CBSE Class 12 Mathematics Integrals MCQs Set 11

Answer: (b) \( 3\sqrt{3} - 2\sqrt{2} - 1 \)

Explanation:
1. f′(x) = 6 cos x − 2 sin x. On [5π/3, 7π/4], cos x > 0 and sin x < 0, so f′ > 0 and f is increasing.
2. Greatest value \( = f\left(\frac{7\pi}{4}\right) = [6\sin t + 2\cos t]_{5\pi/3}^{7\pi/4} \).
3. \( = \left(-3\sqrt{2} + \sqrt{2}\right) - \left(-3\sqrt{3} + 1\right) = 3\sqrt{3} - 2\sqrt{2} - 1 \).

Teacher's Note:
1. The derivative of ∫ₐˣ g(t) dt is g(x); use its sign to locate the maximum.
2. In the fourth quadrant cosine is positive and sine is negative.

Answer: (c) \( -\frac{ka^4}{24} \)

Explanation:
1. Integrate by parts three times: \( \int_{0}^{a} f\,dx = [xf]_{0}^{a} - \int_{0}^{a} xf'\,dx \), then \( \int xf' = \left[\frac{x^2}{2}f'\right] - \int \frac{x^2}{2}f'' \), then \( \int \frac{x^2}{2}f'' = \left[\frac{x^3}{6}f''\right] - \int \frac{x^3}{6}f''' \).
2. So \( \int_{0}^{a} f\,dx = \left\{xf - \frac{x^2}{2!}f' + \frac{x^3}{3!}f''\right\}_{0}^{a} - \int_{0}^{a} \frac{x^3}{6}f'''\,dx \).
3. The required expression \( = -\int_{0}^{a} \frac{kx^3}{6}dx = -\frac{ka^4}{24} \).

Teacher's Note:
1. Repeated integration by parts with x as the 'second' function produces this alternating series.
2. Since f‴ = k is constant, the remaining integral is elementary.

Answer: (b) \( -\frac{3}{5}\left[\frac{2}{\log_e (2/5)} + \frac{1}{2\log_e (5/2)}\right] \)

Explanation:
1. \( \frac{2^{2x + 1} - 5^{2x - 1}}{10^x} = 2\left(\frac{2}{5}\right)^x - \frac{1}{5}\left(\frac{5}{2}\right)^x \).
2. \( \int_{0}^{1} = 2\cdot\frac{2/5 - 1}{\log(2/5)} - \frac{1}{5}\cdot\frac{5/2 - 1}{\log(5/2)} = -\frac{6}{5\log(2/5)} - \frac{3}{10\log(5/2)} \).
3. \( = -\frac{3}{5}\left[\frac{2}{\log_e(2/5)} + \frac{1}{2\log_e(5/2)}\right] \).

Teacher's Note:
1. Split exponentials over 10ˣ = 2ˣ5ˣ into single bases, then use \( \int a^x\,dx = \frac{a^x}{\log a} \).
2. Watch the signs: log(2/5) is negative, so each term needs care.

Answer: (d) Four

Explanation:
1. \( \int_{0}^{\pi/2} \left(\frac{\cos 3x}{4} + \frac{3}{4}\cos x\right)dx = \frac{1}{4}\left(-\frac{1}{3}\right) + \frac{3}{4} = \frac{2}{3} \), and ∫ sin x = ∫ cos x = 1.
2. The condition becomes \( \frac{2a^2}{3} + a - 20 \le -\frac{a^2}{3} \), i.e. \( a^2 + a - 20 \le 0 \).
3. (a + 5)(a − 4) ≤ 0 gives −5 ≤ a ≤ 4; positive integers 1, 2, 3, 4: four values.

Teacher's Note:
1. (cos 3x)/4 + (3/4) cos x = cos³x, whose integral over [0, π/2] is 2/3.
2. After evaluating the integral, the problem is just a quadratic inequality.

Answer: (d) n − 1

Explanation:
1. \( \int_{0}^{n} [x]\,dx = 0 + 1 + \dots + (n - 1) = \frac{n(n - 1)}{2} \).
2. \( \int_{0}^{n} \{x\}\,dx = n\cdot\frac{1}{2} = \frac{n}{2} \).
3. Ratio \( = \frac{n(n - 1)/2}{n/2} = n - 1 \).

