CBSE Class 11 Mathematics Logarithms Competitive Exam MCQs Set 03

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

Explanation:
1. \( \sqrt{x^2 + 1} = a - x \Rightarrow x^2 + 1 = a^2 - 2ax + x^2 \).
2. \( 2ax = a^2 - 1 \Rightarrow x = \frac{1}{2}\left(a - \frac{1}{a}\right) \).

Teacher's Note:
1. Isolate the square root and then square.
2. Alternatively, √(x² + 1) − x = 1/a, and subtracting gives 2x = a − 1/a.

Answer: (b) 368

Explanation:
1. \( y = \frac{4}{\sqrt{7} + \sqrt{3}} = \sqrt{7} - \sqrt{3} \).
2. \( x^2 + y^2 = 2(7 + 3) = 20 \), \( x^2y^2 = 16 \).
3. \( x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 = 400 - 32 = 368 \).

Teacher's Note:
1. x and y are conjugates, which makes symmetric expressions easy.
2. Use x⁴ + y⁴ = (x² + y²)² − 2x²y².

Answer: (a) \( \sqrt{2} \)

Explanation:
1. \( y = 2 - \sqrt{3} \); \( \sqrt{x} = \frac{\sqrt{3} + 1}{\sqrt{2}},\ \sqrt{y} = \frac{\sqrt{3} - 1}{\sqrt{2}} \).
2. First term \( = \frac{(2 + \sqrt{3})\sqrt{2}}{3 + \sqrt{3}} = \frac{\sqrt{2}(2 + \sqrt{3})(3 - \sqrt{3})}{6} = \frac{\sqrt{2}(3 + \sqrt{3})}{6} \).
3. Second term \( = \frac{(2 - \sqrt{3})\sqrt{2}}{3 - \sqrt{3}} = \frac{\sqrt{2}(2 - \sqrt{3})(3 + \sqrt{3})}{6} = \frac{\sqrt{2}(3 - \sqrt{3})}{6} \).
4. Sum \( = \frac{6\sqrt{2}}{6} = \sqrt{2} \).

Teacher's Note:
1. Find √x and √y as simple surds first.
2. The √3 parts cancel when the two terms are added.

Answer: (d) \( 1/\sqrt{5} \)

Explanation:
1. \( \sqrt{x} = \sqrt{3 - \sqrt{5}} = \frac{\sqrt{5} - 1}{\sqrt{2}} \).
2. \( 3x - 2 = 7 - 3\sqrt{5} \Rightarrow \sqrt{3x - 2} = \frac{3 - \sqrt{5}}{\sqrt{2}} \); so the denominator \( = \frac{2 + 3 - \sqrt{5}}{\sqrt{2}} = \frac{5 - \sqrt{5}}{\sqrt{2}} \).
3. Value \( = \frac{\sqrt{5} - 1}{5 - \sqrt{5}} = \frac{\sqrt{5} - 1}{\sqrt{5}(\sqrt{5} - 1)} = \frac{1}{\sqrt{5}} \).

Teacher's Note:
1. Write each root as (p − q)/√2 so that the √2 factors cancel.
2. 5 − √5 = √5(√5 − 1) allows a neat cancellation.

Answer: (d) a < b

Explanation:
1. \( a = \frac{1}{\sqrt{21} + \sqrt{20}},\ b = \frac{1}{\sqrt{18} + \sqrt{17}} \).
2. \( \sqrt{21} + \sqrt{20} > \sqrt{18} + \sqrt{17} \Rightarrow a < b \).

Teacher's Note:
1. Rationalising a difference of roots turns it into a reciprocal of a sum.
2. A bigger denominator gives a smaller fraction.

Answer: (b) 6

Explanation:
1. \( \sqrt{x + 10} = 6 - \sqrt{x - 2} \Rightarrow x + 10 = 36 - 12\sqrt{x - 2} + x - 2 \).
2. \( 12\sqrt{x - 2} = 24 \Rightarrow \sqrt{x - 2} = 2 \Rightarrow x = 6 \).
3. Check: √16 + √4 = 4 + 2 = 6 ✓.

Teacher's Note:
1. Isolate one root before squaring so that only one square root remains.
2. Always verify the answer in the original equation.

Answer: (a) \( u_{n+1} = u_n + u_{n-1} \); (b) \( u_{2n-1} = u_n^2 + u_{n-1}^2 \); (c) \( u_{n+2} = u_n + u_{n+1} \)

Explanation:
1. Let \( \alpha = \frac{1 + \sqrt{5}}{2},\ \beta = \frac{1 - \sqrt{5}}{2} \); they are roots of t² = t + 1, so \( \alpha^{n+1} = \alpha^n + \alpha^{n-1} \) and similarly for β. Hence (a), and (c) is the same relation shifted.
2. αβ = −1, so \( u_n^2 + u_{n-1}^2 = \frac{\alpha^{2n} + \alpha^{2n-2} + \beta^{2n} + \beta^{2n-2}}{5} \) (the cross terms cancel).
3. \( \alpha^{2n} + \alpha^{2n-2} = \alpha^{2n-1}(\alpha + \alpha^{-1}) = \sqrt{5}\,\alpha^{2n-1} \) and \( \beta^{2n} + \beta^{2n-2} = -\sqrt{5}\,\beta^{2n-1} \), giving \( u_{2n-1} \). So (b) holds too.

Teacher's Note:
1. u_n gives the Fibonacci numbers 0, 1, 1, 2, 3, 5, 8, …
2. Check with numbers: u₅ = 5 = u₃² + u₂² = 4 + 1.

Answer: (a) 1

Explanation:
1. \( 6 + 2\sqrt{3} + 2\sqrt{2} + 2\sqrt{6} = (1 + \sqrt{2} + \sqrt{3})^2 \).
2. \( \sqrt{5 - 2\sqrt{6}} = \sqrt{3} - \sqrt{2} \Rightarrow \frac{1}{\sqrt{3} - \sqrt{2}} = \sqrt{3} + \sqrt{2} \).
3. Value \( = 1 + \sqrt{2} + \sqrt{3} - \sqrt{3} - \sqrt{2} = 1 \).

Teacher's Note:
1. (p + q + r)² = p² + q² + r² + 2pq + 2qr + 2rp; here 1 + 2 + 3 = 6.
2. The reciprocal of √3 − √2 is √3 + √2.

Answer: (a) 2, if 1 ≤ x ≤ 2; (d) \( 2\sqrt{x - 1} \), if x > 2

Explanation:
1. \( x \pm 2\sqrt{x - 1} = (\sqrt{x - 1} \pm 1)^2 \).
2. Expression \( = (\sqrt{x - 1} + 1) + |\sqrt{x - 1} - 1| \).
3. For 1 ≤ x ≤ 2, √(x − 1) ≤ 1, so it equals 2; for x > 2 it equals \( 2\sqrt{x - 1} \). Options (a) and (d).

Teacher's Note:
1. √(t²) = |t|, so the sign of √(x − 1) − 1 decides the result.
2. The changeover happens at x = 2, where √(x − 1) = 1.

Answer: (c) 7

Explanation:
1. By the remainder theorem, the remainder is f(1).
2. f(1) = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7.

Teacher's Note:
1. Remainder on division by x − a is f(a).
2. Count the terms carefully: there are seven.

Answer: (b) \( \frac{(-1)^r}{r!(n - r)!} \)

Explanation:
1. Cover-up rule: \( A_r = \frac{1}{\prod_{k \ne r}(k - r)} \) (put x = −r in the other factors).
2. Factors k = 0, …, r − 1 give \( (-r)(-r + 1) \cdots (-1) = (-1)^r r! \); factors k = r + 1, …, n give (n − r)!.
3. \( A_r = \frac{(-1)^r}{r!(n - r)!} \).

Teacher's Note:
1. For non-repeated linear factors, the constant over (x + r) is found by putting x = −r in the rest of the fraction.
2. Check with n = 1: 1/(x(x + 1)) = 1/x − 1/(x + 1).

