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Binomial Theorem question

2020 · 6 Sep · Shift 2 · Q34
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  5. /2020 · 6 Sep · Shift 2 · Q34

Binomial Theorem question

2020 · 6 Sep · Shift 2 · Q34

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If the constant term in the binomial expansion of (x−kx2)10{\left( {\sqrt x - {k \over {{x^2}}}} \right)^{10}}(x​−x2k​)10 is 405, then |k| equals :
  1. A
    3
  2. B
    9
  3. C
    1
  4. D
    2
View written solutionFree

Correct answer: A

  1. Given expression

We need the constant term in (x−kx2)10\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}(x​−x2k​)10 and it is given to be 405405405.

  1. General term in the binomial expansion

For (a+b)10,(a+b)^{10},(a+b)10, 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.

Here, a=x=x1/2,b=−kx2=−kx−2.a=\sqrt{x}=x^{1/2}, \qquad b=-\frac{k}{x^2}=-k x^{-2}.a=x​=x1/2,b=−x2k​=−kx−2.

So, Tr+1=(10r)(x1/2)10−r(−kx−2)r.T_{r+1}=\binom{10}{r}(x^{1/2})^{10-r}(-k x^{-2})^r.Tr+1​=(r10​)(x1/2)10−r(−kx−2)r.

  1. Simplify the power of xxx

Tr+1=(10r)(−k)rx10−r2x−2r.T_{r+1}=\binom{10}{r}(-k)^r x^{\frac{10-r}{2}}x^{-2r}.Tr+1​=(r10​)(−k)rx210−r​x−2r.

Hence power of xxx is 10−r2−2r=5−r2−2r=5−5r2.\frac{10-r}{2}-2r=5-\frac{r}{2}-2r=5-\frac{5r}{2}.210−r​−2r=5−2r​−2r=5−25r​.

For the constant term, power of xxx must be 000: 5−5r2=0.5-\frac{5r}{2}=0.5−25r​=0.

Solving, 10−5r=0  ⟹  r=2.10-5r=0 \implies r=2.10−5r=0⟹r=2.

  1. Find the constant term

Substitute r=2r=2r=2: T3=(102)(x1/2)8(−kx2)2.T_3=\binom{10}{2}(x^{1/2})^8\left(-\frac{k}{x^2}\right)^2.T3​=(210​)(x1/2)8(−x2k​)2.

Now,

(−kx2)2=k2x4.\left(-\frac{k}{x^2}\right)^2=\frac{k^2}{x^4}.(−x2k​)2=x4k2​.

Therefore, T3=45⋅x4⋅k2x4=45k2.T_3=45\cdot x^4\cdot \frac{k^2}{x^4}=45k^2.T3​=45⋅x4⋅x4k2​=45k2.

This is given to be 405405405, so 45k2=405.45k^2=405.45k2=405.

Thus, k2=9  ⟹  ∣k∣=3.k^2=9 \implies |k|=3.k2=9⟹∣k∣=3.

  1. Check options
  • A: 333 ✅
  • B: 999 ❌
  • C: 111 ❌
  • D: 222 ❌

Therefore, the correct answer is A.

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