The length of a longest interval in which the function 3 sin x - 4 sin3x is increasing, is

Question

The length of a longest interval in which the function 3 sin x - 4 sin3x is increasing, is (A) (π/3) (B) (π/2) (C) (3π/2) (D) π

Tardigrade Admin · Accepted Answer

Since 3sin x−4sin 3x=sin3x and sin3x increases when 3x takes values from −π/2 to π/2 or x takes values from −π/6 to π/6. Thus, the length of the longest interval in which 3sinx − 4sin3x increases is π/6−(−π/6)=π/3.

Q. The length of a longest interval in which the function $3 sin x - 4 sin^{3} x$ is increasing, is

$\frac{π}{3}$

$\frac{π}{2}$

$\frac{3 π}{2}$

$π$

Since $3 s in x - 4 s i n^{3} x = s in 3 x$
and $s in 3 x$ increases when $3 x$ takes values from $- π /2$ to $π /2$ or $x$ takes values from $- π /6$ to $π /6$ .
Thus, the length of the longest interval in which $3 s in x - 4 s i n^{3} x$ increases is $π /6 - (- π /6) = π /3$ .

Q. The length of a longest interval in which the function 3sinx−4sin3x is increasing, is

3π​

2π​

23π​

π

Since 3sinx−4sin3x=sin3x and sin3x increases when 3x takes values from −π/2 to π/2 or x takes values from −π/6 to π/6. Thus, the length of the longest interval in which 3sinx−4sin3x increases is π/6−(−π/6)=π/3.

Q. The length of a longest interval in which the function $3 sin x - 4 sin^{3} x$ is increasing, is

$\frac{π}{3}$

$\frac{π}{2}$

$\frac{3 π}{2}$

$π$

Since $3 s in x - 4 s i n^{3} x = s in 3 x$
and $s in 3 x$ increases when $3 x$ takes values from $- π /2$ to $π /2$ or $x$ takes values from $- π /6$ to $π /6$ .
Thus, the length of the longest interval in which $3 s in x - 4 s i n^{3} x$ increases is $π /6 - (- π /6) = π /3$ .