Skip to content
NCERT Exemplar · Q32

Q.The position of the term independent of xx in the expansion of (x3+32x2)10\left(\sqrt{\dfrac{x}{3}} + \dfrac{3}{2x^2}\right)^{10} is ______ .

Manipur CohsemShort· 2mImportance★★★★★est
88% · 56/64 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The exponent of xx in the general term is 10−5r2\dfrac{10-5r}{2}; setting it to 00 gives r=2r=2, so the independent term is at position r+1=3r+1=3.

The general term of (x3+32x2)10\left(\sqrt{\dfrac{x}{3}}+\dfrac{3}{2x^{2}}\right)^{10} is

Tr+1=(10r)(x3)10−r(32x2)r=(10r) 3 r3(10−r)/2 2r  x10−r2−2r.T_{r+1}=\binom{10}{r}\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{3}{2x^{2}}\right)^{r} =\binom{10}{r}\,\frac{3^{\,r}}{3^{(10-r)/2}\,2^{r}}\;x^{\frac{10-r}{2}-2r}.

The power of xx is

10−r2−2r=10−5r2.\frac{10-r}{2}-2r=\frac{10-5r}{2}.

For the term independent of xx, set it to zero: …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.