JEE Mathematics Surds and Indices MCQs Set 01

Answer: (c) 2.646

Explanation:
1. \( 2.645^2 = 6.996 \) and \( 2.646^2 = 7.0013 \), so \( 2.645 < \sqrt{7} < 2.646 \).
2. \( \sqrt{7} = 2.64575\ldots \), which rounds to 2.646 (four figures).

Teacher's Note:
1. Rounding to four figures looks at the fifth figure: 2.6457 becomes 2.646.
2. Check by squaring the rounded value.

Answer: (a) 3.61

Explanation:
1. \( 3.6^2 = 12.96 \) and \( 3.61^2 = 13.0321 \), so \( 3.6 < \sqrt{13} < 3.61 \).
2. \( \sqrt{13} = 3.6055\ldots \), which rounds to 3.61.

Teacher's Note:
1. The third decimal is 5, so round up.
2. Two decimal places means exactly two figures after the point.

Answer: (d) \( 11\sqrt{2} \)

Explanation:
1. \( 242 = 121 \times 2 = 11^2 \times 2 \).
2. \( \sqrt{242} = 11\sqrt{2} \).

Teacher's Note:
1. Take out the largest perfect-square factor.
2. \( 11\sqrt{2} \approx 15.56 \), and \( 15.56^2 \approx 242 \).

Answer: (b) \( 19\sqrt{3} \)

Explanation:
1. \( 1083 = 3 \times 361 = 3 \times 19^2 \).
2. \( \sqrt{1083} = 19\sqrt{3} \).

Teacher's Note:
1. Divide by 3 first to see the square 361.
2. \( 19^2 = 361 \) is worth remembering.

Answer: (b) \( 4\sqrt{3} \), because \( \sqrt{48} > \sqrt{45} \)

Explanation:
1. Write both as a single root: \( 3\sqrt{5} = \sqrt{45} \) and \( 4\sqrt{3} = \sqrt{48} \).
2. 48 > 45, so \( 4\sqrt{3} \) is greater.

Teacher's Note:
1. Bring the coefficient inside the root by squaring it.
2. Compare the numbers under the root once the surds are in the same form.

Answer: (d) \( 7\sqrt{2} \), because \( \sqrt{98} > \sqrt{96} \)

Explanation:
1. \( 7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98} \) and \( 4\sqrt{6} = \sqrt{16 \times 6} = \sqrt{96} \).
2. 98 > 96, so \( 7\sqrt{2} \) is greater.

Teacher's Note:
1. The two values are close (9.899 and 9.798), so use exact squares rather than guessing.
2. Never compare only the numbers under the root.

Answer: (a) \( 6\sqrt{5} \), because \( \sqrt{180} > \sqrt{150} \)

Explanation:
1. \( 5\sqrt{6} = \sqrt{150} \) and \( 6\sqrt{5} = \sqrt{180} \).
2. 180 > 150, so \( 6\sqrt{5} \) is greater.

Teacher's Note:
1. Convert both to the form \( \sqrt{n} \) and compare n.
2. \( 5\sqrt{6} \approx 12.25 \) and \( 6\sqrt{5} \approx 13.42 \).

Answer: (c) a = 4, b = 0

Explanation:
1. Common denominator: \( (\sqrt{3}+1)(\sqrt{3}-1) = 2 \).
2. Numerator: \( (\sqrt{3}-1)^2 + (\sqrt{3}+1)^2 = (4 - 2\sqrt{3}) + (4 + 2\sqrt{3}) = 8 \).
3. The sum is \( \frac{8}{2} = 4 = 4 + \sqrt{3}\cdot 0 \), so a = 4 and b = 0.

Teacher's Note:
1. The surd parts cancel in conjugate pairs.
2. Because \( \sqrt{3} \) is irrational, a and b are determined uniquely.

Answer: (c) 30

Explanation:
1. \( \sqrt{20} \times \sqrt{45} = \sqrt{900} \).
2. \( \sqrt{900} = 30 \).

Teacher's Note:
1. Multiply under one root, then simplify.
2. The product of two surds can be rational.

