MCQ
If the coefficient of x in $\Big(\text{x}^{2}+\frac{\lambda}{\text{x}}\Big)^{5}$ is 270, then $\lambda=$
- A3
- B4
- C5
- DNone of these.
Solution:
The coefficient of x in the given expansion where x occurs at the (r + 1)th term.
We have,
${^\text{15}}\text{C}_{\text{r}}(\text{x}^{2})^{5-\text{r}}\ \Big(\frac{\lambda}{\text{x}}\Big)^{\text{r}}$
$={^\text{15}}\text{C}_{\text{r}}\ \lambda^{\text{r}}\ \text{x}^{10-2\text{r}-\text{r}}$
For it to contain x, we must have
$10-3\text{r}=1$
$\Rightarrow \text{r}=3$
Coefficient of x in the given expansion,
$={^\text{15}}\text{C}_{\text{3}}\ \lambda^{\text{3}}=10\lambda^{3}$
Now, we have
$10\lambda^{3}=270$
$\Rightarrow \lambda^{3}=27$
$\Rightarrow \lambda=3$
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The area of the circle represented by the equation (x + 3)2 + (y + 1)2 = 25 is: