation:
1. \( f_1(x) = [t^2 - 5t]_{2}^{x} = x^2 - 5x + 6 \) and \( f_2(x) = x^2 \).
2. f₁ = f₂ gives −5x + 6 = 0, so x = 6/5.
3. Then \( y = \left(\frac{6}{5}\right)^2 = \frac{36}{25} \), so the point is \( \left(\frac{6}{5}, \frac{36}{25}\right) \).

Teacher's Note:
1. Evaluate the integrals first to turn them into ordinary functions of x.
2. Check by substituting: f₁(6/5) = 36/25 − 6 + 6 = 36/25.

Answer: (c) \( -\frac{3}{8} \)

Explanation:
1. f′(x) = |x| ≥ 0, so f is increasing and its greatest value on [−½, ½] is at x = ½.
2. \( f\left(\frac{1}{2}\right) = \int_{1}^{1/2} t\,dt = \frac{1}{2}\left(\frac{1}{4} - 1\right) = -\frac{3}{8} \).

Teacher's Note:
1. The derivative of ∫ₐˣ g(t) dt is g(x); its sign shows where the function increases.
2. The upper limit ½ is below the lower limit 1, so the value is negative.

Answer: (b) \( 1 - \frac{1}{\sqrt{2}} \)

Explanation:
1. For 0 ≤ x ≤ π/4 (< 1), [x] = 0, so sin(x − [x]) = sin x.
2. \( \int_{0}^{\pi/4} \sin x\,dx = [-\cos x]_{0}^{\pi/4} = 1 - \frac{1}{\sqrt{2}} \).

Teacher's Note:
1. π/4 ≈ 0.785 < 1, so the greatest integer part is zero throughout the interval.
2. Always locate the integers inside the interval before handling [x].

Answer: (b) 11

Explanation:
1. x − [x] = {x} lies in [0, 1); it is 0 only at integers and positive elsewhere.
2. So sgn(x − [x]) = 1 except at finitely many points, which do not affect the integral.
3. \( \int_{-1}^{10} 1\,dx = 11 \).

Teacher's Note:
1. Changing a function at finitely many points does not change its definite integral.
2. sgn(y) = 1 for y > 0, 0 for y = 0 and −1 for y < 0.

Answer: (b) \( \frac{2e^x}{3 - e} + \left(\frac{1 - e}{3 - e}\right) \)

Explanation:
1. Let \( c = \int_{0}^{1} f(x)\,dx \) (a constant). Then f′ = f + c gives f(x) = Aeˣ − c, and f(0) = 1 gives A = 1 + c.
2. \( c = \int_{0}^{1} (Ae^x - c)\,dx = A(e - 1) - c \), so (1 + c)(e − 1) = 2c, giving \( c = \frac{e - 1}{3 - e} \).
3. \( A = 1 + c = \frac{2}{3 - e} \), so \( f(x) = \frac{2e^x}{3 - e} + \frac{1 - e}{3 - e} \).

Teacher's Note:
1. A definite integral of the unknown function is just a constant; name it and solve for it at the end.
2. f′ − f = c is a linear equation whose solution is Aeˣ − c.