The range of f(x) = x - x is - Vidyadip

MCQ

The range of $f(x) = \cos x - \sin x$ is

A
$( - 1,\;1)$
B
$[ - 1,\,\;1)$
C
$\left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right]$
✓
$[ - \sqrt 2 ,\;\sqrt 2 ]$

✓

Answer

Correct option: D.

$[ - \sqrt 2 ,\;\sqrt 2 ]$

d
(d) Since maximum and minimum values of $\cos x - \sin x$ are $\sqrt 2 $ and $ - \sqrt 2 $ respectively,

therefore range of $f(x)$ is $[ - \sqrt 2 ,\,\,\sqrt 2 ].$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If $(x,\,y) \in R$ and $x,\;y \ne 0$; $f(x,\;y) \to \frac{x}{y}$, then this function is a/an

The area enclosed by the curves $y^2+4 x=4$ and $y-2 x=2$ is :

Let $A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha\end{array}\right]$ and $|2 A|^3=2^{21}$ where $\alpha, \beta \in Z,$ Then a value of $\alpha$ is

If $\alpha, \beta, \gamma$ are the real roots of the equation $x^3 -3px^2 + 3qx -1 = 0$, then the centroid of the triangle whose vertices are $(\alpha,\frac{1}{\alpha}),(\beta,\frac{1}{\beta})$ and $(\gamma,\frac{1}{\gamma})$

If $\int_{}^{} {\ln ({x^2} + x)dx = x\ln ({x^2} + x) + A} $, then $A = $

If $\tan \theta = \frac{a}{b},$ then $\frac{{\sin \theta }}{{{{\cos }^8}\theta }} + \frac{{\cos \theta }}{{{{\sin }^8}\theta }} = $

The two circles ${x^2} + {y^2} - 2x + 6y + 6 = 0$ and ${x^2} + {y^2} - 5x + 6y + 15 = 0$

Let $f,\ f',\ f''$ be continuous in $[0, ln\ 2]$ and $f(0) = 0 , f '(0) = 3, f(ln\ 2) = 6 , f'(ln\ 2) = 4$ and $\int\limits_0^{\ln 2} {{e^{ - 2x}}f(x)dx} = 3$ , then $\int\limits_0^{\ln 2} {{e^{ - 2x}}f''(x)dx} $ is

The cubic $\left| {\begin{array}{*{20}{c}}
0&{a - x}&{b - x} \\
{ - a - x}&0&{c - x} \\
{ - b - x}&{ - c - x}&0
\end{array}} \right| = 0$ has a reperated root in $x$ then,

$\mathop {\lim }\limits_{x \to \pi /6} \frac{{{{\cot }^2}\theta - 3}}{{{\rm{cosec}}\theta - 2}} = $