CBSE Class 11 Mathematics Logarithms Competitive Exam MCQs Set 02

Answer: (a) The maximum value of x is \( \frac{1}{\sqrt{10}} \); (b) x lies between \( \frac{1}{100} \) and \( \frac{1}{\sqrt{10}} \); (d) The minimum value of x is \( \frac{1}{100} \)

Explanation:
1. Base 0.1 < 1, so the inequalities reverse: \( (0.1)^2 \le x \le (0.1)^{1/2} \).
2. \( \frac{1}{100} \le x \le \frac{1}{\sqrt{10}} \).
3. So the maximum is 1/√10, the minimum is 1/100 and x lies between them: (a), (b), (d) are correct.

Teacher's Note:
1. For a base between 0 and 1, larger logarithms correspond to smaller numbers.
2. Read each option separately; several can be true together.

Answer: (c) [2, +∞)

Explanation:
1. \( \log_{0.04} (x - 1) = \log_{(0.2)^2} (x - 1) = \frac{1}{2}\log_{0.2} (x - 1) \).
2. So \( \frac{1}{2}\log_{0.2}(x - 1) \ge \log_{0.2}(x - 1) \Rightarrow \log_{0.2}(x - 1) \le 0 \).
3. Base 0.2 < 1 ⇒ x − 1 ≥ 1 ⇒ x ≥ 2, i.e. x ∈ [2, ∞).

Teacher's Note:
1. Convert both logs to the same base (0.04 = 0.2²) before comparing.
2. log_a t ≤ 0 with 0 < a < 1 means t ≥ 1.

Answer: (a) \( \left(-\infty, -\frac{5}{2}\right] \cup (0, +\infty) \)

Explanation:
1. Base 0.2 < 1, so \( \frac{x + 2}{x} \ge 0.2 \) (this also makes the argument positive).
2. \( \frac{x + 2}{x} - \frac{1}{5} \ge 0 \Rightarrow \frac{4x + 10}{5x} \ge 0 \Rightarrow x \le -\frac{5}{2} \) or x > 0.
3. Solution set: \( \left(-\infty, -\frac{5}{2}\right] \cup (0, \infty) \).

Teacher's Note:
1. Never multiply an inequality by x without knowing its sign; bring everything to one side instead.
2. x = 0 is excluded since the fraction is undefined there.

Answer: (c) \( n^m \)

Explanation:
1. \( a^{m\log_a n} = a^{\log_a n^m} \).
2. Since \( a^{\log_a t} = t \), the value is \( n^m \).

Teacher's Note:
1. Move the multiplier m inside the logarithm as a power before using a^(log_a t) = t.
2. Check with numbers: 2^(3 log₂ 5) = 5³ = 125.

Answer: (c) \( n^{1/(n-1)} \)

Explanation:
1. \( (a^m)^n = a^{mn} \), so \( mn = m^n \).
2. \( m^{n-1} = n \Rightarrow m = n^{1/(n-1)} \).

Teacher's Note:
1. (aᵐ)ⁿ = a^(mn) is different from a^(mⁿ); equating the exponents gives the relation.
2. Divide by m (m ≠ 0) before taking the root.

Answer: (d) None of these

Explanation:
1. \( (x^5)^{1/3} = x^{5/3} \); \( (16x^3)^{2/3} = 2^{8/3}x^2 \); \( \left(\frac{1}{4}x^{4/9}\right)^{-3/2} = 8x^{-2/3} \).
2. Product \( = 8 \cdot 2^{8/3} \cdot x^{\frac{5}{3} + 2 - \frac{2}{3}} = 2^{17/3}x^3 = 32\sqrt[3]{4}\,x^3 \).
3. This matches none of (a), (b), (c).

Teacher's Note:
1. Handle the numerical part and the power of x separately.
2. 16^(2/3) is not an integer, which already rules out (a), (b), (c).

Answer: (b) 2y

Explanation:
1. Let each be k: \( a = k^x,\ b = k^y,\ c = k^z \).
2. \( b^2 = ac \Rightarrow k^{2y} = k^{x + z} \Rightarrow x + z = 2y \).