Teacher's Note:
1. Check: [x] + {x} = x, and n(n − 1)/2 + n/2 = n²/2 = ∫₀ⁿ x dx.
2. Over each unit interval [k, k + 1], ∫[x] = k and ∫{x} = ½.

Answer: (c) 4

Explanation:
1. \( \int_{a}^{b} x^3\,dx = \frac{b^4 - a^4}{4} \).
2. Simpson's rule \( \frac{b - a}{6}\left[f(a) + 4f\left(\frac{a + b}{2}\right) + f(b)\right] \) is exact for cubics; check: with a = 0, b = 2, the left side is 4 and the right side is \( \frac{2}{6}[0 + 8 + k] \).
3. So 4 = (8 + k)/3, giving k = 4.

Teacher's Note:
1. Testing simple values of a and b is a quick way to find an unknown constant in an identity.
2. Simpson's rule gives exact answers for all polynomials of degree ≤ 3.

Answer: (c) \( \frac{4}{\pi} \) and 0

Explanation:
1. \( f'(x) = \frac{A\pi}{2}\cos\frac{\pi x}{2} \), so \( f'\left(\frac{1}{2}\right) = \frac{A\pi}{2}\cdot\frac{1}{\sqrt{2}} = \sqrt{2} \), giving \( A = \frac{4}{\pi} \).
2. \( \int_{0}^{1} f\,dx = \left[-\frac{2A}{\pi}\cos\frac{\pi x}{2}\right]_{0}^{1} + B = \frac{2A}{\pi} + B \).
3. Given this equals 2A/π, B = 0. So A = 4/π and B = 0.

Teacher's Note:
1. Use one condition for each unknown constant.
2. cos(π/4) = 1/√2 is needed for the derivative condition.

Answer: (b) \( \frac{1}{(a^2 - b^2)}\left[a \log 2 - 5a + \frac{7}{2}b\right] \)

Explanation:
1. Replace x by 1/x: \( a f\left(\frac{1}{x}\right) + b f(x) = x - 5 \).
2. Eliminate f(1/x): \( (a^2 - b^2)f(x) = a\left(\frac{1}{x} - 5\right) - b(x - 5) \).
3. \( \int_{1}^{2} f\,dx = \frac{1}{a^2 - b^2}\left[a(\log 2 - 5) - b\left(\frac{3}{2} - 5\right)\right] = \frac{1}{a^2 - b^2}\left[a\log 2 - 5a + \frac{7}{2}b\right] \).

Teacher's Note:
1. Functional equations in f(x) and f(1/x) are solved by replacing x with 1/x and solving the pair.
2. a ≠ b (and a ≠ −b) ensures a² − b² ≠ 0.

Answer: (c) \( \frac{5}{3} \)

Explanation:
1. Split at 1: \( \int_{0}^{1} x^2\cdot x\,dx + \int_{1}^{2} x^2(x - 1)\,dx \).
2. \( = \frac{1}{4} + \left[\frac{x^4}{4} - \frac{x^3}{3}\right]_{1}^{2} = \frac{1}{4} + \left(4 - \frac{8}{3}\right) - \left(\frac{1}{4} - \frac{1}{3}\right) \).
3. \( = \frac{1}{4} + \frac{17}{12} = \frac{5}{3} \).

Teacher's Note:
1. For a piecewise function, split the integral at the point where the rule changes.
2. The value of f exactly at x = 1 does not affect the integral.

Answer: (c) \( \frac{1}{6}\left\{f(0) + 4f\left(\frac{1}{2}\right) + f(1)\right\} \)

Explanation:
1. \( \int_{0}^{1} (a + bx + cx^2)\,dx = a + \frac{b}{2} + \frac{c}{3} \).
2. f(0) = a, \( f\left(\frac{1}{2}\right) = a + \frac{b}{2} + \frac{c}{4} \), f(1) = a + b + c.
3. \( \frac{1}{6}\left\{f(0) + 4f\left(\frac{1}{2}\right) + f(1)\right\} = \frac{1}{6}\{6a + 3b + 2c\} = a + \frac{b}{2} + \frac{c}{3} \), which matches.

Teacher's Note:
1. This is Simpson's rule, which is exact for polynomials of degree up to 3.
2. Check any option with f = 1: the weights must add up to 6, which (a), (b) and (d) all fail.

