The term independent of x in the expansion of [ x + 1 \over x^2/3 - x^1/3 + 1 - x - 1 \over x - x^1/2 ]^10 , x e 1, is equal to \underline\hspace2cm .

Binomial Theorem PYQ — JEE Main 2021 · 18 Mar · Shift 2 · Q39 with Solution

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Correct answer: 210

Simplify the expression inside the bracket

We need the constant term in $\left[\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right]^{10}.$

Let $t=x^{1/3}.$ Then $x=t^3$ and $x^{2/3}=t^2$ .

So the first fraction becomes $\frac{x+1}{x^{2/3}-x^{1/3}+1}=\frac{t^3+1}{t^2-t+1}.$ Now use $t^3+1=(t+1)(t^2-t+1).$ Hence, $\frac{t^3+1}{t^2-t+1}=t+1=x^{1/3}+1.$

Now simplify the second fraction: $\frac{x-1}{x-x^{1/2}}.$ Let $u=x^{1/2}$ . Then $x=u^2$ , so $\frac{x-1}{x-x^{1/2}}=\frac{u^2-1}{u^2-u}=\frac{(u-1)(u+1)}{u(u-1)}=\frac{u+1}{u}=1+u^{-1}=1+x^{-1/2}.$

Therefore the whole bracket becomes

=\left(x^{1/3}-x^{-1/2}\right)^{10}.$$ --- 2. **Expand using the binomial theorem** $$\left(x^{1/3}-x^{-1/2}\right)^{10} =\sum_{r=0}^{10} \binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r.$$ General term: $$T_{r+1}=\binom{10}{r}(-1)^r x^{\frac{10-r}{3}-\frac r2}.$$ Simplify the exponent of $x$: $$\frac{10-r}{3}-\frac r2 =\frac{20-2r-3r}{6} =\frac{20-5r}{6}.

For the term independent of $x$ , exponent must be $0$ :

\implies 20-5r=0 \implies r=4.$$ --- 3. **Find the constant term** At $r=4$, $$T_5=\binom{10}{4}(-1)^4 x^0=\binom{10}{4}=210.$$ So the term independent of $x$ is $$\boxed{210}.$$ --- 4. **Compare with stored answer** Stored correct answer = $210$. Our derived answer also equals $210$, so they agree.

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View written solutionFree

Correct answer: 210

Simplify the expression inside the bracket

We need the constant term in $\left[\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right]^{10}.$

Let $t=x^{1/3}.$ Then $x=t^3$ and $x^{2/3}=t^2$ .

So the first fraction becomes $\frac{x+1}{x^{2/3}-x^{1/3}+1}=\frac{t^3+1}{t^2-t+1}.$ Now use $t^3+1=(t+1)(t^2-t+1).$ Hence, $\frac{t^3+1}{t^2-t+1}=t+1=x^{1/3}+1.$

Now simplify the second fraction: $\frac{x-1}{x-x^{1/2}}.$ Let $u=x^{1/2}$ . Then $x=u^2$ , so $\frac{x-1}{x-x^{1/2}}=\frac{u^2-1}{u^2-u}=\frac{(u-1)(u+1)}{u(u-1)}=\frac{u+1}{u}=1+u^{-1}=1+x^{-1/2}.$

Therefore the whole bracket becomes

=\left(x^{1/3}-x^{-1/2}\right)^{10}.$$ --- 2. **Expand using the binomial theorem** $$\left(x^{1/3}-x^{-1/2}\right)^{10} =\sum_{r=0}^{10} \binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r.$$ General term: $$T_{r+1}=\binom{10}{r}(-1)^r x^{\frac{10-r}{3}-\frac r2}.$$ Simplify the exponent of $x$: $$\frac{10-r}{3}-\frac r2 =\frac{20-2r-3r}{6} =\frac{20-5r}{6}.

For the term independent of $x$ , exponent must be $0$ :

\implies 20-5r=0 \implies r=4.$$ --- 3. **Find the constant term** At $r=4$, $$T_5=\binom{10}{4}(-1)^4 x^0=\binom{10}{4}=210.$$ So the term independent of $x$ is $$\boxed{210}.$$ --- 4. **Compare with stored answer** Stored correct answer = $210$. Our derived answer also equals $210$, so they agree.

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