CBSE Class 10 Mathematics Surds and Indices MCQs Set 07

Answer: (a) \( \sqrt[12]{\frac{2}{27}} \)

Explanation:
1. \( 4^{1/6} = 4^{2/12} = 16^{1/12} \) and \( 6^{1/4} = 6^{3/12} = 216^{1/12} \).
2. The quotient is \( \left(\frac{16}{216}\right)^{1/12} = \left(\frac{2}{27}\right)^{1/12} \).

Teacher's Note:
1. Write both surds with the same index (the LCM of 6 and 4 is 12).
2. Then divide the radicands.

Answer: (b) The first is greater than the second

Explanation:
1. \( \sqrt{11} - \sqrt{10} = \frac{1}{\sqrt{11} + \sqrt{10}} \) and \( \sqrt{12} - \sqrt{11} = \frac{1}{\sqrt{12} + \sqrt{11}} \).
2. The first denominator is smaller, so the first difference is larger.

Teacher's Note:
1. Rationalise each difference to compare reciprocals.
2. A smaller denominator means a larger fraction.

Answer: (c) ℓ = 2/k

Explanation:
1. Cross-multiplying, \( \ell k = (\sqrt{10} + \sqrt{14} + \sqrt{15} + \sqrt{21})(\sqrt{10} - \sqrt{14} - \sqrt{15} + \sqrt{21}) \).
2. Writing the factors as \( (\sqrt{10} + \sqrt{21}) + (\sqrt{14} + \sqrt{15}) \) and \( (\sqrt{10} + \sqrt{21}) - (\sqrt{14} + \sqrt{15}) \), the product is \( (31 + 2\sqrt{210}) - (29 + 2\sqrt{210}) = 2 \). So ℓk = 2.

Teacher's Note:
1. Group the surds so that the product is a difference of squares.
2. Compare ℓk with the product obtained.

Answer: (a) \( 2^{2/3} - 1 + 2^{-2/3} \)

Explanation:
1. With \( p = 2^{1/3} \) and \( q = 2^{-1/3} \), use \( p^3 + q^3 = (p + q)(p^2 - pq + q^2) \).
2. Here \( p^2 - pq + q^2 = 2^{2/3} - 1 + 2^{-2/3} \) and the product is \( 2 + \frac{1}{2} \), which is rational.

Teacher's Note:
1. Use the sum of cubes identity to rationalise a sum of cube roots.
2. pq = 1 in this problem.

Answer: (a) 2

Explanation:
1. Let \( t = 2^{1/3} \), so \( t^3 = 2 \) and \( x - 2 = t + t^2 \).
2. Cubing: \( (x - 2)^3 = t^3 + t^6 + 3t^3(t + t^2) = 2 + 4 + 6(t + t^2) = 6 + 6(x - 2) \).
3. So \( x^3 - 6x^2 + 12x - 8 = 6x - 6 \), which gives \( x^3 - 6x^2 + 6x = 2 \).

Teacher's Note:
1. Isolate the surd part, then cube both sides.
2. Use t³ = 2 to reduce powers.

Answer: (c) \( -\sqrt{5} + \sqrt{4} + \sqrt{12} \)

Explanation:
1. Try \( (2 + 2\sqrt{3} - \sqrt{5})^2 = 4 + 12 + 5 + 8\sqrt{3} - 4\sqrt{5} - 4\sqrt{15} = 21 + 8\sqrt{3} - 4\sqrt{5} - 4\sqrt{15} \).
2. The square root is \( 2 + 2\sqrt{3} - \sqrt{5} \) = \( -\sqrt{5} + \sqrt{4} + \sqrt{12} \) (positive, about 3.23).

Teacher's Note:
1. Express the radicand as a perfect square of a three-term surd.
2. Check the sign by evaluating numerically.

