The absolute minimum value of the function f( | vedclass

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

Question

(A) Let $g(x) = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$.For $x \in [-1, 2]$,the range of $g(x)$ is $[\frac{3}{4}, 3]$.Since $g(x) \ge \frac{3

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(A) Let $g(x) = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$.For $x \in [-1, 2]$,the range of $g(x)$ is $[\frac{3}{4}, 3]$.Since $g(x) \ge \frac{3}{4}$,we have $|g(x)| = g(x)$.Thus,$f(x) = g(x) + [g(x)]$.To minimize $f(x)$,we consider the values of $g(x)$ in its range $[\frac{3}{4}, 3]$.If $\frac{3}{4} \le g(x) If $1 \le g(x) If $2 \le g(x) \le 3$,then $[g(x)] = 2$,so $f(x) = g(x) + 2$. The minimum value is $2 + 2 = 4$.Comparing these,the absolute minimum value is $\frac{3}{4}$.

The absolute minimum value of the function $f(x) = |x^2 - x + 1| + [x^2 - x + 1]$,where $[t]$ denotes the greatest integer function,in the interval $[-1, 2]$,is:

Similar Questions

If $f(x)$ is a function such that $f^{\prime}(x)=(x-1)^{2}(4-x),$ then

The maximum value of $f(x) = (7-x)^4 (2+x)^5$ is

$A(1,15), B(3,-12), C(6,12)$ are three consecutive turning points of a continuous curve $y=f(x)$. If $f(x)=0$ only for $x=\alpha$ and $x=\beta$,then $|\beta-\alpha| < $

If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given,show that the area of the triangle is maximum when the angle between them is $\frac{\pi}{3}$.

If $f(x) = 1 + 2 \sin x + 3 \cos^2 x$ for $0 < x < 2\pi / 3$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module