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JEE Main 28-Jan-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 28-Jan-2025 Paper - Shift 24 Marks
MCQ
Let the coefficients of three consecutive terms $T_{r}$, $T_{r+1}$ and $T_{r+2}$ in the binomial expansion of $(a+b)^{12}$ be in a G.P. and let p be the number of all possible values of $r$. Let $q$ be the sum of all rational terms in the binomial expansion of $(\sqrt[4]{3}+\sqrt[3]{4})^{12}$. Then $\mathrm{p}+\mathrm{q}$ is equal to :
  • A
    283
  • B
    295
  • C
    287
  • D
    299
✓

Answer

A.
$(a+b)^{\frac{1}{2}}$
$\mathrm{T}_{\mathrm{r}}, \mathrm{T}_{\mathrm{r}+1}, \mathrm{~T}_{\mathrm{r}+2} \rightarrow \mathrm{GP}$
So, $\frac{T_{r+1}}{T_{r}}=\frac{T_{r+2}}{T_{r+1}}$
$\frac{{ }^{12} \mathrm{C}_{\mathrm{r}}}{{ }^{12} \mathrm{C}_{\mathrm{r}-1}}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}+1}}{{ }^{12} \mathrm{C}_{\mathrm{r}}}$
$\frac{12-\mathrm{r}+1}{\mathrm{r}}=\frac{12-(\mathrm{r}+1)+1}{\mathrm{r}+1}$
$(13-r)(r+1)=(12-r)(r)$
$-\mathrm{r}+12 \mathrm{r}+13=12 \mathrm{r}-\mathrm{r}^{2}$
$
13=0
$
No value of $r$ possible
So $\mathrm{P}-0$
$\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)^{12}=\sum{ }^{12} C_{r}\left(3^{\frac{1}{4}}\right)^{12-r}\left(4^{\frac{1}{3}}\right)^{r}$
Exponent of $\left(3^{\frac{1}{4}}\right)$ exponent of $\left(4^{\frac{1}{3}}\right)$ term
12027
012256

$
\mathrm{q}=27+256=283
$
$\mathrm{p}+\mathrm{q}=0+283=283$

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