JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 7)
If the coefficients of x7 in $${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$$ and x$$-$$7 in $${\left( {{x} - {1 \over {bx^2}}} \right)^{11}}$$, b $$\ne$$ 0, are equal, then the value of b is equal to :
2
$$-$$1
1
$$-$$2
Explanation
Coefficient of x7 in $${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$$ :
General Term = $${}^{11}{C_r}{({x^2})^{11 - r}}.{\left( {{1 \over {bx}}} \right)^r}$$
= $${}^{11}{C_r}{x^{22 - 3r}}.{1 \over {{b^r}}}$$
$$22 - 3r = 7$$
$$r = 5$$
$$\therefore$$ Required Term = $${}^{11}{C_5}.{1 \over {{b^5}}}.{x^7}$$
Coefficient of x$$-$$7 in $${\left( {x - {1 \over {b{x^2}}}} \right)^{11}}$$ :
General Term = $${}^{11}{C_r}{(x)^{11 - r}}.{\left( { - {1 \over {b{x^2}}}} \right)^r}$$
= $${}^{11}{C_r}{x^{11 - 3r}}.{{{{( - 1)}^r}} \over {{b^r}}}$$
$$11 - 3r = - 7$$ $$\therefore$$ $$r = 6$$
$$ \therefore $$ Required Term = $${}^{11}{C_6}.{1 \over {{b^6}}}{x^{ - 7}}$$
According to the question,
$${}^{11}{C_5}.{1 \over {{b^5}}} = {}^{11}{C_6}.{1 \over {{b^6}}}$$
Since, b $$\ne$$ 0 $$\therefore$$ b = 1
General Term = $${}^{11}{C_r}{({x^2})^{11 - r}}.{\left( {{1 \over {bx}}} \right)^r}$$
= $${}^{11}{C_r}{x^{22 - 3r}}.{1 \over {{b^r}}}$$
$$22 - 3r = 7$$
$$r = 5$$
$$\therefore$$ Required Term = $${}^{11}{C_5}.{1 \over {{b^5}}}.{x^7}$$
Coefficient of x$$-$$7 in $${\left( {x - {1 \over {b{x^2}}}} \right)^{11}}$$ :
General Term = $${}^{11}{C_r}{(x)^{11 - r}}.{\left( { - {1 \over {b{x^2}}}} \right)^r}$$
= $${}^{11}{C_r}{x^{11 - 3r}}.{{{{( - 1)}^r}} \over {{b^r}}}$$
$$11 - 3r = - 7$$ $$\therefore$$ $$r = 6$$
$$ \therefore $$ Required Term = $${}^{11}{C_6}.{1 \over {{b^6}}}{x^{ - 7}}$$
According to the question,
$${}^{11}{C_5}.{1 \over {{b^5}}} = {}^{11}{C_6}.{1 \over {{b^6}}}$$
Since, b $$\ne$$ 0 $$\therefore$$ b = 1
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