Discussion Forum



1)

Let f: R→ (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)?


A) $e^{x}-\int_{0}^{x} f(t) \sin t dt $

B) $f(x)+\int_{0}^{\frac{\pi}{2}} f(t)\sin t dt$

C) $x-\int_{0}^{\frac{\pi}{2}-x} f(t)\cos t dt$

D) $x^{9}-f(x)$

Answer:

Option C,D

Explanation:

(a)   $\because e^{x}\epsilon(1,e)in(0,1)$ and

   $\int_{0}^{x} f(t)\sin t dt\in (0,1)$  in (0,1)

$\therefore e^{x}-\int_{0}^{x} f(t) \sin t dt$ cannot be zero

so, option (a) is incorrect

(b)   $f(x)+\int_{0}^{\frac{\pi}{2}}f(t) \sin t dt $ always postive

 option b is incorrect

(c)   Let   $h(x)=x-\int_{0}^{\frac{\pi}{2}-x} f(t) \cos t dt$

$h(0)=-\int_{0}^{\frac{\pi}{2}} f(t) \cos t dt<0$

$h(1)=1-\int_{0}^{\frac{\pi}{2}-1} f(t) \cos t dt>0$

 Option C is correct

(d) Let g(x)= x9-f(x)

  g(0)=-f(0)<0, g(1)=1-f(1)>0

 Option d is correct