Answer: (d) 64

Explanation:
1. Put t = x³: dt = 3x² dx, so \( \frac{3}{x}dx = \frac{3x^2\,dx}{x^3} = \frac{dt}{t} \); x = 1 → t = 1, x = 4 → t = 64.
2. \( \int_{1}^{4} \frac{3}{x}e^{\sin x^3}\,dx = \int_{1}^{64} \frac{e^{\sin t}}{t}dt = f(64) - f(1) \).
3. So k = 64.

Teacher's Note:
1. Since f′(t) = e^(sin t)/t, any integral of that integrand equals a difference of f-values.
2. Multiplying and dividing by x² turns 3/x dx into dt/t.

Answer: (a) \( \frac{1}{2} \log \frac{5}{3} \)

Explanation:
1. Put t = √(x + 1): x = t² − 1, dx = 2t dt, x − 3 = t² − 4; t: 3 → 4.
2. \( \int_{3}^{4} \frac{2t\,dt}{(t^2 - 4)t} = 2\int_{3}^{4} \frac{dt}{t^2 - 4} = \frac{1}{2}\left[\log\frac{t - 2}{t + 2}\right]_{3}^{4} \).
3. \( = \frac{1}{2}\left(\log\frac{1}{3} - \log\frac{1}{5}\right) = \frac{1}{2}\log\frac{5}{3} \).

Teacher's Note:
1. For (linear) × √(linear) in the denominator, substitute t = √(linear).
2. The integrand is positive on [8, 15], so the answer must be positive; this rules out the log(3/5) options.

Answer: (c) \( \frac{1}{2} \)

Explanation:
1. \( \frac{d}{dx}\left[\frac{\sin x}{2 + \cos x}\right] = \frac{\cos x(2 + \cos x) + \sin^2 x}{(2 + \cos x)^2} = \frac{1 + 2\cos x}{(2 + \cos x)^2} \).
2. So the integral \( = \left[\frac{\sin x}{2 + \cos x}\right]_{0}^{\pi/2} = \frac{1}{2} - 0 = \frac{1}{2} \).

Teacher's Note:
1. A squared denominator often signals a quotient-rule derivative; try g/(2 + cos x) with g = sin x.
2. Using sin²x + cos²x = 1 in the numerator makes the match visible.

Answer: (a) \( \frac{1}{20} \log 3 \)

Explanation:
1. Put t = sin x − cos x: dt = (cos x + sin x)dx, sin 2x = 1 − t²; t: −1 → 0.
2. \( \int_{-1}^{0} \frac{dt}{25 - 16t^2} = \frac{1}{40}\left[\log\frac{5 + 4t}{5 - 4t}\right]_{-1}^{0} \).
3. \( = \frac{1}{40}\left(0 - \log\frac{1}{9}\right) = \frac{1}{40}\log 9 = \frac{1}{20}\log 3 \).

Teacher's Note:
1. sin x + cos x in the numerator with sin 2x in the denominator: use t = sin x − cos x.
2. \( \int \frac{dt}{a^2 - b^2t^2} = \frac{1}{2ab}\log\frac{a + bt}{a - bt} \) with a = 5, b = 4.

Answer: (d) 0

Explanation:
1. First integral: t = x + 5 gives \( \int_{1}^{0} e^{t^2}\,dt = -\int_{0}^{1} e^{t^2}\,dt \).
2. Second: t = 3x − 2, dt = 3dx; x = 1/3 → t = −1, x = 2/3 → t = 0: \( 3\int_{1/3}^{2/3} e^{9(x - 2/3)^2}dx = \int_{-1}^{0} e^{t^2}\,dt = \int_{0}^{1} e^{t^2}\,dt \) (even integrand).
3. The two parts cancel, so the value is 0.

Teacher's Note:
1. e^(t²) has no elementary antiderivative; reduce both integrals to the same form instead of evaluating.
2. Notice the first integral runs from −4 down to −5, which gives the minus sign.

