CBSE Class 10 Mathematics Surds and Indices MCQs Set 18

Answer: (c) 21

Explanation:
1. Square both sides: \( 19-4\sqrt{x}=12+7-2\sqrt{84} \)
2. So \( 4\sqrt{x}=2\sqrt{84} \), i.e. \( \sqrt{x}=\tfrac{\sqrt{84}}{2} \)
3. Hence x = 84/4 = 21.

Teacher's Note:
1. Squaring a surd equation and comparing the rational and irrational parts is the standard method.
2. Always check that both sides are positive before squaring.

Answer: (a) 3

Explanation:
1. Let the value be y. Then \( y=\sqrt{6+y} \), so \( y^2-y-6=0 \)
2. \( (y-3)(y+2)=0 \) and y must be positive, so y = 3.

Teacher's Note:
1. For an infinite nested root, set the whole expression equal to y and square.
2. Reject the negative root because a square root is non-negative.

Answer: (b) a < b

Explanation:
1. Both are positive, so compare squares: \( a^2=32+2\sqrt{252} \) and \( b^2=32+2\sqrt{255} \)
2. Since \( \sqrt{252}<\sqrt{255} \), a² < b²
3. Hence a < b.

Teacher's Note:
1. To compare sums of surds, square both and compare the leftover terms.
2. Squaring is safe only for positive numbers.

Answer: (a) \( \sqrt5 \)

Explanation:
1. \( x=8-2\sqrt{15}=(\sqrt5-\sqrt3)^2 \), so \( \sqrt{x}=\sqrt5-\sqrt3 \)
2. \( \tfrac{2}{\sqrt{x}}=\tfrac{2(\sqrt5+\sqrt3)}{2}=\sqrt5+\sqrt3 \)
3. Sum = \( 2\sqrt5 \); half of it is \( \sqrt5 \).

Teacher's Note:
1. Write 8 − 2√15 as a perfect square of a binomial surd.
2. Rationalise 2/√x using the conjugate.

Answer: (a) \( (\sqrt{11}+1)\sqrt[4]{5} \)

Explanation:
1. Take \( \sqrt5 \) out: \( 12\sqrt5+2\sqrt{55}=\sqrt5\,(12+2\sqrt{11}) \)
2. \( 12+2\sqrt{11}=(\sqrt{11}+1)^2 \)
3. So the root is \( \sqrt[4]{5}\,(\sqrt{11}+1) \).

Teacher's Note:
1. Factor out a common surd first, then complete the square.
2. Verify by squaring the answer.

Answer: (d) \( \left(\tfrac{-61}{29},\tfrac{-24}{29}\right) \)

Explanation:
1. Multiply by the conjugate: \( \dfrac{(4+3\sqrt5)^2}{16-45} \)
2. \( =\dfrac{16+45+24\sqrt5}{-29}=-\dfrac{61}{29}-\dfrac{24}{29}\sqrt5 \)
3. So (a, b) = (−61/29, −24/29).

Teacher's Note:
1. Rationalise by multiplying numerator and denominator by the conjugate of the denominator.
2. Watch the sign: 16 − 45 is negative.

Answer: (b) y < z < x

Explanation:
1. The orders are 2, 4, 3, with LCM 12. \( x=5^{6/12}=15625^{1/12} \), \( y=10^{3/12}=1000^{1/12} \), \( z=6^{4/12}=1296^{1/12} \)
2. Since 1000 < 1296 < 15625, y < z < x.

Teacher's Note:
1. Convert surds to the same order (the LCM of the orders) to compare them.
2. Then compare the radicands.

Answer: (a) x > y > z

Explanation:
1. LCM of 3, 4, 6 is 12: \( x=9^{4/12}=6561^{1/12} \), \( y=11^{3/12}=1331^{1/12} \), \( z=17^{2/12}=289^{1/12} \)
2. 6561 > 1331 > 289, so x > y > z.

Teacher's Note:
1. Bring all radicals to a common order before comparing.
2. Using fractional indices makes this quick.

Answer: (b) \( \sqrt{10}+\sqrt2 \)

Explanation:
1. \( 12+4\sqrt5=12+2\sqrt{20}=10+2+2\sqrt{10\cdot2} \)
2. \( =(\sqrt{10}+\sqrt2)^2 \)
3. So the root is \( \sqrt{10}+\sqrt2 \).

Teacher's Note:
1. Split the rational part into two numbers whose product equals the radicand under the surd.
2. Here 10 + 2 = 12 and 10 × 2 = 20.

Answer: (a) x < y

Explanation:
1. x = 1.414 + 1 = 2.414 and y = 4.123 − 1.414 = 2.709
2. So x < y.

Teacher's Note:
1. Use decimal approximations of the surds to compare when the squares are awkward.
2. Keep three decimal places for safe comparison.

Answer: (a) \( \sqrt7+2 \)

Explanation:
1. \( \sqrt{112}=4\sqrt7=2\sqrt{28} \)
2. \( 11+2\sqrt{28}=(7+4)+2\sqrt{7\cdot4}=(\sqrt7+2)^2 \)
3. The root is \( \sqrt7+2 \).

Teacher's Note:
1. Write the surd as 2√(ab) with a + b equal to the rational part.
2. Here a = 7 and b = 4.

Answer: (d) \( \sqrt3+\sqrt2 \)

Explanation:
1. \( 5+2\sqrt6=(3+2)+2\sqrt{3\cdot2}=(\sqrt3+\sqrt2)^2 \)
2. So the root is \( \sqrt3+\sqrt2 \).

Teacher's Note:
1. Look for two numbers with sum 5 and product 6.
2. They are 3 and 2.

