JEE MAIN - Mathematics (2005 - No. 41)
If the coefficient of $${x^7}$$ in $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ equals the coefficient of $${x^{ - 7}}$$ in $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$, then $$a$$ and $$b$$ satisfy the relation
$$a - b = 1$$
$$a + b = 1$$
$${a \over b} = 1$$
$$ab = 1$$
Explanation
General term of $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ is Tr+1.
Tr+1 = $${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$$
= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$$
For the coefficient of x7,
$$ \Rightarrow $$ 22 - 3r = 7
$$ \Rightarrow $$ r = 5
So coefficient of x7 = $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ ......(1)
Now General term of $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$ is Tr+1.
Tr+1 = $${}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}$$
= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$$
For the coefficient of x-7,
11 - 3r = -7
$$ \Rightarrow $$ r = 6
$$\therefore$$ Coefficient of x-7 = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$
According to question,
Coefficient of x7 = Coefficient of x-7
$$ \Rightarrow $$ $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$
$$ \Rightarrow $$ $$ab = 1$$
Tr+1 = $${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$$
= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$$
For the coefficient of x7,
$$ \Rightarrow $$ 22 - 3r = 7
$$ \Rightarrow $$ r = 5
So coefficient of x7 = $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ ......(1)
Now General term of $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$ is Tr+1.
Tr+1 = $${}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}$$
= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$$
For the coefficient of x-7,
11 - 3r = -7
$$ \Rightarrow $$ r = 6
$$\therefore$$ Coefficient of x-7 = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$
According to question,
Coefficient of x7 = Coefficient of x-7
$$ \Rightarrow $$ $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$
$$ \Rightarrow $$ $$ab = 1$$
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