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Binomial Theorem question

2024 · 8 Apr · Shift 2 · Q50
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  5. /2024 · 8 Apr · Shift 2 · Q50

Binomial Theorem question

2024 · 8 Apr · Shift 2 · Q50

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If the term independent of xxx in the expansion of (ax2+12x3)10\left(\sqrt{\mathrm{a}} x^2+\frac{1}{2 x^3}\right)^{10}(a​x2+2x31​)10 is 105 , then a2\mathrm{a}^2a2 is equal to :
  1. A
    6
  2. B
    4
  3. C
    2
  4. D
    9
View written solutionFree

Correct answer: B

  1. Write the general term

For (a x2+12x3)10,\left(\sqrt{a}\,x^2+\frac{1}{2x^3}\right)^{10},(a​x2+2x31​)10, the general term is Tr+1=(10r)(a x2)10−r(12x3)r.T_{r+1}=\binom{10}{r}(\sqrt{a}\,x^2)^{10-r}\left(\frac{1}{2x^3}\right)^r.Tr+1​=(r10​)(a​x2)10−r(2x31​)r.

  1. Simplify the general term

Tr+1=(10r)(a)10−rx2(10−r)⋅12rx3r.T_{r+1}=\binom{10}{r}(\sqrt{a})^{10-r}x^{2(10-r)}\cdot \frac{1}{2^r x^{3r}}.Tr+1​=(r10​)(a​)10−rx2(10−r)⋅2rx3r1​.

So,

=\binom{10}{r}\frac{(\sqrt{a})^{10-r}}{2^r}x^{20-5r}.$$ 3. **Find the term independent of $x$** For the constant term, exponent of $x$ must be zero: $$20-5r=0 \implies r=4.$$ Thus the constant term is the $5$th term: $$T_5=\binom{10}{4}\frac{(\sqrt{a})^6}{2^4}.$$ Since $$(\sqrt{a})^6=a^3,$$ we get $$T_5=\binom{10}{4}\frac{a^3}{16}.$$ Now, $$\binom{10}{4}=210,$$ so $$T_5=\frac{210a^3}{16}=\frac{105a^3}{8}.$$ 4. **Use the given value of the constant term** Given constant term is $105$: $$\frac{105a^3}{8}=105.$$ Divide by $105$: $$\frac{a^3}{8}=1 \implies a^3=8 \implies a=2.$$ 5. **Find $a^2$** $$a^2=2^2=4.$$ 6. **Check options** - A: $6$ ❌ - B: $4$ ✅ - C: $2$ ❌ - D: $9$ ❌ Therefore, the correct option is **B**.
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