Answer: (c) \( \pi/\sqrt{2} \)

Explanation:
1. \( \sqrt{\tan x} + \sqrt{\cot x} = \frac{\sin x + \cos x}{\sqrt{\sin x\cos x}} \).
2. Put t = sin x − cos x: dt = (cos x + sin x)dx and \( \sin x\cos x = \frac{1 - t^2}{2} \); t: −1 → 0.
3. \( \sqrt{2}\int_{-1}^{0} \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2}[\sin^{-1} t]_{-1}^{0} = \sqrt{2}\cdot\frac{\pi}{2} = \frac{\pi}{\sqrt{2}} \).

Teacher's Note:
1. Combine √tan x + √cot x over a common denominator √(sin x cos x).
2. 2 sin x cos x = 1 − (sin x − cos x)² links the denominator to the new variable.

Answer: (c) 0

Explanation:
1. Put t = x/2: \( 2\int_{0}^{\pi} e^t\sin\left(t + \frac{\pi}{4}\right)dt \).
2. \( \int e^t\sin(t + c)\,dt = \frac{e^t}{2}[\sin(t + c) - \cos(t + c)] \), and with c = π/4 the bracket is √2 sin t.
3. So the value \( = 2\left[\frac{\sqrt{2}}{2}e^t\sin t\right]_{0}^{\pi} = 0 \).

Teacher's Note:
1. Use \( \sin\left(t + \frac{\pi}{4}\right) - \cos\left(t + \frac{\pi}{4}\right) = \sqrt{2}\sin t \) to simplify the antiderivative.
2. Since sin t vanishes at both 0 and π, the answer is 0 without any further arithmetic.

Answer: (a) \( \frac{1}{3}\left[\log(\sqrt{2} + 1) + \frac{\pi}{2\sqrt{2}}\right] \)

Explanation:
1. \( \frac{\sec x}{1 + 2\sin^2 x} = \frac{\cos x}{(1 - \sin^2 x)(1 + 2\sin^2 x)} \); put t = sin x (t: 0 → 1/√2).
2. \( \frac{1}{(1 - t^2)(1 + 2t^2)} = \frac{1}{3}\left[\frac{1}{1 - t^2} + \frac{2}{1 + 2t^2}\right] \).
3. \( \frac{1}{3}\left[\frac{1}{2}\log\frac{1 + t}{1 - t} + \sqrt{2}\tan^{-1}(\sqrt{2}\,t)\right]_{0}^{1/\sqrt{2}} = \frac{1}{3}\left[\log(\sqrt{2} + 1) + \frac{\pi}{2\sqrt{2}}\right] \).

Teacher's Note:
1. Multiply sec x by cos x/cos x to create cos x dx for t = sin x.
2. At t = 1/√2: \( \frac{1 + t}{1 - t} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} = (\sqrt{2} + 1)^2 \), so half its log is log(√2 + 1).

Answer: (d) \( a_1 \) and \( a_3 \)

Explanation:
1. The sum is \( a_0\cos^3 x + a_1\cos^2 x\sin x + a_2\cos x\sin^2 x + a_3\sin^3 x \).
2. Under x → π − x, cos x changes sign, so terms with an odd power of cos x (a₀ and a₂ terms) integrate to 0 over [0, π].
3. \( \int_{0}^{\pi} \cos^2 x\sin x\,dx = \frac{2}{3} \) and \( \int_{0}^{\pi} \sin^3 x\,dx = \frac{4}{3} \), so the value is \( \frac{2a_1}{3} + \frac{4a_3}{3} \): it depends on a₁ and a₃.

Teacher's Note:
1. ∫₀^π cosᵐx sinⁿx dx = 0 whenever m is odd.
2. Write out the sum term by term before applying symmetry.

Answer: (d) None of these

Explanation:
1. Rationalise: \( \frac{1}{\sqrt{x + 1} + \sqrt{5x + 1}} = \frac{\sqrt{5x + 1} - \sqrt{x + 1}}{4x} \).
2. Using \( \int \frac{\sqrt{ax + 1}}{x}dx = 2u + \log\frac{u - 1}{u + 1} \) with u = √(ax + 1), the integral from 0 to 3 is \( \frac{1}{4}\left[4 + \log\frac{9}{25}\right] \).
3. So the value \( = 1 - \frac{1}{2}\log\frac{5}{3} \approx 0.745 \), which is not 11/15, 14/15 or 2/5.