Teacher's Note:
1. Setting equal powers to a common k is the standard trick for such chains.
2. b² = ac means a, b, c are in G.P., so the exponents are in A.P.

Answer: (d) x + y + z + 2

Explanation:
1. \( a^{x+1} = abc \Rightarrow \frac{1}{x + 1} = \log_{abc} a \); similarly for y, z.
2. \( \frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} = 1 \).
3. Clearing denominators: (y + 1)(z + 1) + (x + 1)(z + 1) + (x + 1)(y + 1) = (x + 1)(y + 1)(z + 1), which simplifies to xyz = x + y + z + 2.

Teacher's Note:
1. Multiplying both sides by the base makes all three right sides equal to abc.
2. Expand carefully; most terms cancel.

Answer: (a) x = y = z = a/3

Explanation:
1. Multiplying: \( a^{x+y+z} = (x + y + z)^{x+y+z} \Rightarrow a = x + y + z \).
2. Then \( a^x = a^y \), \( a^y = a^z \) ⇒ x = y = z (a ≠ 1).
3. So x = y = z = a/3.

Teacher's Note:
1. Multiplying all the equations often gives a symmetric relation.
2. Once the bases agree, equal powers force equal exponents.

Answer: (a) 1

Explanation:
1. \( a^{x-1} = bc \Rightarrow a^x = abc \Rightarrow \frac{1}{x} = \log_{abc} a \).
2. Similarly \( \frac{1}{y} = \log_{abc} b,\ \frac{1}{z} = \log_{abc} c \).
3. Sum \( = \log_{abc} abc = 1 \).

Teacher's Note:
1. Multiplying by a on both sides makes every equation have the same right side abc.
2. Reciprocal of an exponent becomes a logarithm with base abc.

Answer: (d) 2n

Explanation:
1. Exponent of 2: m(n + 1) + 2n + n − n(m + 1) − 2m = 0.
2. mn + m + 3n − mn − n − 2m = 0 ⇒ 2n − m = 0.
3. m = 2n.

Teacher's Note:
1. When everything is a power of 2, equate the total exponent to 0.
2. Expand the brackets carefully; the mn terms cancel.

Answer: (b) 1

Explanation:
1. \( x^y = y^x \Rightarrow y = x^{y/x} \Rightarrow \frac{x}{y} = \frac{x}{x^{y/x}} = x^{1 - y/x} \).
2. \( (x/y)^{x/y} = x^{(1 - y/x)(x/y)} = x^{x/y - 1} \).
3. So k = 1.

Teacher's Note:
1. Express y as a power of x using the given relation.
2. Then use (aᵖ)^q = a^(pq).

Answer: (a) 1; (d) 64/27

Explanation:
1. \( x^{x^{4/3}} = x^{4x/3} \).
2. If x = 1 both sides are 1, so x = 1 works.
3. Otherwise \( x^{4/3} = \frac{4x}{3} \Rightarrow x^{1/3} = \frac{4}{3} \Rightarrow x = \frac{64}{27} \).
4. x = 0 is not allowed and x = −1 gives −1 ≠ 1, so the answers are 1 and 64/27.

Teacher's Note:
1. In equations of the type f(x)^g(x) = f(x)^h(x), check the base value 1 separately.
2. Equate exponents only when the base is positive and not 1.

Answer: (b) 1

Explanation:
1. Let each be k: \( a = k^{1/x},\ b = k^{1/y},\ ab = k^{1/(xy)} \).
2. \( k^{1/x + 1/y} = k^{1/(xy)} \Rightarrow \frac{x + y}{xy} = \frac{1}{xy} \Rightarrow x + y = 1 \).

Teacher's Note:
1. Put the common value equal to k and express a, b, ab as powers of k.
2. Multiplying a and b adds the exponents.

Answer: (c) 3

Explanation:
1. \( x^3 = 2 - \frac{1}{2} - 3 \cdot 2^{1/3} \cdot 2^{-1/3}(2^{1/3} - 2^{-1/3}) = \frac{3}{2} - 3x \).
2. \( 2x^3 = 3 - 6x \Rightarrow 2x^3 + 6x = 3 \).

