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For natural numbers $$m$$ , $$n$$, if $${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\, = 1 + {a_1}y + {a_2}{y^2} + ..........$$ and $${a_1} = {a_2} = 10,$$ then $$\left( {m,\,n} \right)$$ is

JEE · Math · previous-year question

  1. A.$$\left( {20,\,45} \right)$$
  2. B.$$\left( {35,\,20} \right)$$
  3. C.$$\left( {45,\,35} \right)$$
  4. D.$$\left( {35,\,45} \right)$$correct

Answer

D. $$\left( {35,\,45} \right)$$

Explanation

$${\left( {1 - y} \right)^m}{\left( {1 + y} \right)^n}\,\,$$ = $$\left( {{}^m{C_0} - {}^m{C_1}y + {}^m{C_2}{y^2} + ....} \right)$$ - $$\left( {{}^n{C_0} + {}^n{C_1}y + {}^n{C_2}{y^2} + ....} \right)$$ $${a_1}$$ = Coefficient of y = $${{}^n{C_1}}$$ - $${{}^m{C_1}}$$ = 10 $$ \Rightarrow $$ n - m = 10 $${a_2}$$ = Coefficient of y2 = $${}^n{C_2} + {}^n{C_1} \times {}^m{C_1} + {}^m{C_2} = 10$$ $$ \Rightarrow {{n\left( {n - 1} \right)} \over 2} - nm + {{m\left( {m - 1} \right)} \over 2} = 10$$ $$ \Rightarrow n\left( {n - 1} \right) - 2nm + m\left( {m - 1} \right) = 20$$ $$ \Rightarrow$$ (m + 10)(m + 9) - 2(m + 10)m + m(m - 1) = 20 $$ \Rightarrow$$ 90 + 19m + m2 - 2m2 - 20m + m2 - m - 20 = 0 $$ \Rightarrow$$ 70 - 2m = 0 $$ \Rightarrow$$ m = 35 $$\therefore$$ n = 10 + 35 = 45

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