QJEEM23Mab6fe: JEE Main 2023 Mathematics - Class 11 JEEM23Mab6fe Solution: If the coefficients of and in are 4 and 5 respectively, then is …

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If the coefficients of $x$ and $x^{2}$ in $(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}$ are 4 and $-$ 5 respectively, then $2 p+3 q$ is equal to :

Accepted Answer

We have, coefficient of

x

in

(1+x)^p(1-x)^q=4

and

coefficient of

x^2

in

(1+x)^p(1-x)^q=-5

\begin{aligned}&amp; (1+x)^p(1-x)^q \\\\&amp; =\left(1+p x+\frac{p(p-1)}{2} x^2+\ldots\right)\left(1-q x+\frac{q(q-1)}{2} x^2+\ldots\right) \\\\&amp; =1+(p-q) x+\left(\frac{p(p-1)}{2}+\frac{q(q-1)}{2}-p q\right) x^2+\ldots \ldots\end{aligned}

Coefficient of

x

in

(1+x)^p(1-x)^q=-q+p

\Rightarrow p-q=4

...........(i)

\text { Coefficient of } x^2 \text { in }(1+x)^p(1-x)^q=\frac{q(q-1)}{2}-p q+\frac{p(p-1)}{2}

\begin{aligned}&amp; \Rightarrow \frac{q^2-q-2 p q+p^2-p}{2} =-5 \\\\&amp; \Rightarrow \frac{p^2+q^2-2 p q-(p+q)}{2} =-5 \\\\&amp; \Rightarrow (p-q)^2-(p+q) =-10 \\\\&amp; \Rightarrow (4)^2-(p+q) =-10 \quad[\because \text { From Eq. (i) }] \\\\&amp; \Rightarrow p+q =26 ...........(ii)\end{aligned}

Form Eqs. (i) and (ii), we get

p=15, q=11

\therefore 2 p+3 q=2 \times 15+3 \times 11=30+33=63

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