JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If the term independent of in the expansion of is 105 , then is equal to :
- A6
- B4
- C2
- D9
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Correct answer: B
- Write the general term
For the general term is
- Simplify the general term
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**.More from Binomial Theorem
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