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JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Binomial Theorem

If the term independent of x in the expansion of

(a x^{2} + \frac{1}{2 x ^{3}})^{10}

is 105 , then

a^{2}

is equal to :

(A)4✓ Correct
(B)9
(C)6
(D)2

Answer: (A)

Solution

General term of

(a x^{2} + \frac{1}{2 x ^{3}})^{10}

:

T_{r + 1} = (r 10) (a)^{10 - r} x^{2 (10 - r)} (\frac{1}{2 x ^{3}})^{r} = (r 10) \frac{a ^{(10 - r) /2}}{2 ^{r}} x^{20 - 5 r}

.
Independent of

x

:

20 - 5 r = 0 ⟹ r = 4

.

(4 10) \frac{a ^{3}}{2 ^{4}} = 105 ⟹ 210 \cdot \frac{a ^{3}}{16} = 105 ⟹ a^{3} = 8 ⟹ a = 2 ⟹ a^{2} = 4

JEE Main · 2024 · 08 Apr 2024 · Shift 2 · Single correct option

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