Teacher's Note:
1. When rationalising gives an x in the denominator, substitute u = √(ax + 1) in each part.
2. A rough check helps: the integrand falls from ½ to 1/6 on [0, 3], so the value lies between 0.5 and 1.5.

Answer: (b) \( 3\pi/16\sqrt{2} \)

Explanation:
1. \( \cos 2\theta = 1 - 2\sin^2\theta \); put t = √2 sin θ, dt = √2 cos θ dθ; t: 0 → 1.
2. \( \frac{1}{\sqrt{2}}\int_{0}^{1} (1 - t^2)^{3/2}\,dt \); with t = sin φ: \( \frac{1}{\sqrt{2}}\int_{0}^{\pi/2} \cos^4\varphi\,d\varphi \).
3. \( \int_{0}^{\pi/2} \cos^4\varphi\,d\varphi = \frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3\pi}{16} \), so the value is \( \frac{3\pi}{16\sqrt{2}} \).

Teacher's Note:
1. Express cos 2θ in terms of sin θ so that cos θ dθ becomes the differential.
2. ∫₀^(π/2) cos⁴φ dφ = 3π/16 follows from cos⁴φ = (3 + 4cos 2φ + cos 4φ)/8.

Answer: (c) \( \frac{\pi}{8}(\beta - \alpha)^2 \)

Explanation:
1. Put x = α cos²θ + β sin²θ: x − α = (β − α)sin²θ, β − x = (β − α)cos²θ, dx = 2(β − α) sin θ cos θ dθ; θ: 0 → π/2.
2. \( \int_{0}^{\pi/2} (\beta - \alpha)\sin\theta\cos\theta\cdot 2(\beta - \alpha)\sin\theta\cos\theta\,d\theta = \frac{(\beta - \alpha)^2}{2}\int_{0}^{\pi/2} \sin^2 2\theta\,d\theta \).
3. \( = \frac{(\beta - \alpha)^2}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8}(\beta - \alpha)^2 \).

Teacher's Note:
1. Geometrically, y = √((x − α)(β − x)) is a semicircle with diameter β − α, whose area is π(β − α)²/8.
2. The substitution x = α cos²θ + β sin²θ is the standard one for (x − α)(β − x).

Answer: (a) \( I_1 = I_2 \)

Explanation:
1. In I₁ put x = tan θ: \( I_1 = \int_{0}^{\pi/4} \frac{\theta\sec^2\theta}{\tan\theta}d\theta = \int_{0}^{\pi/4} \frac{\theta}{\sin\theta\cos\theta}d\theta = \int_{0}^{\pi/4} \frac{2\theta}{\sin 2\theta}d\theta \).
2. Put t = 2θ: \( I_1 = \int_{0}^{\pi/2} \frac{t}{\sin t}\cdot\frac{dt}{2} = I_2 \).

Teacher's Note:
1. sec²θ/tan θ = 1/(sin θ cos θ) = 2/sin 2θ is the simplification that links the two integrals.
2. Neither integral has an elementary value, so showing equality by substitution is the intended route.

Answer: (c) \( \frac{1}{2}\left(\log 2 + \frac{\pi}{2} - 1\right) \)

Explanation:
1. By parts with u = log(√(1 + x) + √(1 − x)): \( I = [x\,u]_{0}^{1} - \int_{0}^{1} x\,u'\,dx \), and the bracket is \( \log\sqrt{2} = \frac{1}{2}\log 2 \).
2. Simplifying, \( x\,u' = -\frac{1}{2}\left(\frac{1}{\sqrt{1 - x^2}} - 1\right) \).
3. \( \int_{0}^{1} x u'\,dx = -\frac{1}{2}\left(\frac{\pi}{2} - 1\right) \), so \( I = \frac{1}{2}\log 2 + \frac{1}{2}\left(\frac{\pi}{2} - 1\right) = \frac{1}{2}\left(\log 2 + \frac{\pi}{2} - 1\right) \).

Teacher's Note:
1. For log of an algebraic expression, integrate by parts with 1 as the second function.
2. Rationalising (√(1 − x) − √(1 + x))/(√(1 + x) + √(1 − x)) gives −(1 − √(1 − x²))/x, which simplifies x·u′.

