Q108 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics
Let $\mathrm{n} \in(0, \infty)$. If all the curves $\mathrm{y}=\mathrm{x}^{\mathrm{n}} \log \mathrm{x}$ for distinct values of $n$, always have $y=x-1$ as the tangent drawn at a fixed point $(\alpha, \beta)$, then $\alpha+\beta=$
- A0
- B$\log 2$
- Correct answerC1
- D$\log 3$
Solution
$\because y=x^n \log x$
for $n=1, y=x \log x$
$$
\Rightarrow \frac{d y}{d x}=\log x+1
$$
$\because \mathrm{y}=\mathrm{x}-1$ is tangent line for curve (i) also.
$\therefore \log x+1=1$
$$
\Rightarrow \log \mathrm{x}=0 \Rightarrow \mathrm{x}=1
$$
Now, $\mathrm{y}=\mathrm{x}-1=1-1=0$
$\therefore$ The fixed point is $(x, y) \equiv(1,0)$
$$
\therefore(\alpha, \beta)=(1,0)
$$
Now, $\alpha+\beta=1+0=1$.
Hence, the correct answer is option: 'C'.