Answer: (c) \( \frac{1}{x - 1} - \frac{3}{x - 2} + \frac{2}{x - 3} \)

Explanation:
1. At x = 1: \( \frac{2}{(-1)(-2)} = 1 \); at x = 2: \( \frac{3}{(1)(-1)} = -3 \); at x = 3: \( \frac{4}{(2)(1)} = 2 \).
2. \( \frac{x + 1}{(x - 1)(x - 2)(x - 3)} = \frac{1}{x - 1} - \frac{3}{x - 2} + \frac{2}{x - 3} \).

Teacher's Note:
1. Use the cover-up rule for distinct linear factors.
2. Check: the constants must add to 0 since the numerator has lower degree by 2.

Answer: (a) a = 5; (c) c = 22

Explanation:
1. \( ax^2 + bx + c = 3(x + 2)(2x + 3) + 2(x - 1)(2x + 3) - 5(x - 1)(x + 2) \).
2. = (6x² + 21x + 18) + (4x² + 2x − 6) − (5x² + 5x − 10) = 5x² + 18x + 22.
3. So a = 5, b = 18, c = 22: options (a) and (c).

Teacher's Note:
1. Combine the right side over the common denominator and compare numerators.
2. Watch the sign of b: it is +18, not −18.

Answer: (d) 7

Explanation:
1. Put \( t = e^x \): \( t + 2 = -3(2t - 3) + B(t - 1) \).
2. Put t = 3/2: \( \frac{7}{2} = B \cdot \frac{1}{2} \Rightarrow B = 7 \).

Teacher's Note:
1. Substituting t = eˣ turns it into an ordinary rational function.
2. Choose t to make the other bracket zero.

Answer: (b) (10, 7)

Explanation:
1. 3x + 4 = A(x − 1) − B(x − 2).
2. x = 2: 10 = A. x = 1: 7 = B.
3. (A, B) = (10, 7).

Teacher's Note:
1. Note the minus sign in front of B in the given form.
2. Cover-up: A = (6 + 4)/(2 − 1) = 10.

Answer: (a) \( 2x^2 - 3x + 1 \)

Explanation:
1. The remainder R(x) is of degree ≤ 2 with R(−1) = 6, R(2) = 3, R(−2) = 15.
2. Check \( R(x) = 2x^2 - 3x + 1 \): R(−1) = 2 + 3 + 1 = 6, R(2) = 8 − 6 + 1 = 3, R(−2) = 8 + 6 + 1 = 15.
3. A quadratic is fixed by three values, so R(x) = 2x² − 3x + 1.

Teacher's Note:
1. f(x) and its remainder R(x) take the same values at the zeros of the divisor.
2. Testing options at the three points is quickest.

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

Explanation:
1. Let t = cos x: \( \frac{1 - t}{t(1 + t)} = \frac{A}{t} - \frac{2}{1 + t} \).
2. 1 − t = A(1 + t) − 2t; t = 0 gives A = 1.
3. sin α = 1 ⇒ \( \alpha = \frac{\pi}{2} \).

Teacher's Note:
1. Treat cos x as a single variable and resolve into partial fractions.
2. sin α = 1 gives α = π/2 among the options.

Answer: (d) \( \frac{1}{a^2 - b^2} \)

Explanation:
1. RHS \( = K \cdot \frac{a^2(x^2 + b^2) - b^2(x^2 + a^2)}{(x^2 + a^2)(x^2 + b^2)} = K \cdot \frac{(a^2 - b^2)x^2}{(x^2 + a^2)(x^2 + b^2)} \).
2. Comparing with LHS: \( K(a^2 - b^2) = 1 \Rightarrow K = \frac{1}{a^2 - b^2} \).

Teacher's Note:
1. Combining the right side is quicker than splitting the left side.
2. The a²b² terms cancel in the numerator.

Answer: (c) 5

Explanation:
1. 9 = A(x + 2)² + B(x − 1)(x + 2) + C(x − 1).
2. x = 1: 9 = 9A ⇒ A = 1. x = −2: 9 = −3C ⇒ C = −3.
3. Coefficient of x²: 0 = A + B ⇒ B = −1. A − B − C = 1 + 1 + 3 = 5.

