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Binomial Theorem question

2021 · 27 Jul · Shift 2 · Q24
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  5. /2021 · 27 Jul · Shift 2 · Q24

Binomial Theorem question

2021 · 27 Jul · Shift 2 · Q24

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
A possible value of 'x', for which the ninth term in the expansion of {3log⁡325x−1+7+3(−18)log⁡3(5x−1+1)}10{\left\{ {{3^{{{\log }_3}\sqrt {{{25}^{x - 1}} + 7} }} + {3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}}} \right\}^{10}}{3log3​25x−1+7​+3(−81​)log3​(5x−1+1)}10 in the increasing powers of 3(−18)log⁡3(5x−1+1){3^{\left( { - {1 \over 8}} \right){{\log }_3}({5^{x - 1}} + 1)}}3(−81​)log3​(5x−1+1) is equal to 180, is :
  1. A
    0
  2. B
    −-− 1
  3. C
    2
  4. D
    1
View written solutionFree

Correct answer: D

  1. Rewrite the given expression in simpler form

We have

{3log⁡325x−1+7+3(−18)log⁡3(5x−1+1)}10.\left\{3^{\log_3\sqrt{25^{x-1}+7}}+3^{\left(-\frac18\right)\log_3(5^{x-1}+1)}\right\}^{10}.{3log3​25x−1+7​+3(−81​)log3​(5x−1+1)}10.

Using the identity 3log⁡3A=A,3^{\log_3 A}=A,3log3​A=A, we get

Also,

So the expression becomes

{25x−1+7+(5x−1+1)−1/8}10.\left\{\sqrt{25^{x-1}+7}+(5^{x-1}+1)^{-1/8}\right\}^{10}.{25x−1+7​+(5x−1+1)−1/8}10.

Let

Then we need the 9th term in the expansion of (a+b)10(a+b)^{10}(a+b)10 in increasing powers of bbb.


  1. Find the 9th term

In the expansion of (a+b)10(a+b)^{10}(a+b)10 in increasing powers of bbb, the general term is

Tr+1=(10r)a10−rbr.T_{r+1}=\binom{10}{r}a^{10-r}b^r.Tr+1​=(r10​)a10−rbr.

The 9th term corresponds to r=8.r=8.r=8. Thus,

T9=(108)a2b8.T_9=\binom{10}{8}a^2b^8.T9​=(810​)a2b8.

Now, (108)=45.\binom{10}{8}=45.(810​)=45. So,

T9=45a2b8.T_9=45a^2b^8.T9​=45a2b8.
  1. Substitute aaa and bbb

We have a2=25x−1+7.a^2=25^{x-1}+7.a2=25x−1+7.

Also,

Hence,

T9=45⋅25x−1+75x−1+1.T_9=45\cdot \frac{25^{x-1}+7}{5^{x-1}+1}.T9​=45⋅5x−1+125x−1+7​.

Since 25x−1=(5x−1)225^{x-1}=(5^{x-1})^225x−1=(5x−1)2, let y=5x−1.y=5^{x-1}.y=5x−1. Then 25x−1=y2.25^{x-1}=y^2.25x−1=y2. So,

T9=45⋅y2+7y+1.T_9=45\cdot \frac{y^2+7}{y+1}.T9​=45⋅y+1y2+7​.

We are given that this equals 180180180:

45⋅y2+7y+1=180.45\cdot \frac{y^2+7}{y+1}=180.45⋅y+1y2+7​=180.

Divide by 454545:

y2+7y+1=4.\frac{y^2+7}{y+1}=4.y+1y2+7​=4.

Therefore,

y2+7=4y+4.y^2+7=4y+4.y2+7=4y+4.

So,

y2−4y+3=0.y^2-4y+3=0.y2−4y+3=0.

Factorizing:

(y−1)(y−3)=0.(y-1)(y-3)=0.(y−1)(y−3)=0.

Thus,

Recall that y=5x−1.y=5^{x-1}.y=5x−1. So we get:

  • If 5x−1=15^{x-1}=15x−1=1, then x−1=0x-1=0x−1=0, hence x=1.x=1.x=1.
  • If 5x−1=35^{x-1}=35x−1=3, then x=1+log⁡53,x=1+\log_5 3,x=1+log5​3, which is not among the options.

Thus, the possible value from the given options is 1.\boxed{1}.1​.


  1. Check options
  • A: x=0x=0x=0 gives 5x−1=5−1=155^{x-1}=5^{-1}=\frac155x−1=5−1=51​, not a solution.
  • B: x=−1x=-1x=−1 gives 5x−1=5−2=1255^{x-1}=5^{-2}=\frac1{25}5x−1=5−2=251​, not a solution.
  • C: x=2x=2x=2 gives 5x−1=55^{x-1}=55x−1=5, not a solution.
  • D: x=1x=1x=1 works.

So the correct option is D.


  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.

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