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Mathematical Induction and Binomial Theorem question

2023 · Shift 1 · Q30

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JEE Advanced
Mathematics
Mathematical Induction and Binomial Theorem
2023 · Shift 1 · Q30

Mathematical Induction and Binomial Theorem question

2023 · Shift 1 · Q30

JEE AdvancedMathematicsMathematical Induction and Binomial TheoremNumerical+4 / −1

Let

a

and

b

be two nonzero real numbers. If the coefficient of

x^5

in the expansion of

\left(a x^2+\frac{70}{27 b x}\right)^4

is equal to the coefficient of

x^{-5}

in the expansion of

\left(a x-\frac{1}{b x^2}\right)^7

, then the value of

2 b

is :

Numerical answer

View written solutionFree

Correct answer: 3

Coefficient of $x^5$ in $\left(a x^2+\frac{70}{27 b x}\right)^4$

General term is $T_{r+1}=\binom{4}{r}(a x^2)^{4-r}\left(\frac{70}{27bx}\right)^r$

Simplifying the power of $x$ : $x^{2(4-r)}\cdot x^{-r}=x^{8-3r}$

We need $8-3r=5 \implies r=1$

So the required coefficient is $\binom{4}{1}a^{3}\left(\frac{70}{27b}\right)=4a^3\cdot \frac{70}{27b}=\frac{280a^3}{27b}$
Coefficient of $x^{-5}$ in $\left(ax-\frac{1}{bx^2}\right)^7$

General term is $T_{r+1}=\binom{7}{r}(ax)^{7-r}\left(-\frac{1}{bx^2}\right)^r$

Power of $x$ is $x^{7-r}\cdot x^{-2r}=x^{7-3r}$

We need $7-3r=-5 \implies 3r=12 \implies r=4$

Hence the required coefficient is $\binom{7}{4}a^{3}\left(\frac{1}{b^4}\right)(-1)^4$ $=35\frac{a^3}{b^4}$
Equate the two coefficients

Given, $\frac{280a^3}{27b}=\frac{35a^3}{b^4}$

Since $a\neq 0$ , cancel $a^3$ : $\frac{280}{27b}=\frac{35}{b^4}$

Multiply both sides by $27b^4$ : $280b^3=35\cdot 27$

Divide by $35$ : $8b^3=27$

$b^3=\frac{27}{8}$

$b=\frac{3}{2}$
Find $2b$ $2b=2\cdot \frac{3}{2}=3$

Final Answer: $3$

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