Q108 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics

Let $f: R \rightarrow R$ be a bijection. $A$ curve represented by $y=f(x)$ is such that $f^{\prime}(x)>0 \forall x \in \mathbf{R}$. The tangent and normal drawn at $P(\alpha, 1)$ on the curve cuts the $X$-axis at $A, B$ respectively and $C$ is the foot of the perpendicular from $P$ onto the $X$-axis. If $P(\alpha, \mathrm{l})$ is such a point that $A C+C B$ is minimum, then the tangent at $P$ is parallel to the line

  • Correct answer
    A
    $x-y=0$
  • B
    $\alpha x+y-1=0$
  • C
    $j$
  • D
    $\frac{2 x}{\alpha}-y=\alpha^2$

Solution

Given, $y=f(x)$ Equation of tangent of curve $y=f(x)$ at $P(\alpha, 1)$ is $y-1=f(\alpha)(x-\alpha)$ Equation of normal of curve at $P(\alpha, 1)$ is $y-1=-\frac{1}{f^{\prime}(\alpha)}(x-a)$ $\therefore$ Tangent cut of $X$-axis at $A$ $\therefore \quad A=\left(\alpha-\frac{1}{f^{\prime} \alpha}, 0\right)$ $\therefore$ Normal cut of $X$-axis at $B$ $\begin{aligned} & \therefore \quad B=\left(\alpha+f^{\prime}(\alpha), 0\right) \\ & \text { Point } \quad C=(\alpha, 0) \\ & \end{aligned}$ $\begin{aligned} & A C+B C=\alpha-\alpha+\frac{1}{f^{\prime}(\alpha)}+\alpha+f^{\prime}(\alpha)-\alpha \\ & A C-B C=\frac{1}{f^{\prime}(\alpha)}+f^{\prime}(\alpha) \\ & A C+B C \text { is minimum } \end{aligned}$ $A C+B C$ is minimum When, $\quad f^{\prime}(\alpha)=1$ $\therefore$ Equation of tangent of curve $y-1=1(x-\alpha) \quad \Rightarrow x-y=\alpha+1$ $\therefore$ Equation of tangent of curve is parallel to $x-y=0$.

Hence, the correct answer is option: 'A'.

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