Let [t] denote the greatest integer \leq t . If the constant term in the expansion of (3 x^2-\frac12 x^5)^7 is \alpha , then [\alpha] is equal to \underline\hspace2cm .

Binomial Theorem PYQ — JEE Main 2023 · 8 Apr · Shift 1 · Q45 with Solution

View written solutionFree

Correct answer: 1275

We need the constant term in $\left(3x^2-\frac{1}{2x^5}\right)^7.$
Using the general term of the binomial expansion: $T_{r+1}=\binom{7}{r}(3x^2)^{7-r}\left(-\frac{1}{2x^5}\right)^r, \quad r=0,1,2,\dots,7.$
Simplify the power of $x$ in the general term: $(3x^2)^{7-r}=3^{7-r}x^{2(7-r)}=3^{7-r}x^{14-2r},$ $\left(-\frac{1}{2x^5}\right)^r=\frac{(-1)^r}{2^r}x^{-5r}.$

So,

=\binom{7}{r}3^{7-r}\frac{(-1)^r}{2^r}x^{14-7r}.$$ 4. For the constant term, the exponent of $x$ must be $0$: $$14-7r=0 \implies r=2.$$ 5. Substitute $r=2$ into the coefficient: $$\alpha=\binom{7}{2}3^{5}\frac{(-1)^2}{2^2}.$$ Now compute: $$\binom{7}{2}=21, \qquad 3^5=243, \qquad 2^2=4.$$ Hence, $$\alpha=21\cdot 243\cdot \frac{1}{4}= rac{5103}{4}=1275.75.$$ 6. Therefore, $$[\alpha]=[1275.75]=1275.$$ So the required integer is $$\boxed{1275}.$$ 7. Comparison with stored answer: Stored correct answer = $1275$, which matches our result.

View written solutionFree

Correct answer: 1275

We need the constant term in $\left(3x^2-\frac{1}{2x^5}\right)^7.$
Using the general term of the binomial expansion: $T_{r+1}=\binom{7}{r}(3x^2)^{7-r}\left(-\frac{1}{2x^5}\right)^r, \quad r=0,1,2,\dots,7.$
Simplify the power of $x$ in the general term: $(3x^2)^{7-r}=3^{7-r}x^{2(7-r)}=3^{7-r}x^{14-2r},$ $\left(-\frac{1}{2x^5}\right)^r=\frac{(-1)^r}{2^r}x^{-5r}.$

So,

=\binom{7}{r}3^{7-r}\frac{(-1)^r}{2^r}x^{14-7r}.$$ 4. For the constant term, the exponent of $x$ must be $0$: $$14-7r=0 \implies r=2.$$ 5. Substitute $r=2$ into the coefficient: $$\alpha=\binom{7}{2}3^{5}\frac{(-1)^2}{2^2}.$$ Now compute: $$\binom{7}{2}=21, \qquad 3^5=243, \qquad 2^2=4.$$ Hence, $$\alpha=21\cdot 243\cdot \frac{1}{4}= rac{5103}{4}=1275.75.$$ 6. Therefore, $$[\alpha]=[1275.75]=1275.$$ So the required integer is $$\boxed{1275}.$$ 7. Comparison with stored answer: Stored correct answer = $1275$, which matches our result.