Let $f:[0,2] \rightarrow R$ be the function defined by $f(x)=(3-\sin(2\pi x)) \sin(\pi x-\frac{\pi}{4})-\sin(3\pi x+\frac{\pi}{4})$. If $\alpha, \beta \in[0,2]$ are such that $\{x \in[0,2]: f(x) \geq 0\}=[\alpha, \beta]$,then the value of $\beta-\alpha$ is:

  • A
    $0$
  • B
    $1$
  • C
    $5$
  • D
    $6$

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Answer the following by appropriately matching the lists based on the information given in the paragraph.
Let $f(x) = \sin(\pi \cos x)$ and $g(x) = \cos(2\pi \sin x)$ be two functions defined for $x > 0$. Define the following sets whose elements are written in increasing order:
$X = \{x : f(x) = 0\}, Y = \{x : f'(x) = 0\}$
$Z = \{x : g(x) = 0\}, W = \{x : g'(x) = 0\}$
$List-I$ contains the sets $X, Y, Z$ and $W$. $List-II$ contains some information regarding these sets.
$List-I$$List-II$
$(I) X$$(P) \supseteq \{\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi\}$
$(II) Y$$(Q) \text{ an arithmetic progression}$
$(III) Z$$(R) \text{ NOT an arithmetic progression}$
$(IV) W$$(S) \supseteq \{\frac{\pi}{6}, \frac{7\pi}{6}, \frac{13\pi}{6}\}$
$(T) \supseteq \{\frac{\pi}{3}, \frac{2\pi}{3}, \pi\}$
$(U) \supseteq \{\frac{\pi}{6}, \frac{3\pi}{4}\}$

$(1)$ Which of the following is the only $CORRECT$ combination?
$(1) (II), (R), (S)$ $(2) (I), (P), (R)$ $(3) (II), (Q), (T)$ $(4) (I), (Q), (U)$
$(2)$ Which of the following is the only $CORRECT$ combination?
$(1) (IV), (Q), (T)$ $(2) (IV), (P), (R), (S)$ $(3) (III), (R), (U)$ $(4) (III), (P), (Q), (U)$

Let $f: R \rightarrow (0,1)$ be a continuous function. Then,which of the following function$(s)$ has(have) the value zero at some point in the interval $(0,1)$?

Let $f(x)$ and $g(x)$ be two functions given by $f(x) = \frac{2\sin(\pi x)}{x}$ and $g(x) = f(1 - x) + f(x)$. If $g(x) = k f(\frac{x}{2}) f(\frac{1 - x}{2})$,then the value of $k$ is

Let $f$ and $g$ be increasing and decreasing functions,respectively,from $[0, \infty)$ to $[0, \infty)$. Let $h(x) = f(g(x))$. If $h(0) = 0$,then $h(x) - h(1)$ is:

Let $A = \{1, 2, 3, 5, 8, 9\}$. Then the number of possible functions $f : A \rightarrow A$ such that $f(m \cdot n) = f(m) \cdot f(n)$ for every $m, n \in A$ with $m \cdot n \in A$ is equal to $...............$.

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