Answer: (a) 36

Explanation:
1. \( \sqrt{27} = 3\sqrt{3} \) and \( \sqrt{48} = 4\sqrt{3} \).
2. Product \( = 12 \times 3 = 36 \).

Teacher's Note:
1. Simplify each surd first so that the like surds multiply to a whole number.
2. \( \sqrt{3} \times \sqrt{3} = 3 \).

Answer: (d) 68

Explanation:
1. \( \sqrt{8} \times \sqrt{578} = \sqrt{4624} \).
2. \( 68^2 = 4624 \), so the product is 68. (Or: \( 2\sqrt{2} \times 17\sqrt{2} = 34 \times 2 = 68 \).)

Teacher's Note:
1. Simplify: \( 578 = 2 \times 17^2 \).
2. Both surds are multiples of \( \sqrt{2} \).

Answer: (b) \( \frac{(a+b)\sqrt{ab} + 2ab}{ab} \)

Explanation:
1. \( \sqrt{\frac{a}{b}} - \sqrt{\frac{b}{a}} = \frac{a - b}{\sqrt{ab}} = \frac{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}{\sqrt{ab}} \).
2. Multiplying, the factor \( \sqrt{a}-\sqrt{b} \) cancels: \( \frac{(\sqrt{a}+\sqrt{b})^2}{\sqrt{ab}} = \frac{a + b + 2\sqrt{ab}}{\sqrt{ab}} \).
3. Multiply top and bottom by \( \sqrt{ab} \): \( \frac{(a+b)\sqrt{ab} + 2ab}{ab} \).

Teacher's Note:
1. Write the first bracket as one fraction and factor the difference of squares.
2. Rationalise only at the end.

Answer: (b) \( \frac{2 + \sqrt{2} + \sqrt{6}}{4} \)

Explanation:
1. Multiply by \( \frac{(1+\sqrt{2}) + \sqrt{3}}{(1+\sqrt{2}) + \sqrt{3}} \): the denominator becomes \( (1+\sqrt{2})^2 - 3 = 2\sqrt{2} \).
2. So the expression is \( \frac{1 + \sqrt{2} + \sqrt{3}}{2\sqrt{2}} \). Multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \).
3. \( \frac{\sqrt{2} + 2 + \sqrt{6}}{4} \).

Teacher's Note:
1. Rationalise a three-term denominator in two steps.
2. Numerical check: both sides are about 1.466.

Answer: (d) \( \frac{7\sqrt{6} + 6\sqrt{7} + \sqrt{546}}{84} \)

Explanation:
1. Multiply top and bottom by \( (\sqrt{7} + \sqrt{6}) + \sqrt{13} \): the denominator is \( (\sqrt{7}+\sqrt{6})^2 - 13 = 2\sqrt{42} \).
2. The expression becomes \( \frac{\sqrt{7} + \sqrt{6} + \sqrt{13}}{2\sqrt{42}} \). Multiply by \( \frac{\sqrt{42}}{\sqrt{42}} \): \( \frac{\sqrt{294} + \sqrt{252} + \sqrt{546}}{84} \).
3. \( \sqrt{294} = 7\sqrt{6} \) and \( \sqrt{252} = 6\sqrt{7} \), giving \( \frac{7\sqrt{6} + 6\sqrt{7} + \sqrt{546}}{84} \).

Teacher's Note:
1. Group two terms, use the difference of squares, then rationalise the remaining surd.
2. Numerical check: both sides are about 0.6713.

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

Explanation:
1. Multiply top and bottom by \( \sqrt{a+x} + \sqrt{a-x} \). The denominator becomes \( (a + x) - (a - x) = 2x \).
2. The numerator is \( (\sqrt{a+x} + \sqrt{a-x})^2 = 2a + 2\sqrt{a^2 - x^2} \).
3. So the result is \( \frac{2a + 2\sqrt{a^2 - x^2}}{2x} = \frac{a + \sqrt{a^2 - x^2}}{x} \).

Teacher's Note:
1. Use the conjugate of the denominator.
2. Check with a = 3, x = 1: both forms give about 5.83.

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

Explanation:
1. Multiply by the conjugate \( a\sqrt{1 + b^2} - b\sqrt{1 + a^2} \).
2. Denominator: \( a^2(1 + b^2) - b^2(1 + a^2) = a^2 - b^2 \).
3. So the result is \( \frac{a\sqrt{1 + b^2} - b\sqrt{1 + a^2}}{a^2 - b^2} \).

Teacher's Note:
1. The terms \( a^2b^2 \) cancel in the denominator.
2. Always check the sign of the conjugate.

Answer: (c) \( \frac{5x^2}{y} \)

Explanation:
1. \( (x^4y^8)^{-1/4} = x^{-1}y^{-2} \).
2. \( 5x^3y \cdot x^{-1}y^{-2} = 5x^{3-1}y^{1-2} = 5x^2y^{-1} = \frac{5x^2}{y} \).

Teacher's Note:
1. Multiply the exponents 4, 8 by \( -\frac{1}{4} \).
2. Add exponents of like bases when multiplying.

Answer: (a) \( 3a^3b^4 \)

Explanation:
1. \( (27a^3b^3)^{1/3} = 3ab \).
2. \( a^2b^3 \times 3ab = 3a^3b^4 \).

Teacher's Note:
1. Take the cube root of each factor, including 27.
2. Then add exponents.

Answer: (d) \( 2a^7 \)

Explanation:
1. \( (32a^{10}b^{-15})^{1/5} = 2a^2b^{-3} \).
2. \( a^5b^3 \times 2a^2b^{-3} = 2a^7b^0 = 2a^7 \).

Teacher's Note:
1. The fifth root of 32 is 2.
2. \( b^3 \times b^{-3} = 1 \).

Answer: (b) \( \frac{1}{c\sqrt{a}} \)

Explanation:
1. \( \sqrt{a^4b^{-5}} = a^2b^{-5/2} \) and \( (a^{-4}b^3c^{-1})^{1/2} = a^{-2}b^{3/2}c^{-1/2} \).
2. \( (c^{-2}a^{-2}b^4)^{-1/4} = c^{1/2}a^{1/2}b^{-1} \).
3. Product of the first two: \( a^{0}b^{-1}c^{-1/2} \). Dividing by the third: \( a^{-1/2}b^{-1+1}c^{-1/2-1/2} = a^{-1/2}c^{-1} \).
4. So the result is \( \frac{1}{c\sqrt{a}} \).

Teacher's Note:
1. Reduce each bracket to a product of powers, then add and subtract exponents of each letter.
2. Dividing by a power means subtracting its exponent.

Answer: (b) \( \frac{a^{11/4}c^8}{b^{1/3}} \)

Explanation:
1. \( \left(\frac{a^6}{b^5}\right)^{-1/3} = a^{-2}b^{5/3} \); \( \frac{(c^{-1/2})^4}{a^{-3/4}} = c^{-2}a^{3/4} \); \( (a^{-2}bc^{-5})^2 = a^{-4}b^2c^{-10} \).
2. Combine: \( a^{-2 + 3/4 + 4}\,b^{5/3 - 2}\,c^{-2 + 10} \).
3. \( = a^{11/4}\,b^{-1/3}\,c^{8} = \frac{a^{11/4}c^8}{b^{1/3}} \).

Teacher's Note:
1. Dividing by the last bracket flips the signs of its exponents.
2. A negative exponent moves the letter to the denominator.

Answer: (d) 0.949

Explanation:
1. \( 0.949^2 = 0.900601 \) and \( 0.948^2 = 0.898704 \); 0.9 lies between them and is closer to \( 0.949^2 \).
2. \( \sqrt{0.9} = 0.94868\ldots \), which rounds to 0.949.

Teacher's Note:
1. \( \sqrt{0.9} = \frac{\sqrt{90}}{10} \).
2. Compare the squares of 0.948 and 0.949 with 0.9.

Answer: (a) 2.694

Explanation:
1. \( 2.693^2 = 7.2522 \) and \( 2.694^2 = 7.2576 \), so the root lies between 2.693 and 2.694.
2. \( \sqrt{7.257} = 2.69388\ldots \), which rounds to 2.694.

Teacher's Note:
1. The fourth decimal 8 forces rounding up.
2. Check by squaring the answer.