Visual SolutionPYQ 2024 · Apr 8 Shift 2Standard

Question

If the term independent of

x

in the expansion of

\left(\sqrt{a}\,x^2 + \frac{1}{2x^3}\right)^{10}

is 105, then

a^2

is equal to:

(A)

2

(B)

4

(C)

6

(D)

9

Solution Path

General term:

x^{20-5r}

, independent at

r=4

{}^{10}C_4 \cdot \frac{a^3}{16} = 105

gives

a^3 = 8

, so

a^2 = 4

01Question Setup

1/4

Find the term independent of

x

(\sqrt{a}\,x^2 + \frac{1}{2x^3})^{10} = 105

. Determine

a^2

T_{\text{indep}} = 105

02General Term

2/4

General term has

x^{20-5r}

. For independent term:

20 - 5r = 0

, so

r = 4

r = 4

03Solve for aKEY INSIGHT

3/4

T_5 = {}^{10}C_4 \cdot \frac{a^3}{16} = 210 \cdot \frac{a^3}{16} = 105

. So

a^3 = 8

a = 2

a^3 = 8 \Rightarrow a = 2

04Final Answer

4/4

a^2 = 4

\boxed{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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