Q. For natural numbers $m,n$ if $(1 - y)^{m}(1 + y)^{n} = 1 + a_{1}y^{2} +$….and $a_{1} = a_{2} = 10,\$ then $(m,n)$is

A) (35,20)
B) (45,35)
C) (35,45)
D) (20,45)
Explanation

(c)
$(1 - y)^{m}(1 + y)^{n} = 1 + a_{1}y + a_{2}y^{2} + a_{3}y^{3} +$…
On differentiating w. r. t. $y$, we get
$- m({1 - y)}^{m - 1}(1 + y)^{n} + (1 - y)^{m}\ n(1 + y)^{n - 1}$
$\ \ \ \ \ \ \ \ = a_{1} + 2a_{2}y + 3a_{3}y^{2} +$… …(i)
On putting $y = 0$ in Eq. (i), we get
$- m + n = a_{1} = 10$ $\lbrack\because a_{1} = 10$ given] …(ii)
Again on differentiating Eq. (i) w. r. t. $y$, we get
${- m\lbrack - (m - 1)(1 - y)^{m - 2}(1 + y)^{n} + (1 + y)^{m - 1}n(1 + y)^{n - 1}\rbrack
}{+ n\lbrack - m({1 - y)}^{m - 1}(1 + y)^{n - 1} + (1 - y)^{m}(n - 1)(1 + y)^{n - 2}\rbrack}$
$\ \ \ \ \ \ \ \ = 2a_{2} + 6a_{3}y + ...$ …(iii)
On putting $y = 0\$in Eq. (iii), we get
$- m\left\lbrack - (m - 1) + n \right\rbrack + n\left\lbrack - m + (n - 1) \right\rbrack = 2a_{2} = 20$
$\Longrightarrow m(m - 1) - mn - mn + n(n - 1) = 20$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m^{2} + n^{2} - m - n - 2mn = 20$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (m - n)^{2} - (m + n) = 20$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 100 - (m + n) = 20$
[using Eq. (iii)]
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m + n = 80$ …(iv)
On solving Eqs. (ii) and (iv), we get
$m = 35\$and $n = 45$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.