Answer: (b) 16

Explanation:
1. Put t = x²: dt = 2x dx, so \( \frac{2}{x}dx = \frac{2x\,dx}{x^2} = \frac{dt}{t} \); x = 1 → t = 1, x = 4 → t = 16.
2. \( \int_{1}^{4} \frac{2e^{\sin x^2}}{x}dx = \int_{1}^{16} \frac{e^{\sin t}}{t}dt = f(16) - f(1) \), so k = 16.

Teacher's Note:
1. The integrand e^(sin t)/t is exactly f′(t), so the integral is a difference of f-values.
2. Changing the limits (4 becomes 16) is the whole point of the question.

Answer: (b) \( \pi/4 \)

Explanation:
1. Put x = cos 2θ: \( \sqrt{\frac{1 + x}{1 - x}} = \sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}} = \cot\theta \), so \( 2\tan^{-1}(\cot\theta) = 2\left(\frac{\pi}{2} - \theta\right) = \pi - 2\theta \).
2. Then \( \sin(\pi - 2\theta) = \sin 2\theta = \sqrt{1 - x^2} \).
3. \( \int_{0}^{1} \sqrt{1 - x^2}\,dx = \frac{\pi}{4} \) (a quarter of the unit circle).

Teacher's Note:
1. Simplify an inverse trigonometric expression with a suitable substitution before integrating.
2. ∫₀¹ √(1 − x²) dx is the area of a quarter circle of radius 1.

Answer: (c) \( \frac{\pi}{1 - a^2} \)

Explanation:
1. Put t = tan(x/2): \( 1 - 2a\cos x + a^2 = \frac{(1 - a)^2 + (1 + a)^2 t^2}{1 + t^2} \), dx = 2dt/(1 + t²); t: 0 → ∞.
2. \( \int_{0}^{\infty} \frac{2\,dt}{(1 - a)^2 + (1 + a)^2 t^2} = \frac{2}{(1 + a)(1 - a)}\left[\tan^{-1}\frac{(1 + a)t}{1 - a}\right]_{0}^{\infty} \).
3. \( = \frac{2}{1 - a^2}\cdot\frac{\pi}{2} = \frac{\pi}{1 - a^2} \).

Teacher's Note:
1. |a| < 1 makes both 1 − a and 1 + a positive, so the arctangent goes to π/2.
2. Check a = 0: the integral is ∫₀^π dx = π, matching π/(1 − 0).

Answer: (b) 2

Explanation:
1. |x − 3| ranges over [0, 2] on [1, 5]; [|x − 3|] = 1 when 1 ≤ |x − 3| < 2 and 0 when |x − 3| < 1.
2. So the integrand is 1 on (1, 2] and [4, 5), and 0 on (2, 4).
3. Value = 1 × 1 + 1 × 1 = 2.

Teacher's Note:
1. Find where the inner expression crosses integers; those are the break points.
2. The graph is symmetric about x = 3, so you can compute ∫₃⁵ and double it.

Answer: (d) 4

Explanation:
1. |[x]| = 2 on [−2, −1), 1 on [−1, 0), 0 on [0, 1), 1 on [1, 2).
2. Each piece has length 1, so the value is 2 + 1 + 0 + 1 = 4.

Teacher's Note:
1. Take the greatest integer first, then the modulus.
2. Do not use symmetry here: |[x]| is not an even function ([−0.5] = −1 but [0.5] = 0).

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

Explanation:
1. On [π/2, 3π/2], 2 sin x falls from 2 to −2: [2 sin x] = 1 on (π/2, 5π/6), 0 on [5π/6, π], −1 on (π, 7π/6], −2 on (7π/6, 3π/2).
2. Value \( = 1\cdot\frac{\pi}{3} + 0 - 1\cdot\frac{\pi}{6} - 2\cdot\frac{\pi}{3} \).
3. \( = \frac{\pi}{3} - \frac{\pi}{6} - \frac{2\pi}{3} = -\frac{\pi}{2} \).

Teacher's Note:
1. Solve 2 sin x = 1, 0, −1 in the interval to locate the jumps of the greatest integer.
2. sin x = ½ at 5π/6 and sin x = −½ at 7π/6 in this interval.

