Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Binomial Theorem question

2023 · 30 Jan · Shift 1 · Q27
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.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Binomial Theorem
  5. /2023 · 30 Jan · Shift 1 · Q27

Binomial Theorem question

2023 · 30 Jan · Shift 1 · Q27

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If the coefficient of x15x^{15}x15 in the expansion of (ax3+1 bx1/3)15\left(\mathrm{a} x^{3}+\frac{1}{\mathrm{~b} x^{1 / 3}}\right)^{15}(ax3+ bx1/31​)15 is equal to the coefficient of x−15x^{-15}x−15 in the expansion of (ax1/3−1bx3)15\left(a x^{1 / 3}-\frac{1}{b x^{3}}\right)^{15}(ax1/3−bx31​)15, where aaa and bbb are positive real numbers, then for each such ordered pair (a,b)(\mathrm{a}, \mathrm{b})(a,b) :
  1. A
    a = 3b
  2. B
    ab = 1
  3. C
    ab = 3
  4. D
    a = b
View written solutionFree

Correct answer: B

  1. General term of the first expansion

Consider

(ax3+1bx1/3)15.\left(ax^3+\frac{1}{bx^{1/3}}\right)^{15}.(ax3+bx1/31​)15.

Its general term is

Tr+1=(15r)(ax3)15−r(1bx1/3)r,r=0,1,2,…,15.T_{r+1}=\binom{15}{r}(ax^3)^{15-r}\left(\frac{1}{bx^{1/3}}\right)^r, \qquad r=0,1,2,\dots,15.Tr+1​=(r15​)(ax3)15−r(bx1/31​)r,r=0,1,2,…,15.

So,

Tr+1=(15r)a15−rb−rx3(15−r)−r/3.T_{r+1}=\binom{15}{r}a^{15-r}b^{-r}x^{3(15-r)-r/3}.Tr+1​=(r15​)a15−rb−rx3(15−r)−r/3.

Exponent of xxx is

45−3r−r3=45−10r3.45-3r-\frac r3=45-\frac{10r}{3}.45−3r−3r​=45−310r​.

We need coefficient of x15x^{15}x15, so

45−10r3=15.45-\frac{10r}{3}=15.45−310r​=15.

Thus,

10r3=30  ⟹  10r=90  ⟹  r=9.\frac{10r}{3}=30 \implies 10r=90 \implies r=9.310r​=30⟹10r=90⟹r=9.

Hence the required coefficient is

\binom{15}{9}a^{15-9}b^{-9}=inom{15}{9}\frac{a^6}{b^9}.
  1. General term of the second expansion

Now consider

(ax1/3−1bx3)15.\left(ax^{1/3}-\frac{1}{bx^3}\right)^{15}.(ax1/3−bx31​)15.

Its general term is

Tr+1=(15r)(ax1/3)15−r(−1bx3)r.T_{r+1}=\binom{15}{r}(ax^{1/3})^{15-r}\left(-\frac{1}{bx^3}\right)^r.Tr+1​=(r15​)(ax1/3)15−r(−bx31​)r.

So,

Tr+1=(15r)(−1)ra15−rb−rx(15−r)/3−3r.T_{r+1}=\binom{15}{r}(-1)^r a^{15-r}b^{-r}x^{(15-r)/3-3r}.Tr+1​=(r15​)(−1)ra15−rb−rx(15−r)/3−3r.

Exponent of xxx is

15−r3−3r=5−r3−3r=5−10r3.\frac{15-r}{3}-3r=5-\frac r3-3r=5-\frac{10r}{3}.315−r​−3r=5−3r​−3r=5−310r​.

We need coefficient of x−15x^{-15}x−15, so

5−10r3=−15.5-\frac{10r}{3}=-15.5−310r​=−15.

Thus,

10r3=20  ⟹  10r=60  ⟹  r=6.\frac{10r}{3}=20 \implies 10r=60 \implies r=6.310r​=20⟹10r=60⟹r=6.

Hence the required coefficient is

\binom{15}{6}(-1)^6 a^{15-6}b^{-6}=inom{15}{6}\frac{a^9}{b^6}.

Since (−1)6=1(-1)^6=1(−1)6=1.

  1. Equate the coefficients

Given these two coefficients are equal:

(159)a6b9=(156)a9b6.\binom{15}{9}\frac{a^6}{b^9}=\binom{15}{6}\frac{a^9}{b^6}.(915​)b9a6​=(615​)b6a9​.

Now,

(159)=(156),\binom{15}{9}=\binom{15}{6},(915​)=(615​),

so they cancel out. Therefore,

a6b9=a9b6.\frac{a^6}{b^9}=\frac{a^9}{b^6}.b9a6​=b6a9​.

Multiply by b9b^9b9:

a6=a9b3.a^6=a^9b^3.a6=a9b3.

Since a,b>0a,b>0a,b>0, divide by a6a^6a6:

1=a3b3=(ab)3.1=a^3b^3=(ab)^3.1=a3b3=(ab)3.

Hence,

ab=1.ab=1.ab=1.
  1. Check options
  • A: a=3ba=3ba=3b — not necessary from the condition, so false.
  • B: ab=1ab=1ab=1 — true.
  • C: ab=3ab=3ab=3 — false.
  • D: a=ba=ba=b — not necessary; for example a=2,b=12a=2, b=\tfrac12a=2,b=21​ works, so false.

Therefore the correct option is:

B\boxed{B}B​
PreviousNext

More from Binomial Theorem

  • Let x=(83​+13)13 and y=(72​+9)9. If [t] denotes the greatest integer ≤t, then :2023 · MCQ
  • 50th  root of a number x is 12 and 50th  root of another number y is 18 . Then the remainder obtained on dividing (x+y) by 25 is ​.2023 · Numerical
  • The remainder on dividing 599 by 11 is ​.2023 · Numerical
  • Let α>0, be the smallest number such that the expansion of (x32​+x32​)30 has a term βx−α,β∈N. Then α is equal to ​.2023 · Numerical
  • The coefficient of x−6, in the expansion of (54x​+2x25​)9, is2023 · Numerical
  • If the constant term in the binomial expansion of (2x25​​−xl4​)9 is −84 and the coefficient of x−3l is 2αβ, where β<0 is an odd number, then ∣αl−β∣…2023 · Numerical
  • The remainder when 32022 is divided by 5 is :2022 · MCQ
  • The remainder on dividing 1 + 3 + 32 + 33 + ..... + 32021 by 50 is ​.2022 · Numerical