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

Explanation:
1. Rearrange: \( x - a = \sqrt{a^2 + 1} \). Square both sides: \( x^2 - 2ax + a^2 = a^2 + 1 \).
2. So \( 2ax = x^2 - 1 \) and \( a = \frac{x^2 - 1}{2x} \).

Teacher's Note:
1. Isolate the root and square to remove it.
2. a² cancels on both sides.

Answer: (b) \( \sqrt[3]{4} - 1 \)

Explanation:
1. Let \( t = \sqrt[3]{4} \); then the expression is \( t^2 + t + 1 \).
2. Multiplying by \( t - 1 \) gives \( t^3 - 1 = 3 \), a rational number.

Teacher's Note:
1. Use (t − 1)(t² + t + 1) = t³ − 1.
2. Note that the cube root of 16 is t².

Answer: (a) \( 2a^2 + 2a\sqrt{a^2 - 1} - 1 \)

Explanation:
1. Multiply numerator and denominator by \( a + \sqrt{a^2 - 1} \). The denominator becomes \( a^2 - (a^2 - 1) = 1 \).
2. Numerator = \( (a + \sqrt{a^2 - 1})^2 = a^2 + (a^2 - 1) + 2a\sqrt{a^2 - 1} = 2a^2 - 1 + 2a\sqrt{a^2 - 1} \).

Teacher's Note:
1. Use the conjugate of the denominator.
2. The denominator collapses to 1.

Answer: (b) \( x + \sqrt{x^2 - 1} \)

Explanation:
1. Write \( 2x^2 - 1 = x^2 + (x^2 - 1) \).
2. So the expression is \( \sqrt{(x + \sqrt{x^2 - 1})^2} = x + \sqrt{x^2 - 1} \) for x ≥ 1.

Teacher's Note:
1. Split 2x² − 1 into x² and x² − 1 to form (A + B)².
2. Both terms under the root are positive for x ≥ 1.

Answer: (d) None of these

Explanation:
1. Let \( s = \sqrt{a - 1} \) so that \( a = s^2 + 1 \). Then the expression is \( \sqrt{(s - 1)^2} - \sqrt{(s + 1)^2} = |s - 1| - (s + 1) \).
2. For 1 ≤ a ≤ 2 we have 0 ≤ s ≤ 1, so |s − 1| = 1 − s and the value is \( -2s = -2\sqrt{a - 1} \), which is not among the first three options.

Teacher's Note:
1. Write a as (a − 1) + 1 to see the perfect squares.
2. The modulus must be resolved using the given range.

Answer: (d) \( -\frac{1}{\sqrt{3}} \)

Explanation:
1. \( \sqrt{5 - \sqrt{24}} = \sqrt{3} - \sqrt{2} \) and \( \sqrt{5 + \sqrt{24}} = \sqrt{3} + \sqrt{2} \).
2. The sum is \( \frac{1}{\sqrt{5} - \sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{5} - \sqrt{3} - \sqrt{2}} = \frac{2(\sqrt{5} - \sqrt{3})}{(\sqrt{5} - \sqrt{3})^2 - 2} \).
3. The denominator is \( 8 - 2\sqrt{15} - 2 = 6 - 2\sqrt{15} \), so the value is \( \frac{2(\sqrt{5} - \sqrt{3})}{2\sqrt{3}(\sqrt{3} - \sqrt{5})} = -\frac{1}{\sqrt{3}} \).

Teacher's Note:
1. Denest the square roots using (√3 ± √2)² = 5 ± √24.
2. Combine the two fractions over a common denominator.

Answer: (a) \( \sqrt[3]{4} + 1 \)

Explanation:
1. With \( t = \sqrt[3]{4} \) the expression is \( t^2 - t + 1 \).
2. Multiplying by \( t + 1 \) gives \( t^3 + 1 = 5 \).

Teacher's Note:
1. Use (t + 1)(t² − t + 1) = t³ + 1.
2. The factor is the conjugate-type partner of the given trinomial.

