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 this as $f(x) = \cos^2 x + \sin^2 x \cdot \sin^2 x$. Since $\sin^2 x = 1 - \cos^2 x$,we have $f

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $f(x) = \cos^2 x + \sin^4 x$. We can rewrite this as $f(x) = \cos^2 x + \sin^2 x \cdot \sin^2 x$. Since $\sin^2 x = 1 - \cos^2 x$,we have $f(x) = \cos^2 x + \sin^2 x(1 - \cos^2 x)$. $f(x) = \cos^2 x + \sin^2 x - \sin^2 x \cos^2 x$. Since $\cos^2 x + \sin^2 x = 1$,we get $f(x) = 1 - \sin^2 x \cos^2 x$. Multiplying and dividing by $4$,we get $f(x) = 1 - \frac{1}{4}(4 \sin^2 x \cos^2 x) = 1 - \frac{1}{4} \sin^2(2x)$. Since $0 \le \sin^2(2x) \le 1$,we have $1 - \frac{1}{4}(1) \le f(x) \le 1 - \frac{1}{4}(0)$. Thus,$\frac{3}{4} \le f(x) \le 1$. Therefore,$f(R) \in \left[ \frac{3}{4}, 1 \right]$.

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

The domain of $\sqrt{|x|-x}$ is

The domain of the function $f(x) = \log_e(x - [x])$ is

$A$ and $B$ are subsets of $R$. Every element $x$ of $A$ is mapped to an element of $B$ by the rule, $y(x) = \begin{cases} \frac{5x}{(x-3)(x+3)} & \text{if } x \neq -1 \\ -1 & \text{if } x = -1 \end{cases}$, then $A =$

If the domain of the function $f(x) = \log_{(10x^{2}-17x+7)}(18x^{2}-11x+1)$ is $(-\infty, a) \cup (b, c) \cup (d, \infty) - \{e\}$, then $90(a+b+c+d+e)$ equals:

The largest interval lying in $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ for which the function $f(x) = 4^{-x^2} + \cos^{-1}\left( \frac{x}{2} - 1 \right) + \log(\cos x)$ is defined is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module