JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 5)

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 :
66
60
69
63

Explanation

We have, coefficient of $x$ in $(1+x)^p(1-x)^q=4$ and

coefficient of $x^2$ in $(1+x)^p(1-x)^q=-5$

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

Coefficient of $x$ in $(1+x)^p(1-x)^q=-q+p$

$\Rightarrow p-q=4$ ...........(i)

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

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

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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