Teacher's Note:
1. Find A and C by substitution, then B by comparing the highest power.
2. For a repeated factor, two constants are needed.

Answer: (b) b = 1; (c) a = 3

Explanation:
1. ax + b = (3x + 4) − 3 = 3x + 1.
2. So a = 3 and b = 1: options (b) and (c).

Teacher's Note:
1. Multiply both sides by (3x + 4)² and compare coefficients.
2. Two options can be correct at the same time.

Answer: (a) \( \frac{1}{x + 3} - \frac{1}{2x + 3} + \frac{5}{(x + 3)^2} \)

Explanation:
1. x² + 13x + 15 = A(x + 3)² + B(2x + 3)(x + 3) + C(2x + 3).
2. x = −3/2: \( \frac{9}{4} - \frac{39}{2} + 15 = -\frac{9}{4} = \frac{9}{4}A \Rightarrow A = -1 \). x = −3: 9 − 39 + 15 = −15 = −3C ⇒ C = 5.
3. x²: 1 = A + 2B ⇒ B = 1. Result: 1/(x + 3) − 1/(2x + 3) + 5/(x + 3)².

Teacher's Note:
1. Substitute the zeros of the linear factors first, then compare the x² coefficient.
2. Double-check by putting x = 0: 15/27 = 5/9 on both sides.

Answer: (b) \( \frac{3}{x - 1} + \frac{1}{(x - 1)^2} - \frac{7}{(x - 1)^3} + \frac{5}{(x - 1)^4} \)

Explanation:
1. Put x − 1 = y: numerator \( = 3(y + 1)^3 - 8(y + 1)^2 + 10 = 3y^3 + y^2 - 7y + 5 \).
2. Divide by y⁴: \( \frac{3}{y} + \frac{1}{y^2} - \frac{7}{y^3} + \frac{5}{y^4} \).
3. So the result is \( \frac{3}{x - 1} + \frac{1}{(x - 1)^2} - \frac{7}{(x - 1)^3} + \frac{5}{(x - 1)^4} \).

Teacher's Note:
1. For a single repeated factor (x − a)ⁿ, substitute y = x − a and expand.
2. Each power of y in the numerator gives one partial fraction.

Answer: (a) \( \frac{1}{x^2 + 1} + \frac{22}{(x^2 + 1)^2} + \frac{5}{(x^2 + 1)^3} \)

Explanation:
1. Put y = x² + 1: numerator \( = (y - 1)^2 + 24(y - 1) + 28 = y^2 + 22y + 5 \).
2. \( \frac{y^2 + 22y + 5}{y^3} = \frac{1}{y} + \frac{22}{y^2} + \frac{5}{y^3} \).

Teacher's Note:
1. Since only x² appears, treat x² + 1 as a single variable.
2. This avoids solving for six unknowns.

Answer: (b) A = 1, B = 0, C = −2

Explanation:
1. (x − 1)² = A(x² + 1) + (Bx + C)x.
2. x = 0: A = 1.
3. x²: 1 = A + B ⇒ B = 0; x: −2 = C.

Teacher's Note:
1. Put x = 0 to get A, then compare coefficients for B and C.
2. x³ + x = x(x² + 1), and x² + 1 is irreducible.

Answer: (d) A ≠ B ≠ C

Explanation:
1. 2x = A(x² + x + 1) + (Bx + C)(x − 1).
2. x = 1: 2 = 3A ⇒ \( A = \frac{2}{3} \). x²: 0 = A + B ⇒ \( B = -\frac{2}{3} \). Constant: 0 = A − C ⇒ \( C = \frac{2}{3} \).
3. So A ≠ B and B ≠ C (with A = C), which matches option (d).

Teacher's Note:
1. x³ − 1 = (x − 1)(x² + x + 1).
2. Read the options as written: (d) only says A ≠ B and B ≠ C.

Answer: (b) \( \frac{-5}{3(2x - 1)} + \frac{1}{3(x + 1)} + \frac{1}{x - 1} \)

Explanation:
1. x = 1/2: \( \frac{5/4}{(1/4 - 1)} = -\frac{5}{3} \) ⇒ term \( \frac{-5}{3(2x - 1)} \).
2. x = −1: \( \frac{2}{(-3)(-2)} = \frac{1}{3} \) ⇒ \( \frac{1}{3(x + 1)} \); x = 1: \( \frac{2}{(1)(2)} = 1 \) ⇒ \( \frac{1}{x - 1} \).
3. This is option (b).

Teacher's Note:
1. For a factor (2x − 1), the cover-up value goes over (2x − 1) after dividing by the leading 2, or compute directly as shown.
2. x² − 1 = (x − 1)(x + 1) must be factorised first.

Answer: (b) 3

Explanation:
1. ax − 1 = x(2 + x) − (1 − x + x²) = 2x + x² − 1 + x − x² = 3x − 1.
2. So a = 3.

Teacher's Note:
1. Combine the right side and compare numerators.
2. The x² terms cancel, as they must.

Answer: (a) (1, −1, 0)

Explanation:
1. 1 = A(x² + 1) + (Bx + C)x.
2. x = 0 ⇒ A = 1; x²: 0 = A + B ⇒ B = −1; x: 0 = C.

Teacher's Note:
1. 1/(x(x² + 1)) = 1/x − x/(x² + 1) is a standard result.
2. Compare coefficients after finding A.

Answer: (d) \( \frac{1}{x^2 - x + 1} - \frac{1}{x^2 + x + 1} \)

Explanation:
1. \( x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2 = (x^2 - x + 1)(x^2 + x + 1) \).
2. \( (x^2 + x + 1) - (x^2 - x + 1) = 2x \).
3. So \( \frac{2x}{x^4 + x^2 + 1} = \frac{1}{x^2 - x + 1} - \frac{1}{x^2 + x + 1} \).

Teacher's Note:
1. x⁴ + x² + 1 factorises as a difference of squares after adding and subtracting x².
2. The difference of the two factors is exactly the numerator 2x.

Answer: (b) (3, 2)

Explanation:
1. 3x² + 5 = 3(x² + 1) + 2.
2. So a = 3, b = 2.

Teacher's Note:
1. Write the numerator in powers of x² + 1.
2. Then divide term by term.

Answer: (d) 1

Explanation:
1. The fraction is improper (degree 2 over degree 2), so C is the quotient on division.
2. Leading coefficients are both 1, so the quotient is 1.
3. Hence C = 1.

Teacher's Note:
1. For equal degrees, the constant term in the partial fraction form is the ratio of leading coefficients.
2. Letting x → ∞ on both sides also gives C = 1.

Answer: (d) \( 1 + \frac{4}{x - 1} - \frac{1}{x - 2} \)

Explanation:
1. \( \frac{x^2 - 5}{x^2 - 3x + 2} = 1 + \frac{3x - 7}{(x - 1)(x - 2)} \).
2. x = 1: \( \frac{-4}{-1} = 4 \); x = 2: \( \frac{-1}{1} = -1 \).
3. Result: \( 1 + \frac{4}{x - 1} - \frac{1}{x - 2} \).

Teacher's Note:
1. Divide first when the fraction is improper.
2. x² − 3x + 2 = (x − 1)(x − 2) has no repeated factor, so squared terms cannot appear.

Answer: (c) \( p = \frac{1}{2} \); (d) 6q − 3r + 2s = 3

Explanation:
1. Leading coefficients: \( p = \frac{1}{2} \).
2. x = 1/2: \( q = \frac{1/8}{(5/2)(-5/2)} = -\frac{1}{50} \); x = −2: \( r = \frac{-8}{(-5)(-5)} = -\frac{8}{25} \); x = 3: \( s = \frac{27}{5 \cdot 5} = \frac{27}{25} \).
3. \( 6q - 3r + 2s = -\frac{3}{25} + \frac{24}{25} + \frac{54}{25} = 3 \). Options (c) and (d).

Teacher's Note:
1. The polynomial part p is the ratio of the leading coefficients: 1/2.
2. For (2x − 1), multiply through and put x = 1/2.

