JEE Main 2022 · 29 Jun · Shift 1 · Mathematics Q16

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MathematicsSingle correct+4 / -1Question 16 of 90

If the constant term in the expansion of $\left(3x^3-2x^2+\dfrac5{x^5}\right)^{10}$ is $2^k\cdot l$ , where $l$ is an odd integer, then the value of $k$ is equal to :

Answer and worked solution

Official answer: D

Constant term. $(3x^3)^a(-2x^2)^b(5x^{-5})^c$ with $a+b+c=10$ and $3a+2b=5c$ , i.e. $8a+7b=50$ : only $(a,b,c)=(1,6,3)$ .

Value. $\dfrac{10!}{1!\,6!\,3!}\cdot3\cdot2^6\cdot5^3=840\cdot3\cdot64\cdot125=20160000=2^9\cdot39375$ .

$k=9$ .

(D)

Topics:General and middle terms Combinations

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JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 16