Teacher's Note:
1. Use (a − b)³ = a³ − b³ − 3ab(a − b); here ab = 1.
2. The bracket a − b is x itself, which is why x reappears.

Answer: (b) 2

Explanation:
1. \( x - 2 = 2^{2/3} + 2^{1/3} \). Cube: \( (x - 2)^3 = 4 + 2 + 3 \cdot 2^{2/3} \cdot 2^{1/3}(2^{2/3} + 2^{1/3}) = 6 + 6(x - 2) \).
2. \( x^3 - 6x^2 + 12x - 8 = 6x - 6 \).
3. \( x^3 - 6x^2 + 6x = 2 \).

Teacher's Note:
1. Isolate the surd part before cubing.
2. (a + b)³ = a³ + b³ + 3ab(a + b); here ab = 2.

Answer: (a) 9/4; (b) 1

Explanation:
1. \( x^{x^{3/2}} = x^{3x/2} \) with x > 0.
2. x = 1 satisfies it (both sides 1).
3. Otherwise \( x^{3/2} = \frac{3x}{2} \Rightarrow \sqrt{x} = \frac{3}{2} \Rightarrow x = \frac{9}{4} \).
4. √x is not real for x = −1 and x = 0 makes the expression undefined; answers 9/4 and 1.

Teacher's Note:
1. Base 1 must always be tested separately in such equations.
2. √x requires x ≥ 0, which removes the negative option.

Answer: (b) 1; (c) 3

Explanation:
1. \( 5(0.2)^{x-2} = 5 \cdot 5^{2-x} = 5^{3-x} \). Let \( t = 5^{x-1} \); then \( 5^{3-x} = \frac{25}{t} \).
2. \( t + \frac{25}{t} = 26 \Rightarrow t^2 - 26t + 25 = 0 \Rightarrow t = 1 \) or 25.
3. \( 5^{x-1} = 1 \Rightarrow x = 1 \); \( 5^{x-1} = 25 \Rightarrow x = 3 \).

Teacher's Note:
1. Write 0.2 as 5⁻¹ so that both terms use the same base.
2. A substitution t = 5^(x − 1) reduces the equation to a quadratic.

Answer: (a) 1

Explanation:
1. Multiply numerator and denominator of the first term by \( x^{-a} \): \( \frac{x^{-a}}{x^{-a} + x^{-b} + x^{-c}} \).
2. Similarly the other terms become \( \frac{x^{-b}}{x^{-a} + x^{-b} + x^{-c}} \) and \( \frac{x^{-c}}{x^{-a} + x^{-b} + x^{-c}} \).
3. Adding, the sum is 1.

Teacher's Note:
1. Making all denominators the same by multiplying by a suitable power is the key idea.
2. The cyclic sum means the other two terms are 1/(1 + x^(b−c) + x^(b−a)) and 1/(1 + x^(c−a) + x^(c−b)).

Answer: (a) A = 1; (c) C = 2; (d) D = 1

Explanation:
1. Let \( t = 2^{1/4} \). Then \( \frac{7}{t^2 + t + 1} = \frac{7(t - 1)}{t^3 - 1} = \frac{7(t - 1)(t^3 + 1)}{t^6 - 1} \).
2. \( t^6 - 1 = 2t^2 - 1 \), and \( (2t^2 - 1)(2t^2 + 1) = 4t^4 - 1 = 7 \), so the value is \( (t - 1)(t^3 + 1)(2t^2 + 1) \).
3. Using t⁴ = 2, this expands to \( 1 - 3t + 2t^2 + t^3 \).
4. So A = 1, B = −3, C = 2, D = 1: options (a), (c), (d) are correct.

Teacher's Note:
1. Rationalise in stages, using t⁴ = 2 to reduce higher powers.
2. Since 1, t, t², t³ are independent over rationals, the coefficients are unique.

