JEE Mathematics Surds and Indices MCQs Set 02

Answer: (a) 125

Explanation:
1. \( \frac{a^m}{a^n}=a^{m-n} \).
2. \( 5^{7-4}=5^3=125 \).

Teacher's Note:
1. Subtract the indices when dividing.
2. \( 25^3=5^6 \) not \( 5^3 \).

Answer: (d) 3

Explanation:
1. For any non-zero a, \( a^0=1 \).
2. \( 1+1+1=3 \).

Teacher's Note:
1. The law \( \frac{a^n}{a^n}=a^{n-n}=a^0 \) with the left side equal to 1 proves it.
2. The sum of three ones is 3.

Answer: (b) \( \frac18 \)

Explanation:
1. \( a^{-n}=\frac{1}{a^n} \).
2. \( 2^{-3}=\frac1{2^3}=\frac18 \).

Teacher's Note:
1. A negative index means reciprocal, not a negative number.
2. -6 would come from multiplying 2 by -3.

Answer: (b) 1

Explanation:
1. \( (ab)^n=a^nb^n \), so \( 20^3=(4\times5)^3=4^3\times5^3=2^6\times5^3 \).
2. The numerator equals the denominator, so the value is 1.

Teacher's Note:
1. Write 20 as 4 x 5 and 4 as 2 squared.
2. \( 20^3=8000 \) and \( 2^6\cdot5^3=64\times125=8000 \).

Answer: (d) \( \frac{81}{16} \)

Explanation:
1. \( \left(\frac ab\right)^n=\frac{a^n}{b^n} \).
2. \( \frac{3^4}{2^4}=\frac{81}{16} \).

Teacher's Note:
1. Raise both numerator and denominator to the power.
2. 12/8 would come from multiplying by 4 instead.

Answer: (a) 4096

Explanation:
1. \( (a^m)^n=a^{mn} \).
2. \( (2^3)^4=2^{12}=4096 \).

Teacher's Note:
1. Multiply the indices for a power of a power.
2. Adding would give \( 2^7 \), which is the rule for products.

Answer: (c) 16

Explanation:
1. \( 64=4^3 \), so \( 64^{2/3}=4^{2}=16 \).

Teacher's Note:
1. Take the cube root first, then square.
2. 128/3 would come from multiplying 64 by 2/3.

Answer: (d) \( \frac{9}{3125} \)

Explanation:
1. Numerator base: \( 5^3\cdot2^6\cdot(2\cdot3)^2=5^3\cdot2^8\cdot3^2 \); cubed: \( 5^9\cdot2^{24}\cdot3^6 \).
2. Denominator: \( (5^7\cdot3^2\cdot2^{12})^2=5^{14}\cdot3^4\cdot2^{24} \).
3. Ratio \( =\frac{3^{6-4}}{5^{14-9}}=\frac{3^2}{5^5}=\frac{9}{3125} \).

Teacher's Note:
1. The powers of 2 cancel exactly: 2^24 on top and bottom.
2. 3125 = 5^5.

Answer: (a) 27

Explanation:
1. \( 2^x=2^{3(y+1)} \) gives \( x=3y+3 \).
2. \( 3^{2y}=3^{x-9} \) gives \( 2y=x-9 \).
3. Substitute: \( 2y=3y-6 \), so y = 6 and x = 21.
4. x + y = 27.

Teacher's Note:
1. Reduce both sides to the same base before equating the indices.
2. Check: 2^21 = 8^7 and 9^6 = 3^12 = 3^(21-9).

Answer: (c) \( \frac78 \)

Explanation:
1. Divide numerator and denominator by \( 2^n \).
2. Numerator: \( 2^4-2=14 \); denominator: \( 2\cdot2^3=16 \).
3. Value \( =\frac{14}{16}=\frac78 \).

Teacher's Note:
1. Factor out the lowest power 2^n.
2. The value does not depend on n.

Answer: (c) 2

Explanation:
1. \( 125=5^3 \), so \( x+1=3 \).
2. x = 2.

Teacher's Note:
1. Do not equate x to 3; the index is x + 1.
2. Check: 5^3 = 125.

Answer: (a) -3

Explanation:
1. \( \frac1{64}=4^{-3} \).
2. So x = -3.

Teacher's Note:
1. 64 is 4 cubed.
2. Reciprocal means a negative index.

Answer: (d) 9

Explanation:
1. \( \left(\frac1{27}\right)^{-2/3}=27^{2/3} \).
2. \( 27^{2/3}=(3^3)^{2/3}=9 \).

Teacher's Note:
1. A negative index flips the fraction.
2. Take the cube root first: 3, then square it.

Answer: (b) 1

Explanation:
1. Each bracket simplifies: \( x^{(a-b)(a+b)}\,x^{(b-c)(b+c)}\,x^{(c-a)(c+a)} \).
2. Total index: \( (a^2-b^2)+(b^2-c^2)+(c^2-a^2)=0 \).
3. So the value is \( x^0=1 \).

Teacher's Note:
1. The indices telescope to zero.
2. Any non-zero x to the power 0 is 1.

Answer: (c) 4

Explanation:
1. Factor out \( 3^x \): \( 3^x\left(\frac13+3\right)=270 \).
2. \( 3^x\cdot\frac{10}{3}=270 \), so \( 3^x=81 \) and x = 4.

Teacher's Note:
1. Take the smallest power out as a common factor.
2. Check: 3^3 + 3^5 = 27 + 243 = 270.

Answer: (c) 8

Explanation:
1. \( 2^{10}\times5^7=2^3\times(2^7\times5^7)=8\times10^7 \).
2. 8 followed by seven zeros has 8 digits.

Teacher's Note:
1. Pair equal numbers of 2s and 5s to make powers of 10.
2. Remaining factor is 2^3 = 8.

Answer: (a) 8

Explanation:
1. \( \left(\frac14\right)^{-3/2}=4^{3/2} \).
2. \( 4^{3/2}=(\sqrt4)^3=8 \).

Teacher's Note:
1. Flip the fraction for a negative index.
2. Square root first, then cube.

Answer: (d) 0

Explanation:
1. Let \( a^x=b^y=c^z=k \). Then \( a=k^{1/x},\ b=k^{1/y},\ c=k^{1/z} \).
2. \( abc=k^{\frac1x+\frac1y+\frac1z}=1 \).
3. Since \( k\ne1 \), the index must be 0.

Teacher's Note:
1. Express each of a, b, c as a power of the common value.
2. Product of powers adds the indices.

Answer: (b) \( 3^{1/3} \)

Explanation:
1. Raise each to the power 6 (the L.C.M. of 2, 3, 6).
2. \( (2^{1/2})^6=8 \), \( (3^{1/3})^6=9 \), \( (6^{1/6})^6=6 \).
3. 9 is largest, so \( 3^{1/3} \) is largest.

Teacher's Note:
1. Bring the numbers to a common index to compare roots.
2. Raising to a positive power preserves the order of positive numbers.

Answer: (d) \( \sqrt{x} \)

Explanation:
1. Multiply the indices: \( \frac23\times\frac34=\frac12 \).
2. \( x^{1/2}=\sqrt x \).

Teacher's Note:
1. Power of a power means multiply the indices.
2. 17/12 comes from adding 2/3 and 3/4 by mistake.