Let f: [- (π)/(2), (π)/(2)] → ℝ be a continuous function such that...

(A) Let $g\left(x\right) = f\left(x\right) - 3\cos 3x$

Now

${∫}_{0}^{\dfrac{\pi}{3}}g\left(x\right)dx = {∫}_{0}^{\dfrac{\pi}{3}}f\left(x\right)dx - 3{∫}_{0}^{\dfrac{\pi}{3}}\cos 3xdx$

$= 0 - {\left[\sin 3x\right]}_{0}^{\dfrac{\pi}{3}} = 0$

Hence $g\left(x\right) = 0$ has a root in \left(0, \dfrac{\pi}{3}\right)$</p><p>(B) Let$h\left(x\right) = f\left(x\right) - 3\sin 3x + \dfrac{6}{\pi}$</p><p>Now</p><p>${∫}_{0}^{\dfrac{\pi}{3}}h\left(x\right)dx = {∫}_{0}^{\dfrac{\pi}{3}}f\left(x\right)dx - 3{∫}_{0}^{\dfrac{\pi}{3}}\sin 3xdx + {∫}_{0}^{\dfrac{\pi}{3}}\dfrac{6}{\pi}dx$</p><p>$= 0 - {\left[- \cos 3x\right]}_{0}^{\dfrac{\pi}{3}} + {\left[\dfrac{6x}{\pi}\right]}_{0}^{\dfrac{\pi}{3}}$</p><p>$= 0 - 2 + 2 = 0$</p><p>Hence$h\left(x\right) = 0$has a root in \left(0, \dfrac{\pi}{3}\right)$

(C) $\lim_{x \to 0}\dfrac{x{∫}_{0}^{x}f ( t ) dt}{1 - {e}^{{x}^{2}}}$

$= \lim_{x \to 0}\left(\mathop{\mathop{\left(\dfrac{{x}^{2}}{1 - {e}^{{x}^{2}}}\right) \times}_{⏟}}_{- 1}\mathop{\mathop{\left(\dfrac{{∫}_{0}^{x}f( t )dt}{x}\right)}_{⏟}}_{Apply{L}^{'}hospitalrule}\right)$

$= - 1\lim_{x \to 0}\dfrac{f\left(x\right)}{1} = - 1$

(D) $\lim_{x \to 0}\dfrac{( \sin x ) {∫}_{0}^{x}f\left(t\right)dt}{{x}^{2}}$

$= \lim_{x \to 0}\left(\mathop{\mathop{\left(\dfrac{\sin x}{x}\right)}_{⏟}}_{1} \times \mathop{\mathop{\left(\dfrac{{∫}_{0}^{x}f( t )dt}{x}\right)}_{⏟}}_{Apply{L}^{'}hospitalrule}\right)$

$= 1\lim_{x \to 0}\dfrac{f\left(x\right)}{1} = 1$