Answer: (c) 3/2

Explanation:
1. \( 4 \cdot 3^{2x-2} = 3 \cdot 2^{x + \frac{1}{2}} \Rightarrow 3^{2x-3} = 2^{x - \frac{3}{2}} \).
2. \( 3^{2x-3} = 2^{\frac{1}{2}(2x - 3)} \Rightarrow (3/\sqrt{2})^{2x-3} = 1 \).
3. 2x − 3 = 0 ⇒ x = 3/2.

Teacher's Note:
1. Bring all powers of 3 to one side and powers of 2 to the other.
2. aᵗ = bᵗ with a ≠ b (positive) forces t = 0.

Answer: (b) \( \log_{(9/2)} (9/\sqrt{8}) \)

Explanation:
1. \( 9^x + \frac{9^x}{3} = 2^x(\sqrt{2} + 2\sqrt{2}) \Rightarrow \frac{4}{3} \cdot 9^x = 3\sqrt{2} \cdot 2^x \).
2. \( \left(\frac{9}{2}\right)^x = \frac{9\sqrt{2}}{4} = \frac{9}{\sqrt{8}} \).
3. \( x = \log_{(9/2)} (9/\sqrt{8}) \).

Teacher's Note:
1. Group powers of 9 on one side and powers of 2 on the other.
2. Dividing gives a single power (9/2)ˣ, solved by a logarithm.

Answer: (b) \( \frac{1}{2}[1 + \sqrt{4a + 1}] \)

Explanation:
1. Let y be the value; then \( y = \sqrt{a + y} \Rightarrow y^2 - y - a = 0 \).
2. \( y = \frac{1 \pm \sqrt{1 + 4a}}{2} \).
3. y > 0, so \( y = \frac{1}{2}[1 + \sqrt{4a + 1}] \).

Teacher's Note:
1. An infinite nested radical contains itself, which gives a quadratic.
2. Reject the negative root since a square root is non-negative.

Answer: (b) 7/13

Explanation:
1. \( \sqrt{4 \pm \sqrt{15}} = \frac{\sqrt{5} \pm \sqrt{3}}{\sqrt{2}} \), \( \sqrt{6 \pm \sqrt{35}} = \frac{\sqrt{7} \pm \sqrt{5}}{\sqrt{2}} \).
2. Numerator \( = \frac{(\sqrt{5} + \sqrt{3})^3 + (\sqrt{5} - \sqrt{3})^3}{2\sqrt{2}} = \frac{2(5\sqrt{5} + 9\sqrt{5})}{2\sqrt{2}} = \frac{14\sqrt{5}}{\sqrt{2}} \).
3. Denominator \( = \frac{(\sqrt{7} + \sqrt{5})^3 - (\sqrt{7} - \sqrt{5})^3}{2\sqrt{2}} = \frac{2(21\sqrt{5} + 5\sqrt{5})}{2\sqrt{2}} = \frac{26\sqrt{5}}{\sqrt{2}} \).
4. Ratio = 14/26 = 7/13.

Teacher's Note:
1. Write each a ± √b as a perfect square first; then the 3/2 power is a cube of a simple surd.
2. (p + q)³ ± (p − q)³ keeps only alternate terms of the expansion.

Answer: (b) \( \frac{1}{3}[56\sqrt{10} + 12] \)

Explanation:
1. \( x = \frac{(\sqrt{5} + \sqrt{2})^2}{3} = \frac{7 + 2\sqrt{10}}{3},\ y = \frac{7 - 2\sqrt{10}}{3} \), xy = 1.
2. \( x + y = \frac{14}{3},\ x - y = \frac{4\sqrt{10}}{3} \Rightarrow x^2 - y^2 = \frac{56\sqrt{10}}{9} \).
3. \( 3(x^2 - y^2) + 4xy = \frac{56\sqrt{10}}{3} + 4 = \frac{1}{3}[56\sqrt{10} + 12] \).

Teacher's Note:
1. x and y are reciprocals, so xy = 1 at once.
2. Group 3x² − 3y² as 3(x + y)(x − y).

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

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

Teacher's Note:
1. With three terms, rationalise in two steps: first remove √2, then √5.
2. Check numerically: both sides are about 4.48.

