MCQ
The function $\sin x - bx + c$ will be increasing in the interval $( - \infty ,\,\,\infty )$, if
  • A
    $b \le 1$
  • B
    $b \le 0$
  • $b < - 1$
  • D
    $b \ge 0$

Answer

Correct option: C.
$b < - 1$
c
(c) Let $f(x) = \sin x - bx + c$

$\therefore f'(x) = \cos x - b > 0$ or $\cos x > b$ or $b < - 1$.

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

Similar questions

If the function f(x) defined by $\text{f(x)}=\begin{cases}\frac{\log(1+3\text{x})-\log(1-2\text{x})}{\text{x}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuous at x = 0, then k =
If the point of minima of the function, $f(x) = 1 + a^2x -x^3 $ satisfy the inequality $\frac{{{x^2} + x + 2}}{{{x^2} + 5x + 6}} < 0$, then $'a'$ must lie in the interval
A speaks truth in 75% cases and B seaks truth in 80% cases. Probability that they contradict each other in a statement, is
The differential equation satisfied by $\text{ax}^{2}+\text{by}^{2}=1$ is:
If for the matrix, $A =\left[\begin{array}{cc}1 & -\alpha \\ \alpha & \beta\end{array}\right], AA ^{ T }= I _{2},$ then the value of $\alpha^{4}+\beta^{4}$ is ....... .
If $y = \frac{{5x}}{{\sqrt[3]{{{{(1 - x)}^2}}}}} + {\cos ^2}(2x + 1)$, then $\frac{{dy}}{{dx}} = $
$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$ is equal to:
For $x > 0$, let $h(x) = \begin{array}{*{20}{c}}
{\frac{1}{q}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} x{\mkern 1mu} {\mkern 1mu}  = {\mkern 1mu} {\mkern 1mu} \frac{p}{q}}\\
{0\,\,\,\,\,\,\,\,if{\mkern 1mu} {\mkern 1mu} x\,{\mkern 1mu} is{\mkern 1mu} irrational\,\,\,}
\end{array}$ are relativily prime integer then which one does not hold good ?
A letter is known to have come either from $LONDON$ or $CLIFTON$; on the postmark only the two consecutive letters $ON$ are legible. The probability that it came from $LONDON$ is
If $A, B, C$ are the angles of a triangle and $\left| {\begin{array}{*{20}{c}}1&1&1\\{1 + \sin A}&{1 + \sin B}&{1 + \sin C}\\{\sin A + {{\sin }^2}A}&{\sin B + {{\sin }^2}B}&{\sin C + {{\sin }^2}C} \end{array}} \right|$ $= 0$, then the triangle is