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Binomial Theorem question

2021 · 26 Feb · Shift 1 · Q23
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  5. /2021 · 26 Feb · Shift 1 · Q23

Binomial Theorem question

2021 · 26 Feb · Shift 1 · Q23

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
The maximum value of the term independent of 't' in the expansion of (tx15+(1−x)110t)10{\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}}(tx51​+t(1−x)101​​)10 where x ∈\in∈(0, 1) is :
  1. A
    10!3(5!)2{{10!} \over {\sqrt 3 {{(5!)}^2}}}3​(5!)210!​
  2. B
    2.10!33(5!)2{{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}33​(5!)22.10!​
  3. C
    10!3(5!)2{{10!} \over {3{{(5!)}^2}}}3(5!)210!​
  4. D
    2.10!3(5!)2{{2.10!} \over {3{{(5!)}^2}}}3(5!)22.10!​
View written solutionFree

Correct answer: B

  1. Find the term independent of ttt

We need the constant term in ttt in

(tx1/5+(1−x)1/10t)10.\left(t x^{1/5} + \frac{(1-x)^{1/10}}{t}\right)^{10}.(tx1/5+t(1−x)1/10​)10.

Using the general term:

Tr+1=(10r)(tx1/5)10−r((1−x)1/10t)r.T_{r+1} = \binom{10}{r} (t x^{1/5})^{10-r} \left(\frac{(1-x)^{1/10}}{t}\right)^r.Tr+1​=(r10​)(tx1/5)10−r(t(1−x)1/10​)r.

Simplify the power of ttt:

Tr+1=(10r)t10−rx(10−r)/5⋅(1−x)r/10t−rT_{r+1} = \binom{10}{r} t^{10-r} x^{(10-r)/5} \cdot (1-x)^{r/10} t^{-r}Tr+1​=(r10​)t10−rx(10−r)/5⋅(1−x)r/10t−r =(10r)t10−2rx(10−r)/5(1−x)r/10.= \binom{10}{r} t^{10-2r} x^{(10-r)/5} (1-x)^{r/10}.=(r10​)t10−2rx(10−r)/5(1−x)r/10.

For the term independent of ttt, we need

10−2r=0  ⟹  r=5.10-2r=0 \implies r=5.10−2r=0⟹r=5.

So the constant term is

T6=(105)x(10−5)/5(1−x)5/10.T_6 = \binom{10}{5} x^{(10-5)/5}(1-x)^{5/10}.T6​=(510​)x(10−5)/5(1−x)5/10.

That is,

T6=(105)x(1−x)1/2.T_6 = \binom{10}{5} x(1-x)^{1/2}.T6​=(510​)x(1−x)1/2.
  1. Maximize x1−xx\sqrt{1-x}x1−x​ for x∈(0,1)x\in(0,1)x∈(0,1)

Let

f(x)=x1−x,0<x<1.f(x)=x\sqrt{1-x}, \qquad 0<x<1.f(x)=x1−x​,0<x<1.

We maximize this.

Differentiate:

f(x)=x(1−x)1/2f(x)=x(1-x)^{1/2}f(x)=x(1−x)1/2 f′(x)=1−x+x⋅12(1−x)−1/2(−1)f'(x)=\sqrt{1-x}+x\cdot \frac{1}{2}(1-x)^{-1/2}(-1)f′(x)=1−x​+x⋅21​(1−x)−1/2(−1) =1−x−x21−x.=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}.=1−x​−21−x​x​.

Set f′(x)=0f'(x)=0f′(x)=0:

1−x−x21−x=0.\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}=0.1−x​−21−x​x​=0.

Multiply by 21−x2\sqrt{1-x}21−x​:

2(1−x)−x=02(1-x)-x=02(1−x)−x=0 2−3x=02-3x=02−3x=0 x=23.x=\frac{2}{3}.x=32​.

Now,

f(23)=231−23=2313=233.f\left(\frac{2}{3}\right)=\frac{2}{3}\sqrt{1-\frac{2}{3}}=\frac{2}{3}\sqrt{\frac{1}{3}}=\frac{2}{3\sqrt{3}}.f(32​)=32​1−32​​=32​31​​=33​2​.

Hence the maximum constant term is

(105)⋅233.\binom{10}{5}\cdot \frac{2}{3\sqrt{3}}.(510​)⋅33​2​.

Since

(105)=10!(5!)2,\binom{10}{5}=\frac{10!}{(5!)^2},(510​)=(5!)210!​,

we get

Maximum value=10!(5!)2⋅233=2⋅10!33(5!)2.\text{Maximum value} = \frac{10!}{(5!)^2}\cdot \frac{2}{3\sqrt{3}} = \frac{2\cdot 10!}{3\sqrt{3}(5!)^2}.Maximum value=(5!)210!​⋅33​2​=33​(5!)22⋅10!​.
  1. Compare with options

This matches:

2⋅10!33(5!)2\boxed{\frac{2\cdot 10!}{3\sqrt{3}(5!)^2}}33​(5!)22⋅10!​​

which is Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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