JEE Main 2022 · 28 Jun · Shift 2 · Mathematics Q4

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MathematicsSingle correct+4 / -1Question 4 of 90

The term independent of $x$ in the expansion of $\left(1-x^2+3x^3\right)\left(\dfrac52x^3-\dfrac1{5x^2}\right)^{11}$ , $x\ne0$ is :

\dfrac7{40}

\dfrac{33}{200}

\dfrac{39}{200}

\dfrac{11}{50}

Answer and worked solution

Official answer: B

General term of the bracket: ${}^{11}C_r\left(\frac52\right)^{11-r}\left(-\frac15\right)^rx^{33-5r}$ . Exponents $0$ and $-3$ are impossible; $-2$ needs $r=7$ , which pairs with $-x^2$ .

Term. $-{}^{11}C_7\left(\frac52\right)^4\left(-\frac15\right)^7=\dfrac{330\cdot625}{16\cdot78125}=\dfrac{33}{200}$ .

(B)

Topics:General and middle terms

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JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 4