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Binomial Theorem question

2022 · 25 Jul · Shift 1 · Q38
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  5. /2022 · 25 Jul · Shift 1 · Q38

Binomial Theorem question

2022 · 25 Jul · Shift 1 · Q38

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
If the maximum value of the term independent of ttt in the expansion of (t2x15+(1−x)110t)15,x⩾0\left(\mathrm{t}^{2} x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{\mathrm{t}}\right)^{15}, x \geqslant 0(t2x51​+t(1−x)101​​)15,x⩾0, is K\mathrm{K}K, then 8 K8 \mathrm{~K}8 K is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 6006

  1. General term in the expansion

For

(t2x1/5+(1−x)1/10t)15,\left(t^2 x^{1/5} + \frac{(1-x)^{1/10}}{t}\right)^{15},(t2x1/5+t(1−x)1/10​)15,

the general term is

Tr+1=(15r)(t2x1/5)15−r((1−x)1/10t)r.T_{r+1} = \binom{15}{r} \left(t^2 x^{1/5}\right)^{15-r} \left(\frac{(1-x)^{1/10}}{t}\right)^r.Tr+1​=(r15​)(t2x1/5)15−r(t(1−x)1/10​)r.

So,

Tr+1=(15r) t2(15−r)x(15−r)/5 t−r(1−x)r/10.T_{r+1} = \binom{15}{r} \, t^{2(15-r)} x^{(15-r)/5} \, t^{-r} (1-x)^{r/10}.Tr+1​=(r15​)t2(15−r)x(15−r)/5t−r(1−x)r/10.

Thus the power of ttt is

2(15−r)−r=30−3r.2(15-r)-r = 30-3r.2(15−r)−r=30−3r.
  1. Find the term independent of ttt

For the term independent of ttt,

30−3r=0  ⟹  r=10.30-3r=0 \implies r=10.30−3r=0⟹r=10.

Hence the required term is

T11=(1510)x(15−10)/5(1−x)10/10.T_{11} = \binom{15}{10} x^{(15-10)/5}(1-x)^{10/10}.T11​=(1015​)x(15−10)/5(1−x)10/10.

That is,

T11=(1510)x(1−x).T_{11} = \binom{15}{10} x(1-x).T11​=(1015​)x(1−x).

Since

(1510)=(155)=3003,\binom{15}{10} = \binom{15}{5} = 3003,(1015​)=(515​)=3003,

we get

T11=3003 x(1−x).T_{11} = 3003\,x(1-x).T11​=3003x(1−x).
  1. Maximize this expression for x≥0x\ge 0x≥0

Because (1−x)1/10(1-x)^{1/10}(1−x)1/10 appears in the original expression, for real values we must also have

1−x≥0  ⟹  0≤x≤1.1-x \ge 0 \implies 0\le x\le 1.1−x≥0⟹0≤x≤1.

So we maximize

3003 x(1−x)3003\,x(1-x)3003x(1−x)

on [0,1][0,1][0,1].

Now,

x(1−x)=x−x2.x(1-x)=x-x^2.x(1−x)=x−x2.

This is a downward parabola whose maximum occurs at

x=12.x=\frac12.x=21​.

Then

max⁡x(1−x)=12(1−12)=14.\max x(1-x)=\frac12\left(1-\frac12\right)=\frac14.maxx(1−x)=21​(1−21​)=41​.

Therefore,

K=3003⋅14.K=3003\cdot \frac14.K=3003⋅41​.
  1. Compute 8K8K8K
8K=8⋅3003⋅14=2⋅3003=6006.8K=8\cdot 3003\cdot \frac14=2\cdot 3003=6006.8K=8⋅3003⋅41​=2⋅3003=6006.

So the required integer is

6006.\boxed{6006}.6006​.
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