Question:(more than one correct answer)

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

(a) $e^{x}-\displaystyle \int_{0}^{x} f(t)\ sint\ dt$

(b) $f(x)+\displaystyle \int_{0}^{\dfrac{\pi}{2}} f(t)\ sint\ dt$

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

(d) $x-\displaystyle \int_{0}^{\dfrac{\pi-2x}{2}} f(t)\ cost\ dt$

my attempt: since,

$0<f(x)<1\implies0<\displaystyle\int_{0}^\dfrac{\pi}{2} f(x)\ sinx\ dx<\displaystyle\int_{0}^{\dfrac{\pi}{2}}sinx\ dx=1$


$f(x)+\displaystyle\int_{0}^\dfrac{\pi}{2} f(x)\ sinx\ dx$ this can never be zero so option (b) is ruled out

now coming to option (a) let $ y=e^{x}-\displaystyle \int_{0}^{x} f(t)\ sint\ dt\implies \dfrac{dy}{dx}=e^{x}-f(x) sinx\ $ which implies $'y'$ is increasing $\forall x>0$ and at $x=0; y = 1$ therefore for any $1>x>0$ it will always be greater than 1 therefore this option too can never be zero

so , thus i assured myself that answer will be from either option (c),(d) both correct


only(c) or only (d) correct

but i'm unable to check for them

i'm looking forward to eliminate any of those options(if possible) to reach at correct answer ......

any hint or solution is appreciated because i'm stuck from so long.......thanks in advance


Both (C) (D)would be the correct answer.
Set $g(x)=x^{9}-f(x)$, $h(x)=x-\displaystyle \int_{0}^{\dfrac{\pi-2x}{2}} f(t)\ cost\ dt$, $g(x)$ and $h(x)$ are both continuous

$g(x)$ must have value $0$ at at least a point in $(0,1)$

$h(0)=0-\displaystyle \int_{0}^{\dfrac{\pi}{2}} f(t)\ cost\ dt<0$
$h(1)=1-\displaystyle \int_{0}^{\dfrac{\pi}{2}-1} f(t)\ cost\ dt>1-(\dfrac{\pi}{2}-1)=2-\dfrac{\pi}{2}>0$
$h(x)$ also can reach value $0$


$(c): $

$0<x<1 \to 0<x^{9}<1$


$\to 0\leq|f(x)-x^9|<1$


$\dfrac{d}{dx}\left( x-\displaystyle \int_{0}^{\dfrac{\pi-2x}{2}} f(t)\ cost\ dt \right)=1-(-1)f(\dfrac{\pi-2x}{2})cos(\dfrac{\pi-2x}{2})=1+f(\dfrac{\pi-2x}{2})sin(x)>1$

Can you take it from here?

  • $\begingroup$ @ Mehrdad Zandigohar :yes .thank you $\endgroup$ – Faraday Pathak Feb 24 '18 at 14:55
  • $\begingroup$ @veereshpandey You're welcome veeresh. $\endgroup$ – Mehrdad Zandigohar Feb 24 '18 at 14:56

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