Let f(x)=3 sqrt{x-2}+sqrt{4-x} be a real-valu | vedclass

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

Question

(B) Given the function $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$.The domain of the function is determined by $x-2 \geq 0$ and $4-x \geq 0$,which implies $x \in [

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(B) Given the function $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$.The domain of the function is determined by $x-2 \geq 0$ and $4-x \geq 0$,which implies $x \in [2, 4]$.To find the range,we use the Cauchy-Schwarz inequality or trigonometric substitution. Let $u = \sqrt{x-2}$ and $v = \sqrt{4-x}$. Then $u^2 + v^2 = (x-2) + (4-x) = 2$.The function is $f(x) = 3u + v$. By the Cauchy-Schwarz inequality,$(3u + v)^2 \leq (3^2 + 1^2)(u^2 + v^2) = (9+1)(2) = 20$.Thus,$f(x)^2 \leq 20$,so $f(x) \leq \sqrt{20} = \beta$.At the boundaries of the domain:If $x=2$,$f(2) = 3(0) + \sqrt{2} = \sqrt{2}$.If $x=4$,$f(4) = 3\sqrt{2} + 0 = 3\sqrt{2} = \sqrt{18}$.Since $\sqrt{2} Therefore,$\alpha^2 + 2\beta^2 = (\sqrt{2})^2 + 2(\sqrt{20})^2 = 2 + 2(20) = 2 + 40 = 42$.

Let $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$ be a real-valued function. If $\alpha$ and $\beta$ are respectively the minimum and the maximum values of $f$,then $\alpha^2+2 \beta^2$ is equal to

Similar Questions

If $y = a \log x + bx^2 + x$ has extreme values at $x = -1$ and $x = 2$,then:

The figure shown below is the graph of the derivative of some function $y=f(x)$. Then,

If $f(x) = 1 + 2x^2 + 2^2 x^4 + \dots + 2^{10} x^{20}$,then $f(x)$ has:

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in $sq.m$) of the flowerbed is

The sum of the maximum and minimum values of the function $f(x) = |5x - 7| + [x^2 + 2x]$ in the interval $[\frac{5}{4}, 2]$,where $[t]$ denotes the greatest integer function $\leq t$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module