Answer: (a) 0

Explanation:
1. \( \sqrt{11 - 2\sqrt{30}} = \sqrt{6} - \sqrt{5} \), \( \sqrt{7 - 2\sqrt{10}} = \sqrt{5} - \sqrt{2} \), \( \sqrt{8 + 4\sqrt{3}} = \sqrt{6} + \sqrt{2} \).
2. Terms: \( \sqrt{6} + \sqrt{5} \), \( \sqrt{5} + \sqrt{2} \), \( \sqrt{6} - \sqrt{2} \).
3. \( (\sqrt{6} + \sqrt{5}) - (\sqrt{5} + \sqrt{2}) - (\sqrt{6} - \sqrt{2}) = 0 \).

Teacher's Note:
1. a − 2√(pq) with p + q = a has square root √p − √q (p > q).
2. 8 + 4√3 = 8 + 2√12, and 6 + 2 = 8, 6·2 = 12.

Answer: (a) Rational

Explanation:
1. \( \sqrt{7 - 3\sqrt{5}} = \sqrt{\frac{14 - 6\sqrt{5}}{2}} = \frac{3 - \sqrt{5}}{\sqrt{2}} \), so numerator \( = \frac{\sqrt{5} + 3 - \sqrt{5}}{\sqrt{2}} = \frac{3}{\sqrt{2}} \).
2. \( \sqrt{16 - 5\sqrt{7}} = \sqrt{\frac{32 - 10\sqrt{7}}{2}} = \frac{5 - \sqrt{7}}{\sqrt{2}} \), so denominator \( = \frac{5}{\sqrt{2}} \).
3. Value = 3/5, a rational number.

Teacher's Note:
1. Multiply inside the root by 2/2 to create the form a − 2√b.
2. The surd parts cancel completely, leaving a rational number.

Answer: (b) 1

Explanation:
1. \( \sqrt{2 \pm \sqrt{3}} = \frac{\sqrt{3} \pm 1}{\sqrt{2}} \).
2. Difference \( = \frac{2}{\sqrt{2}} = \sqrt{2} \).
3. Value \( = \frac{\sqrt{2}}{\sqrt{2}} = 1 \).

Teacher's Note:
1. √(2 ± √3) = (√3 ± 1)/√2 is a very common simplification.
2. Squaring (√3 ± 1)/√2 confirms it: (4 ± 2√3)/2.

Answer: (a) \( 2 + \sqrt{2} + \sqrt{6} \)

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

Teacher's Note:
1. Group (1 + √2) together and use (p − q)(p + q) = p² − q².
2. Multiplying by √2/√2 finishes the rationalisation.

Answer: (d) 0

Explanation:
1. \( \frac{3\sqrt{2}}{\sqrt{6} + \sqrt{3}} = \frac{3\sqrt{2}(\sqrt{6} - \sqrt{3})}{3} = 2\sqrt{3} - \sqrt{6} \).
2. \( \frac{4\sqrt{3}}{\sqrt{6} + \sqrt{2}} = \frac{4\sqrt{3}(\sqrt{6} - \sqrt{2})}{4} = 3\sqrt{2} - \sqrt{6} \); \( \frac{\sqrt{6}}{\sqrt{3} + \sqrt{2}} = \sqrt{6}(\sqrt{3} - \sqrt{2}) = 3\sqrt{2} - 2\sqrt{3} \).
3. Sum \( = (2\sqrt{3} - \sqrt{6}) - (3\sqrt{2} - \sqrt{6}) + (3\sqrt{2} - 2\sqrt{3}) = 0 \).

Teacher's Note:
1. Rationalise each term separately; the denominators become simple numbers.
2. Collect like surds at the end.

Answer: (c) (−1, 12)

Explanation:
1. \( \sqrt{7 + 4\sqrt{3}} = 2 + \sqrt{3} \).
2. \( \frac{4 + 3\sqrt{3}}{2 + \sqrt{3}} = (4 + 3\sqrt{3})(2 - \sqrt{3}) = 8 - 4\sqrt{3} + 6\sqrt{3} - 9 = -1 + 2\sqrt{3} \).
3. \( -1 + 2\sqrt{3} = -1 + \sqrt{12} \Rightarrow (a, b) = (-1, 12) \).

