Visual SolutionPYQ 2024 · Jan (1 Feb) Shift 2Standard

Question

Let

m

and

n

be the coefficients of seventh and thirteenth terms respectively in the expansion of

\left(\frac{1}{3}x^{\frac{1}{3}} + \frac{1}{2x^{\frac{2}{3}}}\right)^{18}

. Then

\left(\frac{n}{m}\right)^{\frac{1}{3}}

is:

(A)

\frac{4}{9}

(B)

\frac{1}{9}

(C)

\frac{1}{4}

(D)

\frac{9}{4}

Solution Path

General term gives

m = \binom{18}{6}/(3^{12} \cdot 2^6)

n = \binom{18}{12}/(3^6 \cdot 2^{12})

. Since

\binom{18}{6} = \binom{18}{12}

n/m = (3/2)^6

(n/m)^{1/3} = 9/4

01Question Setup

1/4

Coefficients

m

(7th term) and

n

(13th term) in the expansion of

\left(\frac{x^{1/3}}{3} + \frac{1}{2x^{2/3}}\right)^{18}

. Find

\left(\frac{n}{m}\right)^{1/3}

\left(\frac{n}{m}\right)^{1/3} = \;?

02General Term

2/4

Apply the general term formula. Since

\binom{18}{6} = \binom{18}{12}

, the binomial coefficients cancel in the ratio.

T_{r+1} = \binom{18}{r} \cdot \frac{1}{3^{18-r} \cdot 2^r} \cdot x^{(18-3r)/3}

03RatioKEY INSIGHT

3/4

\frac{n}{m} = \left(\frac{3}{2}\right)^6

, so

\left(\frac{n}{m}\right)^{1/3} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}

\left(\frac{n}{m}\right)^{1/3} = \frac{9}{4}

04Final Answer

4/4

The ratio simplifies cleanly because the binomial coefficients are equal.

\boxed{\dfrac{9}{4}}

Concepts from this question1 concepts unlocked

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EASY

T(r+1) = C(n,r) * a^(n-r) * b^r gives any specific term without expanding fully

T_{r+1} = {}^nC_r\, a^{n-r}\, b^r

Core formula for finding specific terms, term independent of x, rational terms, etc. Used in 4-5 questions every JEE paper.

Specific termIndependent termRational/integral termsCoefficient finding

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