JEE Main 2022 — Binomial Theorem Question with Solution

Given Binomial expression is

$${\left(2x^3+\frac{3}{x^k}\right)}^{12}$$

General term,

$$T_{r+1}={}^{12}{C}_r(2x^3{)}^r.{\left(\frac{3}{x^k}\right)}^{12-r}$$

$$=\left({}^{12}{C}_r.2^r.3^{12-r}\right).x^{3r-12k+kr}$$

For constant term,

$$3r-12k+kr=0$$

$$\Rightarrow k(12-r)=3r$$

$$\Rightarrow k=\frac{3r}{12-r}$$

For r = 1, $$k=\frac{3}{11}$$ (not integer)

For r = 2, $$k=\frac{6}{10}$$ (not integer)

For r = 3, $$k=\frac{9}{9}=1$$ (integer)

For r = 6, $$k=\frac{18}{6}=3$$ (integer)

For r = 8, $$k=\frac{24}{4}=6$$ (integer)

For r = 9, $$k=\frac{27}{3}=9$$ (integer)

For r = 10, $$k=\frac{30}{2}=15$$ (integer)

For r = 11, $$k=\frac{33}{1}=33$$ (integer)

So, for r = 3, 6, 8, 9, 10 and 11 k is positive integer.

When k = 1 then r = 3 and constant term is

$$={}^{12}{C}_3.2^3.3^9$$

$$=\frac{12.11.10}{3.2.1}.2^3.3^9$$

$$=2.11.2.5.2^3.3^9$$

$$=11.5.2^5.3^9$$

$$=2^5.(55.3^9)$$

$$=2^5(l)$$

$$\ne 2^8.l$$

When x = 3 then r = 6 and constant term

$$={}^{12}{C}_6.2^6.3^6$$

$$=\frac{12.11.10.9.8.7}{6.5.4.3.2.1}.2^6.3^6$$

$$=2^8.231.3^6$$

$$=2^8(l)$$

When k = 6 then r = 8 and constant term

$$={}^{12}{C}_8.2^8.3^4$$

$$=\frac{12.11.10.9}{4.3.2.1}.2^8.3^4$$

$$=2^8.55.3^6$$

$$=2^8(l)$$

When x = 9 then r = 9 and constant term

$$={}^{12}{C}_9.2^9.3^3$$

$$=\frac{12.11.10}{3.2.1}.2^9.3^3$$

$$=2^{11}.55.3^3$$

Here power of 2 is 11 which is greater than 8. So, k = 9 is not possible.

Similarly for k = 15 and k = 33, $$2^8.l$$ form is not possible.

$$\therefore$$ k = 3 and k = 6 is accepted.

$$\therefore$$ For 2 positive integer value of k, $$2^8.l$$ form of constant term possible.