JEE MAIN - Mathematics (2004 - No. 39)
The coefficient of the middle term in the binomial expansion in powers of $$x$$ of $${\left( {1 + \alpha x} \right)^4}$$ and $${\left( {1 - \alpha x} \right)^6}$$ is the same if $$\alpha $$ equals
$${3 \over 5}$$
$${10 \over 3}$$
$${{ - 3} \over {10}}$$
$${{ - 5} \over {3}}$$
Explanation
For $${\left( {1 + \alpha x} \right)^4}$$ the middle term $${T_{{4 \over 2} + 1}}$$ = $${}^4{C_2}.{\alpha ^2}{x^2}$$
$$\therefore$$ Coefficient of middle term = $${}^4{C_2}.{\alpha ^2}$$
For $${\left( {1 - \alpha x} \right)^6}$$ the middle term $${T_{{6 \over 2} + 1}}$$ = $${}^6{C_3}.-{\alpha ^3}{x^3}$$
$$\therefore$$ Coefficient of middle term = $${}^6{C_3}.{-\alpha ^3}$$
$$\therefore$$ According to question,
$${}^4{C_2}.{\alpha ^2}$$ = $${}^6{C_3}.{-\alpha ^3}$$
$$ \Rightarrow 6 = 20 \times - \alpha $$
$$ \Rightarrow \alpha = - {3 \over {10}}$$
$$\therefore$$ Coefficient of middle term = $${}^4{C_2}.{\alpha ^2}$$
For $${\left( {1 - \alpha x} \right)^6}$$ the middle term $${T_{{6 \over 2} + 1}}$$ = $${}^6{C_3}.-{\alpha ^3}{x^3}$$
$$\therefore$$ Coefficient of middle term = $${}^6{C_3}.{-\alpha ^3}$$
$$\therefore$$ According to question,
$${}^4{C_2}.{\alpha ^2}$$ = $${}^6{C_3}.{-\alpha ^3}$$
$$ \Rightarrow 6 = 20 \times - \alpha $$
$$ \Rightarrow \alpha = - {3 \over {10}}$$
Comments (0)
