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

Let $f : R \rightarrow R$ be a function defined by $f(x) = \log_{\sqrt{m}}\{\sqrt{2}(\sin x - \cos x) + m - 2\}$,for some $m$,such that the range of $f$ is $[0, 2]$. Then the value of $m$ is $............$

The domain of the function $y=f(x)$,where $x$ and $y$ are related by $2^x+2^y=2$ is

If $[x]^2-5[x]+6=0$,where $[\cdot]$ denotes the greatest integer function,then $x$ belongs to:

If $[\cdot]$ denotes the greatest integer function,then the domain and range of the function $f(x) = \frac{\sin([x]\pi) + \tan([x]\pi)}{1 + [x]^2 + [x]^4}$ are respectively

For the function $f(x) = (1 + \frac{1}{x})^x$,the domain of $f(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module