Answer: (a) 9

Explanation:
1. Let \( p=\sqrt{4+\sqrt{15}}=\dfrac{\sqrt{10}+\sqrt6}{2} \) and \( q=\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{10}-\sqrt6}{2} \)
2. The expression is \( p^3-q^3=(p-q)(p^2+pq+q^2) \) with \( p-q=\sqrt6,\ pq=1,\ p^2+q^2=8 \)
3. So it equals \( \sqrt6\,(8+1)=9\sqrt6 \), giving k = 9.

Teacher's Note:
1. Fractional powers of binomial surds can be evaluated through their square roots.
2. A numerical check confirms k = 9.

Answer: (a) 1

Explanation:
1. \( 7\pm4\sqrt3=(2\pm\sqrt3)^2 \), so \( x=\dfrac{2+\sqrt3}{2-\sqrt3}=(2+\sqrt3)^2=7+4\sqrt3 \)
2. \( x-14=4\sqrt3-7=-\dfrac1x \), so \( x(x-14)=-1 \)
3. Hence \( x^2(x-14)^2=1 \).

Teacher's Note:
1. Express both radicands as perfect squares to simplify x.
2. Note that 7 − 4√3 is the reciprocal of 7 + 4√3.

Answer: (a) 6

Explanation:
1. Square the right side: \( 18+5+2\sqrt{90}=23+6\sqrt{10} \)
2. Compare with \( 23+x\sqrt{10} \): x = 6.

Teacher's Note:
1. Equate the irrational parts after squaring.
2. Simplify √90 as 3√10.

Answer: (b) \( 5+\sqrt2+\sqrt5 \)

Explanation:
1. \( \dfrac1{2-\sqrt3}=2+\sqrt3,\quad \dfrac1{\sqrt3+\sqrt2}=\sqrt3-\sqrt2,\quad \dfrac4{3-\sqrt5}=3+\sqrt5 \)
2. Sum = \( (2+\sqrt3)-(\sqrt3-\sqrt2)+(3+\sqrt5) \) = \( 5+\sqrt2+\sqrt5 \).

Teacher's Note:
1. Rationalise each term separately with its own conjugate.
2. Then add the rational and irrational parts.

Answer: (b) p > q

Explanation:
1. p = 5.657 − 4.899 = 0.758 and q = 7.071 − 6.928 = 0.143
2. So p > q.

Teacher's Note:
1. Differences of close surds are small, so use accurate decimals.
2. Alternatively rationalise: p = 8/(√32+√24), q = 2/(√50+√48).

Answer: (c) 0

Explanation:
1. First term \( =\sqrt2(\sqrt6+\sqrt3)=2\sqrt3+\sqrt6 \); second \( =\sqrt3(\sqrt6+\sqrt2)=3\sqrt2+\sqrt6 \); third \( =\sqrt3(\sqrt6-2)=3\sqrt2-2\sqrt3 \)
2. Total = \( (2\sqrt3+\sqrt6)-(3\sqrt2+\sqrt6)+(3\sqrt2-2\sqrt3)=0 \).

Teacher's Note:
1. Rationalise each fraction, then cancel the like surds.
2. Check each conjugate pair carefully.

Answer: (c) \( 9\sqrt7 \)

Explanation:
1. \( 567=81\times7 \)
2. \( \sqrt{567}=9\sqrt7 \).

Teacher's Note:
1. Factor out the largest perfect square.
2. 567 = 3⁴ × 7.

Answer: (d) \( 13+7\sqrt3 \)

Explanation:
1. Multiply by \( 2+\sqrt3 \): numerator \( (5+\sqrt3)(2+\sqrt3)=13+7\sqrt3 \), denominator \( 4-3=1 \)
2. Value = \( 13+7\sqrt3 \).

Teacher's Note:
1. Rationalise using the conjugate of the denominator.
2. Here the denominator becomes 1.

Answer: (c) 14

Explanation:
1. As before \( x=7+4\sqrt3 \), so \( \dfrac1x=7-4\sqrt3 \)
2. \( x+\dfrac1x=14 \).

Teacher's Note:
1. x and 1/x are conjugates here, so their sum is rational.
2. Their product is 1.

Answer: (b) 3, 8

Explanation:
1. \( \sqrt{27}=3\sqrt3,\ \sqrt{75}=5\sqrt3 \)
2. Sum = \( 3+8\sqrt3 \), so a = 3, b = 8.

Teacher's Note:
1. Reduce each surd to a multiple of √3.
2. Then compare rational and irrational parts.

Answer: (c) 6

Explanation:
1. Cube both sides: 6y − 9 = 27
2. 6y = 36, so y = 6.

Teacher's Note:
1. Cube both sides to remove a cube root.
2. Verify: 6(6) − 9 = 27 and ∛27 = 3.

Answer: (b) 6

Explanation:
1. a = 3 + 2√2 and b = 3 − 2√2, so ab = 1 and a + b = 6
2. \( a^2b+ab^2=ab(a+b)=6 \).

Teacher's Note:
1. Factorise as ab(a + b) before substituting.
2. Rationalising shows a and b are conjugates.

Answer: (a) 194

Explanation:
1. y = 1/x = 7 − 4√3, so 1/x² + 1/y² = y² + x²
2. x + y = 14 and xy = 1, so \( x^2+y^2=14^2-2=194 \).

Teacher's Note:
1. Use the identity x² + y² = (x + y)² − 2xy.
2. Here xy = 1.

Answer: (b) \( 2\sqrt2 \)

Explanation:
1. x = (√3 + √2)², so √x = √3 + √2 and x − 1 = 4 + 2√6 = 2√2(√2 + √3)
2. Therefore the value is \( 2\sqrt2 \).

Teacher's Note:
1. Perfect-square form of x makes √x easy.
2. Factor x − 1 to cancel with √x.