Answer: (c) \( -\frac{5\pi}{3} \)

Explanation:
1. On (π, 2π), sin x < 0. [2 sin x] = −1 when −1 ≤ 2 sin x < 0, i.e. on (π, 7π/6] and [11π/6, 2π).
2. [2 sin x] = −2 when 2 sin x < −1, i.e. on (7π/6, 11π/6).
3. Value \( = -1\cdot\frac{\pi}{6} - 2\cdot\frac{2\pi}{3} - 1\cdot\frac{\pi}{6} = -\frac{\pi}{3} - \frac{4\pi}{3} = -\frac{5\pi}{3} \).

Teacher's Note:
1. sin x = −½ at 7π/6 and 11π/6; these are the break points.
2. For negative values, [y] is the integer below y, e.g. [−0.3] = −1.

Answer: (d) None of these

Explanation:
1. f(x) is the distance from x to the nearest of −1, 0, 1: f = x + 1 on [−1, −½], −x on [−½, 0], x on [0, ½], 1 − x on [½, 1].
2. The graph forms four triangles, each with base ½ and height ½, area \( \frac{1}{8} \).
3. Total \( = 4\cdot\frac{1}{8} = \frac{1}{2} \), which is not 1, 0 or 2.

Teacher's Note:
1. Sketch |x − 1|, |x| and |x + 1| together; the minimum is the lowest graph at each point.
2. The switch points ±½ are midway between neighbouring zeros.

Answer: (a) \( \frac{5 - \sqrt{5}}{2} \)

Explanation:
1. x² − x + 1 ≥ 3/4 on [0, 2]; it equals 1 at x = 0, 1 and 2 at x = φ = (1 + √5)/2.
2. So [x² − x + 1] = 0 on (0, 1), 1 on [1, φ), 2 on [φ, 2), and the integral is \( 0 + (\varphi - 1) + 2(2 - \varphi) = 3 - \varphi \).
3. \( = 3 - \frac{1 + \sqrt{5}}{2} = \frac{5 - \sqrt{5}}{2} \).

Teacher's Note:
1. Solve x² − x + 1 = 1, 2, 3 to find where the greatest integer jumps; x² − x + 1 = 3 gives x = 2.
2. On (0, 1) the quadratic dips below 1, so its integer part is 0 there.

Answer: (b) 1

Explanation:
1. For non-integer x, −x − [−x] = 1 − (x − [x]), so the integrand is min({x}, 1 − {x}).
2. This is a triangular wave of period 1 with height ½: each period contributes ½ × 1 × ½ = ¼.
3. [−2, 2] contains 4 periods, so the value is 4 × ¼ = 1.

Teacher's Note:
1. {−x} = 1 − {x} for non-integer x; at integers both are 0.
2. Reduce to one period and multiply by the number of periods.

Answer: (b) At least one root in (0, 2)

Explanation:
1. The condition gives \( \int_{1}^{2} (1 + \cos^8 x)(ax^2 + bx + c)\,dx = 0 \).
2. Since 1 + cos⁸x > 0, the quadratic ax² + bx + c cannot keep one sign throughout (1, 2); otherwise the integral would be non-zero.
3. So ax² + bx + c changes sign in (1, 2) and has a root there, i.e. at least one root in (0, 2).

Teacher's Note:
1. A continuous function with zero integral over an interval (with a positive weight) must vanish somewhere inside it.
2. The question says nothing about a double root; only existence of a root follows.

Answer: (a) Question is true

Explanation:
1. x → π/2 − x leaves sin 2x unchanged and swaps sin x and cos x, so the first two integrals are equal; call each I.
2. Adding: \( 2I = \int_{0}^{\pi/2} f(\sin 2x)(\sin x + \cos x)\,dx = \sqrt{2}\int_{0}^{\pi/2} f(\sin 2x)\cos\left(x - \frac{\pi}{4}\right)dx \).
3. Put u = π/4 − x: sin 2x = cos 2u and the integrand is even in u on [−π/4, π/4], so \( 2I = 2\sqrt{2}\int_{0}^{\pi/4} f(\cos 2u)\cos u\,du \).
4. Hence both equalities hold, and the statement is true.

Teacher's Note:
1. sin x + cos x = √2 cos(x − π/4) is the identity that brings in the factor √2.
2. Shifting the variable to the midpoint π/4 makes the symmetry of the integrand visible.