Teacher's Note:
1. 7 + 4√3 = 7 + 2√12 = (2 + √3)².
2. Write 2√3 as √12 to match the form √b.

Answer: (b) \( 2\sqrt{3} + \sqrt{7} \)

Explanation:
1. \( (2\sqrt{3} - \sqrt{7})(2\sqrt{3} + \sqrt{7}) = 12 - 7 = 5 \), which is rational.
2. So the rationalising factor is \( 2\sqrt{3} + \sqrt{7} \).

Teacher's Note:
1. The conjugate of a binomial quadratic surd is its rationalising factor.
2. Change only the sign between the two terms.

Answer: (b) \( 11 + \sqrt{13} \)

Explanation:
1. \( \sqrt{6292} = \sqrt{4 \cdot 121 \cdot 13} = 22\sqrt{13} \).
2. \( 134 + 22\sqrt{13} = 121 + 13 + 2 \cdot 11\sqrt{13} = (11 + \sqrt{13})^2 \).
3. Square root \( = 11 + \sqrt{13} \).

Teacher's Note:
1. Factorise the number under the inner root to bring it to the form 2pq.
2. Then look for p² + q² equal to the rational part.

Answer: (b) \( 2 - \sqrt{2} \)

Explanation:
1. \( 68 + 48\sqrt{2} = 36 + 32 + 2 \cdot 6 \cdot 4\sqrt{2} = (6 + 4\sqrt{2})^2 \).
2. \( 12 - (6 + 4\sqrt{2}) = 6 - 4\sqrt{2} = (2 - \sqrt{2})^2 \).
3. Value \( = 2 - \sqrt{2} \).

Teacher's Note:
1. Simplify the innermost root first.
2. 6 − 4√2 = 4 + 2 − 2·2·√2 = (2 − √2)².

Answer: (d) \( 2^{1/4}(\sqrt{3} + \sqrt{2}) \)

Explanation:
1. \( \sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3} = \sqrt{2}(5 + 2\sqrt{6}) \).
2. \( 5 + 2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2 \).
3. Square root \( = 2^{1/4}(\sqrt{3} + \sqrt{2}) \).

Teacher's Note:
1. Take out √2 so that the bracket becomes a perfect square.
2. The square root of √2 is 2^(1/4).

Answer: (b) \( \sqrt{5/2} - \sqrt{3/2} \)

Explanation:
1. \( \sqrt{3 + \sqrt{5}} = \frac{\sqrt{5} + 1}{\sqrt{2}} \), \( \sqrt{2 + \sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{2}} \).
2. Difference \( = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{2}} = \sqrt{5/2} - \sqrt{3/2} \).

Teacher's Note:
1. Multiply inside by 2/2: 3 + √5 = (6 + 2√5)/2 = (√5 + 1)²/2.
2. The 1/√2 terms cancel in the difference.

Answer: (c) \( 5^{1/4}[\sqrt{11} + 1] \)

Explanation:
1. \( 12\sqrt{5} + 2\sqrt{55} = \sqrt{5}(12 + 2\sqrt{11}) \).
2. \( 12 + 2\sqrt{11} = (\sqrt{11} + 1)^2 \).
3. Value \( = 5^{1/4}(\sqrt{11} + 1) \).

Teacher's Note:
1. Take out the common surd √5 first.
2. √(√5) = 5^(1/4).

Answer: (d) \( \sqrt{3} + \sqrt{2} \)

Explanation:
1. \( (\sqrt{3} + \sqrt{2})^3 = 3\sqrt{3} + 3 \cdot 3 \cdot \sqrt{2} + 3\sqrt{3} \cdot 2 + 2\sqrt{2} \).
2. \( = 9\sqrt{3} + 11\sqrt{2} \).
3. So the cube root is \( \sqrt{3} + \sqrt{2} \).

Teacher's Note:
1. Checking the options by cubing is often fastest for cube-root questions.
2. (a + b)³ = a³ + 3a²b + 3ab² + b³.