Answer: (a) a = 1, b = 4/3, c = 7/3

Explanation:
1. Rationalise: \( \frac{4}{(2 + \sqrt{3}) + \sqrt{7}} \cdot \frac{(2 + \sqrt{3}) - \sqrt{7}}{(2 + \sqrt{3}) - \sqrt{7}} = \frac{4(2 + \sqrt{3} - \sqrt{7})}{4\sqrt{3}} \) since \( (2 + \sqrt{3})^2 - 7 = 4\sqrt{3} \).
2. This equals \( \frac{2 + \sqrt{3} - \sqrt{7}}{\sqrt{3}} = \frac{2}{\sqrt{3}} + 1 - \frac{\sqrt{7}}{\sqrt{3}} = \sqrt{\frac{4}{3}} + \sqrt{1} - \sqrt{\frac{7}{3}} \), so a = 1, b = 4/3, c = 7/3.

Teacher's Note:
1. Group 2 + √3 as one term and use the difference of squares.
2. Bring each term under the root sign at the end.

Answer: (d) \( \frac{71}{84} \)

Explanation:
1. \( \frac{5}{6} = \frac{70}{84} \) and \( \frac{6}{7} = \frac{72}{84} \).
2. The fraction \( \frac{71}{84} \) lies strictly between them.

Teacher's Note:
1. Write both fractions with the common denominator 84.
2. Pick a numerator between 70 and 72.

Answer: (b) 2

Explanation:
1. x = (√3 + √2)², so \( \sqrt{x} = \sqrt{3} + \sqrt{2} \) and \( \frac{1}{\sqrt{x}} = \sqrt{3} - \sqrt{2} \).
2. The expression is \( \frac{1}{\sqrt{2}}\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right) = \frac{2\sqrt{2}}{\sqrt{2}} = 2 \).

Teacher's Note:
1. Recognise 5 + 2√6 as (√3 + √2)².
2. Factor out 1/√2.

Answer: (b) x > y

Explanation:
1. \( x = \frac{2}{\sqrt{7} + \sqrt{5}} \) and \( y = \frac{2}{\sqrt{13} + \sqrt{11}} \).
2. Since \( \sqrt{7} + \sqrt{5} < \sqrt{13} + \sqrt{11} \), x > y.

Teacher's Note:
1. Rationalise to compare using reciprocals.
2. The larger denominator gives the smaller number.

Answer: (b) \( \sqrt{3} \)

Explanation:
1. Simplify: \( \sqrt{75} - \sqrt{48} = 5\sqrt{3} - 4\sqrt{3} = \sqrt{3} \) and \( -\sqrt{32} + \sqrt{50} = -4\sqrt{2} + 5\sqrt{2} = \sqrt{2} \).
2. The expression is \( \frac{3 + \sqrt{6}}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3}(\sqrt{3} + \sqrt{2})}{\sqrt{3} + \sqrt{2}} = \sqrt{3} \).

Teacher's Note:
1. Reduce each root to a multiple of a simple surd.
2. Notice 3 + √6 = √3(√3 + √2).

Answer: (b) 8

Explanation:
1. Rationalise the fraction inside: \( \frac{6 + 2\sqrt{3}}{33 - 19\sqrt{3}} \cdot \frac{33 + 19\sqrt{3}}{33 + 19\sqrt{3}} = \frac{198 + 114\sqrt{3} + 66\sqrt{3} + 114}{1089 - 1083} \).
2. This is \( \frac{312 + 180\sqrt{3}}{6} = 52 + 30\sqrt{3} \) and \( (5 + 3\sqrt{3})^2 = 25 + 27 + 30\sqrt{3} = 52 + 30\sqrt{3} \).
3. So a = 5, b = 3 and a + b = 8.

Teacher's Note:
1. Rationalise inside the root first.
2. Spot the perfect square 52 + 30√3.

