The function f:R to R is defined by f(x) = co | vedclass

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

Question

(C) Given $f(x) = \cos^2 x + \sin^4 x$.We can rewrite the expression as:$f(x) = \cos^2 x + \sin^2 x \cdot \sin^2 x$$f(x) = \cos^2 x + (1 - \cos^2 x) \

Vedclass · Accepted Answer

$\left[ \frac{3}{4}, 1 \right]$

Vedclass · Answer

(C) Given $f(x) = \cos^2 x + \sin^4 x$.We can rewrite the expression as:$f(x) = \cos^2 x + \sin^2 x \cdot \sin^2 x$$f(x) = \cos^2 x + (1 - \cos^2 x) \sin^2 x$Alternatively,using $\sin^2 x = 1 - \cos^2 x$:$f(x) = \cos^2 x + (1 - \cos^2 x)^2$$f(x) = \cos^2 x + 1 - 2\cos^2 x + \cos^4 x$$f(x) = \cos^4 x - \cos^2 x + 1$Let $t = \cos^2 x$,where $t \in [0, 1]$.Then $g(t) = t^2 - t + 1$.This is a parabola opening upwards with vertex at $t = -(-1)/(2 \cdot 1) = 1/2$.Since $1/2 \in [0, 1]$,the minimum value is $g(1/2) = (1/4) - (1/2) + 1 = 3/4$.The maximum value occurs at the boundaries $t=0$ or $t=1$:$g(0) = 1$ and $g(1) = 1 - 1 + 1 = 1$.Thus,the range is $[3/4, 1]$.

The function $f:R \to R$ is defined by $f(x) = \cos^2 x + \sin^4 x$ for $x \in R$,then $f(R) \in $

Similar Questions

If $f: R \to R$,then the range of the function $f(x) = \frac{x^2}{x^2 + 1}$ is

Which of the following statements is false for the sets $A, B$ and $C,$ where
$A = \{x \mid x \text{ is a letter of the word } 'BOWL'\}$
$B = \{x \mid x \text{ is a letter of the word } 'ELBOW'\}$
$C = \{x \mid x \text{ is a letter of the word } 'BELLOW'\}$

If $A$ and $B$ are two sets,then $A \cap (A \cup B)'$ is equal to

The range of $f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}$ is

The domain of the function $\sqrt{\log \left( \frac{5x - x^2}{6} \right)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module