Answer: (a) \( x^2 + \frac{1}{1 + 2x} + \frac{1}{1 + 3x} \)

Explanation:
1. \( 6x^4 + 5x^3 + x^2 = x^2(6x^2 + 5x + 1) \), so the fraction \( = x^2 + \frac{5x + 2}{(1 + 2x)(1 + 3x)} \).
2. 5x + 2 = A(1 + 3x) + B(1 + 2x): x = −1/2 ⇒ A = 1; x = −1/3 ⇒ B = 1.
3. Result: \( x^2 + \frac{1}{1 + 2x} + \frac{1}{1 + 3x} \).

Teacher's Note:
1. Spot that the first three terms of the numerator are x² times the denominator.
2. Then resolve the proper part as usual.

Answer: (a) \( A = \frac{13}{2} \); (d) A + B + C = 5

Explanation:
1. Put t = sin x: \( \frac{t^2 + 1}{(2t - 3)(t - 1)} \). Leading coefficients give \( C = \frac{1}{2} \).
2. \( t^2 + 1 = A(t - 1) + B(2t - 3) + C(2t - 3)(t - 1) \). t = 3/2: \( \frac{13}{4} = \frac{A}{2} \Rightarrow A = \frac{13}{2} \). t = 1: 2 = −B ⇒ B = −2.
3. A + B + C = \( \frac{13}{2} - 2 + \frac{1}{2} = 5 \). Options (a) and (d).

Teacher's Note:
1. Treat sin x as the variable t; the fraction is improper, so a constant C appears.
2. Find A and B by putting the zeros of the linear factors.

Answer: (b) \( \frac{15}{16} \)

Explanation:
1. \( \frac{3x}{(x - 2)(x + 1)} = \frac{2}{x - 2} + \frac{1}{x + 1} \).
2. \( \frac{2}{x - 2} = -\left(1 - \frac{x}{2}\right)^{-1} \), coefficient of x⁴: \( -\frac{1}{16} \). \( \frac{1}{1 + x} \): coefficient of x⁴ is 1.
3. Total \( = 1 - \frac{1}{16} = \frac{15}{16} \).

Teacher's Note:
1. Split into partial fractions and expand each as a geometric series.
2. Write 2/(x − 2) as −1/(1 − x/2) to get the standard form.

Answer: (d) \( -\frac{1}{256} \)

Explanation:
1. \( \frac{x^2 + 1}{(x^2 + 4)(x - 2)} = \frac{5/8}{x - 2} + \frac{\frac{3}{8}x + \frac{3}{4}}{x^2 + 4} \).
2. \( \frac{5/8}{x - 2} = -\frac{5}{16}\left(1 - \frac{x}{2}\right)^{-1} \): x⁵ coefficient \( -\frac{5}{16} \cdot \frac{1}{32} = -\frac{5}{512} \).
3. \( \frac{\frac{3}{8}x}{4(1 + x^2/4)} \): x⁵ coefficient \( \frac{3}{32} \cdot \frac{1}{16} = \frac{3}{512} \); the constant part gives only even powers.
4. Total \( = -\frac{5}{512} + \frac{3}{512} = -\frac{1}{256} \).

Teacher's Note:
1. Only the odd part of the quadratic-factor fraction contributes to x⁵.
2. Expand 1/(1 + x²/4) as 1 − x²/4 + x⁴/16 − …

Answer: (c) \( \frac{-1}{2^n} + \frac{1}{3^{n+1}} \)

Explanation:
1. \( \frac{x - 4}{(x - 2)(x - 3)} = \frac{2}{x - 2} - \frac{1}{x - 3} \).
2. \( \frac{2}{x - 2} = -\left(1 - \frac{x}{2}\right)^{-1} \): coefficient \( -\frac{1}{2^n} \). \( -\frac{1}{x - 3} = \frac{1}{3}\left(1 - \frac{x}{3}\right)^{-1} \): coefficient \( \frac{1}{3^{n+1}} \).
3. Total \( = \frac{-1}{2^n} + \frac{1}{3^{n+1}} \).

Teacher's Note:
1. Turn each term into a(1 − x/k)⁻¹ before expanding.
2. Check with n = 0: the constant term −4/6 = −2/3 = −1 + 1/3.