Answer: (b) \( 1 + \sqrt{5} + \sqrt{2} - \sqrt{10} \)

Explanation:
1. Multiply by \( (3 + \sqrt{5}) - 2\sqrt{2} \): the denominator is \( (3 + \sqrt{5})^2 - 8 = 6 + 6\sqrt{5} \).
2. Value = \( \frac{12(3 + \sqrt{5} - 2\sqrt{2})}{6(1 + \sqrt{5})} = \frac{2(3 + \sqrt{5} - 2\sqrt{2})(\sqrt{5} - 1)}{4} \) = \( 1 + \sqrt{5} + \sqrt{2} - \sqrt{10} \) (numerically 1.488).

Teacher's Note:
1. Rationalise in two steps with conjugates.
2. Verify with a decimal check: the value is about 1.488.

Answer: (d) 2, −2, √2, −√2

Explanation:
1. Since \( (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 1 \), put \( y = (5 + 2\sqrt{6})^{x^2 - 3} \) so that the equation is \( y + \frac{1}{y} = 10 \).
2. Then \( y = 5 \pm 2\sqrt{6} \), so \( x^2 - 3 = \pm 1 \), which gives \( x^2 = 4 \) or \( x^2 = 2 \).
3. x = ±2 or ±√2.

Teacher's Note:
1. The two surds are reciprocals of each other.
2. Solve both the positive and negative exponent cases.

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

Explanation:
1. y = 2 − √3. Also \( \sqrt{x} = \frac{\sqrt{3} + 1}{\sqrt{2}} \) and \( \sqrt{y} = \frac{\sqrt{3} - 1}{\sqrt{2}} \).
2. Then \( \sqrt{2} + \sqrt{x} = \frac{3 + \sqrt{3}}{\sqrt{2}} \) and \( \sqrt{2} - \sqrt{y} = \frac{3 - \sqrt{3}}{\sqrt{2}} \).
3. The sum is \( \sqrt{2}\left[\frac{2 + \sqrt{3}}{3 + \sqrt{3}} + \frac{2 - \sqrt{3}}{3 - \sqrt{3}}\right] = \sqrt{2}\cdot\frac{(2 + \sqrt{3})(3 - \sqrt{3}) + (2 - \sqrt{3})(3 + \sqrt{3})}{6} \) = \( \sqrt{2}\cdot\frac{6}{6} \) = \( \sqrt{2} \).

Teacher's Note:
1. Express √x and √y using (√3 ± 1)²/2 = 2 ± √3.
2. Combine the two fractions over the common denominator 9 − 3 = 6.

Answer: (b) p + q

Explanation:
1. \( p + \frac{1}{q} = \frac{pq + 1}{q} \) and \( q + \frac{1}{p} = \frac{pq + 1}{p} \), so their ratio is \( \frac{p}{q} \) raised to the power p.
2. Similarly \( \frac{p - 1/q}{q - 1/p} = \frac{p}{q} \) raised to the power q.
3. The whole expression is \( \left(\frac{p}{q}\right)^{p + q} \).

Teacher's Note:
1. Combine each bracket into a single fraction.
2. Add the exponents of the same base.

Answer: (d) 27abc

Explanation:
1. If p + q + r = 0 then \( p^3 + q^3 + r^3 = 3pqr \). Take \( p = \sqrt[3]{a} \) etc.: a + b + c = 3·∛(abc).
2. Cubing, \( (a + b + c)^3 = 27abc \).

Teacher's Note:
1. The identity p³ + q³ + r³ − 3pqr = (p + q + r)(…) is used.
2. Cube both sides at the end.

Answer: (b) 1

Explanation:
1. Multiply the radicands: \( \frac{b}{a} \times \frac{c}{b} \times \frac{a}{c} = 1 \).
2. So the product is \( \sqrt{1} = 1 \).

Teacher's Note:
1. Combine the square roots into one root.
2. Everything cancels to 1.