New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Binomial Theorem question

2005 · Shift 0 · Q99

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Mathematics
Binomial Theorem
2005 · Shift 0 · Q99

Binomial Theorem question

2005 · Shift 0 · Q99

JEE MainMathematicsBinomial TheoremMCQ+4 / −1

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
$a - b = 1$
B
$a + b = 1$
C
${a \over b} = 1$
D
$ab = 1$

View written solutionFree

Correct answer: D

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

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

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

For coefficient of $x^7$ , $22-3r=7$ $3r=15 \implies r=5$

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

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

Simplifying, $T_{r+1}=\binom{11}{r}a^{11-r}(-1)^rb^{-r}x^{(11-r)-2r}$ $=\binom{11}{r}a^{11-r}(-1)^rb^{-r}x^{11-3r}$

For coefficient of $x^{-7}$ , $11-3r=-7$ $3r=18 \implies r=6$

Hence the coefficient of $x^{-7}$ is $\binom{11}{6}a^5(-1)^6b^{-6}=\binom{11}{6}\frac{a^5}{b^6}$
Equating the two coefficients

Given, $\binom{11}{5}\frac{a^6}{b^5}=\binom{11}{6}\frac{a^5}{b^6}$

Since $\binom{11}{5}=\binom{11}{6}$ we get $\frac{a^6}{b^5}=\frac{a^5}{b^6}$

Multiply both sides by $b^6$ : $a^6b=a^5$

Assuming nonzero $a,b$ (necessary because of terms involving $1/b$ ), divide by $a^5$ : $ab=1$
Checking options

A: $a-b=1$ ❌
B: $a+b=1$ ❌
C: $\dfrac{a}{b}=1$ ❌
D: $ab=1$ ✅

Therefore, the correct option is D.

More from Binomial Theorem

The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1 + α x)^{4}$ and $(1 - α x)^{6}$ is the same if $α$ equals2004 · MCQ
The coefficient of $x^{n}$ in expansion of $(1 + x) (1 - x)^{n}$ is2004 · MCQ
If $x$ is positive, the first negative term in the expansion of $(1 + x)^{27/5}$ is2003 · MCQ
The number of integral terms in the expansion of $(3 + 85)^{256}$ is2003 · MCQ
The coefficients of $x^{p}$ and $x^{q}$ in the expansion of $(1 + x)^{p + q}$ are2002 · MCQ
The positive integer just greater than $(1 + 0.0001)^{10000}$ is2002 · MCQ
The largest $n \in N$ such that $3^{n}$ divides 50 ! is :2025 · MCQ
The term independent of $x$ in the expansion of $(\frac{( x + 1 )}{( x ^{2/3} + 1 - x ^{1/3} )} - \frac{( x - 1 )}{( x - x ^{1/2} )})^{10}, x > 1$ , is :2025 · MCQ