Let f: R rightarrow R be a bijection. A curve | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $y=f(x)$. The point $P$ is $(\alpha, 1)$.Equation of the tangent at $P(\alpha, 1)$ is $y-1=f^{\prime}(\alpha)(x-\alpha)$.Setting $y=0$,the $

Vedclass · Accepted Answer

$x-y=0$

Vedclass · Answer

(A) Given $y=f(x)$. The point $P$ is $(\alpha, 1)$.Equation of the tangent at $P(\alpha, 1)$ is $y-1=f^{\prime}(\alpha)(x-\alpha)$.Setting $y=0$,the $X$-intercept $A$ is $\alpha - \frac{1}{f^{\prime}(\alpha)}$. So $A = (\alpha - \frac{1}{f^{\prime}(\alpha)}, 0)$.Equation of the normal at $P(\alpha, 1)$ is $y-1=-\frac{1}{f^{\prime}(\alpha)}(x-\alpha)$.Setting $y=0$,the $X$-intercept $B$ is $\alpha + f^{\prime}(\alpha)$. So $B = (\alpha + f^{\prime}(\alpha), 0)$.The point $C$ is $(\alpha, 0)$.Then $AC = |\alpha - (\alpha - \frac{1}{f^{\prime}(\alpha)})| = \frac{1}{f^{\prime}(\alpha)}$ and $CB = |(\alpha + f^{\prime}(\alpha)) - \alpha| = f^{\prime}(\alpha)$.We want to minimize $AC+CB = \frac{1}{f^{\prime}(\alpha)} + f^{\prime}(\alpha)$.By $AM$-$GM$ inequality,$\frac{1}{f^{\prime}(\alpha)} + f^{\prime}(\alpha) \geq 2 \sqrt{\frac{1}{f^{\prime}(\alpha)} \cdot f^{\prime}(\alpha)} = 2$.The minimum occurs when $f^{\prime}(\alpha) = 1$.The slope of the tangent at $P$ is $f^{\prime}(\alpha) = 1$.$A$ line parallel to the tangent must have slope $1$. The line $x-y=0$ has slope $1$.

Similar Questions

The equation of the tangent to the curve $y^{2}=ax^{2}+b$ at the point $(2,3)$ is $y=4x-5$. Then the values of $a$ and $b$ are:

If the normal to the curve $x^{2/3} + y^{2/3} = a^{2/3}$ makes an angle $\phi$ with the $X$-axis,then the equation of that normal is

Let $n \in (0, \infty)$. If all the curves $y = x^n \log x$ for distinct values of $n$ always have $y = x - 1$ as the tangent drawn at a fixed point $(\alpha, \beta)$,then $\alpha + \beta =$

The angle at which the curve $y = K e^{Kx}$ intersects the $y$-axis is

Find the slope of the tangents to the curve $y = (x + 1)(x - 3)$ at the points where it meets the $x$-axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module