If P is a point on the segment AB of length 1 | vedclass

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

Question

(C) Let $d(AP) = x$. Then $d(BP) = 12 - x$. Define the function $f(x) = AP^{2} + BP^{2} = x^{2} + (12 - x)^{2}$. Expanding this,we get $f(x) = x^{2} +

Vedclass · Accepted Answer

$P$ is the midpoint of segment $AB$

Vedclass · Answer

(C) Let $d(AP) = x$. Then $d(BP) = 12 - x$. Define the function $f(x) = AP^{2} + BP^{2} = x^{2} + (12 - x)^{2}$. Expanding this,we get $f(x) = x^{2} + 144 - 24x + x^{2} = 2x^{2} - 24x + 144$. To find the minimum,we calculate the first derivative: $f'(x) = 4x - 24$. Setting $f'(x) = 0$,we get $4x = 24$,which implies $x = 6$. Checking the second derivative,$f''(x) = 4$. Since $f''(x) > 0$,the function $f(x)$ attains a minimum at $x = 6$. Since $x = 6$ is exactly half of the total length $12 	ext{ cm}$,$P$ is the midpoint of segment $AB$.

If $P$ is a point on the segment $AB$ of length $12 \text{ cm}$,then the position of $P$ for $AP^{2} + BP^{2}$ to be minimum is such that

Similar Questions

The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius $R = \sqrt{3}$ is

The shortest distance between the point $\left( \frac{3}{2}, 0 \right)$ and the curve $y = \sqrt{x}, (x > 0)$ is:

Find the local maximum and local minimum values for the function $h(x) = \sin x + \cos x$ in the interval $0 < x < \frac{\pi}{2}$.

Let $f(x)$ be a polynomial of degree $4$ having extreme values at $x = 1$ and $x = 2$. If $\mathop {\lim }\limits_{x \to 0} \left( {\frac{{f(x)}}{{{x^2}}} + 1} \right) = 3$,then $f(-1)$ is equal to

If $x$ is real,then the minimum value of $\frac{x^2-x+1}{x^2+x+1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module