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Binomial Theorem question

2023 · 6 Apr · Shift 2 · Q24

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JEE Main
Mathematics
Binomial Theorem
2023 · 6 Apr · Shift 2 · Q24

Binomial Theorem question

2023 · 6 Apr · Shift 2 · Q24

JEE MainMathematicsBinomial TheoremMCQ+4 / −1

If the coefficient of

{x^7}

in

{\left( {a{x^2} + {1 \over {2bx}}} \right)^{11}}

and

{x^{ - 7}}

in

{\left( {ax - {1 \over {3b{x^2}}}} \right)^{11}}

are equal, then :

A
$243ab = 64$
B
$32ab = 729$
C
$64ab = 243$
D
$729ab = 32$

View written solutionFree

Correct answer: D

Find the coefficient of $x^7$ in $\left(ax^2+\frac{1}{2bx}\right)^{11}$

The general term is $T_{r+1}=\binom{11}{r}(ax^2)^{11-r}\left(\frac{1}{2bx}\right)^r.$

Simplifying: $T_{r+1}=\binom{11}{r}a^{11-r}\frac{1}{(2b)^r}x^{2(11-r)-r}$ $=\binom{11}{r}a^{11-r}\frac{1}{(2b)^r}x^{22-3r}.$

For the term containing $x^7$ , $22-3r=7$ $3r=15\Rightarrow r=5.$

So the coefficient of $x^7$ is $\binom{11}{5}a^{6}\frac{1}{(2b)^5}.$
Find the coefficient of $x^{-7}$ in $\left(ax-\frac{1}{3bx^2}\right)^{11}$

The general term is $T_{r+1}=\binom{11}{r}(ax)^{11-r}\left(-\frac{1}{3bx^2}\right)^r.$

Simplifying: $T_{r+1}=\binom{11}{r}a^{11-r}\frac{(-1)^r}{(3b)^r}x^{11-r-2r}$ $=\binom{11}{r}a^{11-r}\frac{(-1)^r}{(3b)^r}x^{11-3r}.$

For the term containing $x^{-7}$ , $11-3r=-7$ $3r=18\Rightarrow r=6.$

So the coefficient of $x^{-7}$ is
$\binom{11}{5}a^6\frac{1}{(2b)^5}=\binom{11}{6}\frac{a^5}{(3b)^6}.$
Equate the two coefficients

Since they are equal, $\binom{11}{5}a^6\frac{1}{(2b)^5}=\binom{11}{6}\frac{a^5}{(3b)^6}.$

Now, $\binom{11}{5}=\binom{11}{6},$ so they cancel.

Thus, $\frac{a^6}{(2b)^5}=\frac{a^5}{(3b)^6}.$

Cancel $a^5$ : $\frac{a}{(2b)^5}=\frac{1}{(3b)^6}.$

Cross-multiplying: $a(3b)^6=(2b)^5.$

$a\cdot 3^6 b^6=2^5 b^5$ $729ab=32.$
Match with the options

This is option D.
Compare with stored correct answer

Stored correct answer: D

Our derived answer: D

Hence, the stored answer is correct.

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