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JEE Main 20238 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2023 · 8 April · Morning Shift · Mathematics · Binomial Theorem

Let [t] denote the greatest integer ≤ t . if the constant term in...

Let [t]\left[t\right] denote the greatest integer ≤t\le t. if the constant term in the expansion of (3x2−12x5)7is{\left(3{x}^{2} - \dfrac{1}{2{x}^{5}}\right)}^{7} is\alphathen[α]then \left[\alpha\right] is equal to _____

Correct answer

0

Step-by-step solution

The given expansion is (3x2−12x5)7{\left(3{x}^{2} - \dfrac{1}{2{x}^{5}}\right)}^{7}.

The general term in the binomial expansion of (x+a)n{\left(x + a\right)}^{n} is given by Tr+1=Crnxn−rar{T}_{r + 1} = Cr n {x}^{n - r}{a}^{r}.

⇒Tr+1=Cr7(3x2)7−r(−12x5)r⇒ {T}_{r + 1} = Cr 7 {\left(3{x}^{2}\right)}^{7 - r}{\left(\dfrac{- 1}{2{x}^{5}}\right)}^{r}

=Cr7(−1)r(37−r2r)x14−2r−5r= Cr 7 {\left(- 1\right)}^{r}\left(\dfrac{{3}^{7 - r}}{{2}^{r}}\right){x}^{14 - 2r - 5r}

Now for term independent of xx⇒14−2r−5r=0⇒ 14 - 2r - 5r = 0

⇒r=2⇒ r = 2

Coefficient of x0{x}^{0} is C27(−1)2×(37−222)C2 7 {\left(- 1\right)}^{2} \times \left(\dfrac{{3}^{7 - 2}}{{2}^{2}}\right)

=7×62×3522= \dfrac{7 \times 6}{2} \times \dfrac{{3}^{5}}{{2}^{2}}.

⇒α=51034⇒ \alpha = \dfrac{5103}{4}

⇒[α]=1275⇒ \left[\alpha\right] = 1275

Hence this is the required answer.

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