MCQ
If $f(x) = \sin x - \cos x,$ the function decreasing in $0 \le x \le 2\pi $ is
- A$[5\pi /6,\,3\pi /4]$
- B$[\pi /4,\,\pi /2]$
- C$[3\pi /2,\,5\pi /2]$
- ✓None of these
✓
Answer
Correct option: D.
None of these
d
(d) $f(x) = \sin x - \cos x$
(d) $f(x) = \sin x - \cos x$
$f'(x) = \cos x + \sin x = \sqrt 2 \left[ {\cos \,\,\left( {x - \frac{\pi }{4}} \right)} \right]$
$=\sqrt 2 \cos \left( {x - \frac{\pi }{4}} \right)$
For $f(x)$ decreasing, $f'(x) < 0$
$\frac{\pi }{2} < \left( {x - \frac{\pi }{4}} \right) < \frac{{3\pi }}{2}$, (within $0 \le x \le 2\pi $).
==> $\frac{{3\pi }}{4} < x \le \frac{{7\pi }}{4}$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$\int_{}^{} {\frac{{x\;dx}}{{({x^2} - {a^2})({x^2} - {b^2})}} = } $
→The area (in square units) of the region bounded by the curves $y + 2x^2 = 0$ and $y + 3x^2 = 1$ , is equal to
→If the function $f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}$, $x \in R, $ is continuous at $x=0, $ then $f(0)$ is equal to :
→The differential equation whose solution is $(x -h)^2 + (y -k)^2 = a^2$ is ($a$ is a constant)
→ $(i)$ $f (x)$ is continuous and defined for all real numbers
→$(ii)$ $f '(-5) = 0 \,; \,f '(2)$ is not defined and $f '(4) = 0$
$(iii)$ $(-5, 12)$ is a point which lies on the graph of $f (x)$
$(iv)$ $f ''(2)$ is undefined, but $f ''(x)$ is negative everywhere else.
$(v)$ the signs of $f '(x)$ is given below
Possible graph of $y = f (x)$ is
Locus of the point $z$ satisfying the equation $|iz - 1|$+ $|z - i| = 2$ is
→From the point $(-1, 2)$ tangent lines are drawn to the parabola ${y^2} = 4x$, the area of triangle formed by chord of contact and the tangents is given by
→If $y = {\cot ^{ - 1}}\left[ {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right]$, then ${{dy} \over {dx}} = $
→The point at which the maximum value of $Z=3 x+2 y$ subject to the constraints $x+2 y \leq 2, x \geq 0, y \geq 0$ is $.....$
→$1 + \frac{1}{3}x + \frac{{1.4}}{{3.6}}{x^2} + \frac{{1.4.7}}{{3.6.9}}{x^3} + .... $ is equal to
→