Answer: (a) g(x) + g(π)

Explanation:
1. \( g(x + \pi) = \int_{0}^{\pi} \cos^4 t\,dt + \int_{\pi}^{x + \pi} \cos^4 t\,dt \).
2. cos⁴t has period π, so \( \int_{\pi}^{x + \pi} \cos^4 t\,dt = \int_{0}^{x} \cos^4 t\,dt = g(x) \).
3. So g(x + π) = g(π) + g(x).

Teacher's Note:
1. For an integrand of period T, the integral over [T, x + T] equals the integral over [0, x].
2. cos⁴(t + π) = (−cos t)⁴ = cos⁴t, confirming the period π.

Answer: (d) \( \frac{\pi^2}{2ab} \)

Explanation:
1. Using x → π − x: \( I = \frac{\pi}{2}\int_{0}^{\pi} \frac{dx}{a^2\cos^2 x + b^2\sin^2 x} \) (the integrand depends only on cos² and sin²).
2. By symmetry about π/2, \( \int_{0}^{\pi} = 2\int_{0}^{\pi/2} = 2\cdot\frac{\pi}{2ab} = \frac{\pi}{ab} \).
3. So \( I = \frac{\pi}{2}\cdot\frac{\pi}{ab} = \frac{\pi^2}{2ab} \).

Teacher's Note:
1. ∫₀^π x f(x) dx = (π/2)∫₀^π f(x) dx whenever f(π − x) = f(x).
2. ∫₀^(π/2) dx/(a²cos²x + b²sin²x) = π/(2ab) is a standard result (divide by cos²x, t = tan x).

Answer: (c) \( \frac{1}{n + 1} - \frac{1}{n + 2} \)

Explanation:
1. Replace x by 1 − x: \( I = \int_{0}^{1} (1 - x)x^n\,dx = \int_{0}^{1} (x^n - x^{n + 1})\,dx \).
2. \( = \frac{1}{n + 1} - \frac{1}{n + 2} \).

Teacher's Note:
1. ∫₀¹ f(x) dx = ∫₀¹ f(1 − x) dx moves the power onto x, where integration is trivial.
2. The value simplifies to \( \frac{1}{(n + 1)(n + 2)} \).

Answer: (b) \( \pi^2 \)

Explanation:
1. Split: \( \frac{2x}{1 + \cos^2 x} \) is odd (integral 0), and \( \frac{2x\sin x}{1 + \cos^2 x} \) is even.
2. So \( I = 2\int_{0}^{\pi} \frac{2x\sin x}{1 + \cos^2 x}dx = 4\cdot\frac{\pi}{2}\int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x}dx \) (by x → π − x).
3. With t = cos x, \( \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x}dx = \int_{-1}^{1} \frac{dt}{1 + t^2} = \frac{\pi}{2} \), so \( I = 2\pi\cdot\frac{\pi}{2} = \pi^2 \).

Teacher's Note:
1. Separate the odd and even parts first; then use ∫₀^π x f(sin x) dx = (π/2)∫₀^π f(sin x) dx.
2. x sin x is even (odd × odd), which is why that part survives.

Answer: (a) 0

Explanation:
1. \( \sin mx\sin nx = \frac{1}{2}[\cos(m - n)x - \cos(m + n)x] \).
2. Since |m| ≠ |n|, both m − n and m + n are non-zero integers, and \( \int_{-\pi}^{\pi} \cos kx\,dx = \frac{2\sin k\pi}{k} = 0 \) for every non-zero integer k.
3. So the integral is 0.

Teacher's Note:
1. This is the orthogonality of sines on [−π, π], the basis of Fourier series.
2. If m = −n ≠ 0, the integrand becomes −sin²nx and the integral is −π; that is why |m| ≠ |n| is needed.

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

Explanation:
1. Replace x by −x: \( I = \int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^{-x}}dx = \int_{-\pi}^{\pi} \frac{a^x\cos^2 x}{a^x + 1}dx \).
2. Adding: \( 2I = \int_{-\pi}^{\pi} \cos^2 x\,dx = \pi \).
3. So \( I = \frac{\pi}{2} \).

Teacher's Note:
1. For ∫₋ₐᵃ g(x)/(1 + bˣ) dx with g even, the answer is ∫₀ᵃ g(x) dx.
2. 1/(1 + aˣ) + 1/(1 + a⁻ˣ) = 1 is the identity behind the method.