If the solution curve $f(x, y)=0$ of the differential equation $\left(1+\log _e x\right) \frac{d x}{d y}-x \log _e x=e^y, x>0, y$ passes through the points $(1,0)$ and $(\alpha, 2)$ then $\alpha^\alpha$ is equal to
Select the correct option:
A
Only (S1) is true
B
both (S1) and (S2) are true
C
neither (S1) nor (S2) is true
D
only (S2) is true
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol. $f(x)=\frac{1}{\sqrt{[x]-x}}$
If $\mathrm{x} \in \mathrm{I}\lceil\mathrm{x}\rceil=[\mathrm{x}]$ (greatest integer function)
If $x \notin I\lceil x\rceil=[x]+1$
$$
\begin{aligned}
& \Rightarrow f(x)=\left\{\begin{array}{c}
\frac{4}{\sqrt{x\rceil-x}}, x \in I \\
\frac{4}{\sqrt{x\rceil+1-x}}, x \notin I
\end{array}\right. \\
& \Rightarrow f(x)=\left\{\begin{array}{c}
\frac{1}{\sqrt{-\{x\}}}, x \in I, \text { (does not exist) } \\
\frac{1}{\sqrt{1-\{x\}}}, x \notin I
\end{array}\right. \\
& \Rightarrow \text { domain of } f(x)=R-I \\
& \text { Now, } f(x)=\frac{1}{\sqrt{1-\{x\}}}, x \notin I \\
& \Rightarrow 0<\{x\}<1
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow 0<\sqrt{1-\{x\}}<1 \\
& \Rightarrow \text { Range }(1, \infty) \\
& \Rightarrow A=R-1 \\
& B=(1, \infty)
\end{aligned}
$$
So, $A \cap B=(1, \infty)-N$
$A \cup B(1, \infty)$
$\Rightarrow \mathrm{S} 1$ is only correct
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇
