Let f: R rightarrow (0,1) be a continuous fun | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For option $[A]$,let $g(x) = e^x - \int_0^x f(t) \sin t \, dt$. Since $f(t) \in (0,1)$ and $\sin t \in [0,1]$ for $t \in (0,1)$,the integral $\int

Vedclass · Accepted Answer

$B, D$

Vedclass · Answer

(A) For option $[A]$,let $g(x) = e^x - \int_0^x f(t) \sin t \, dt$. Since $f(t) \in (0,1)$ and $\sin t \in [0,1]$ for $t \in (0,1)$,the integral $\int_0^x f(t) \sin t \, dt  e^x - x$. Since $e^x - x > 0$ for all $x$,$g(x)$ cannot be zero.For option $[B]$,let $h(x) = x^9 - f(x)$. Since $f(x) \in (0,1)$,$h(0) = 0^9 - f(0) = -f(0)  0$. By the Intermediate Value Theorem $(IVT)$,there exists $c \in (0,1)$ such that $h(c) = 0$.For option $[C]$,let $k(x) = f(x) + \int_0^{\pi/2} f(t) \sin t \, dt$. Since $f(x) > 0$ and the integral is positive,$k(x) > 0$ for all $x \in (0,1)$. Thus,it cannot be zero.For option $[D]$,let $m(x) = x - \int_0^{\pi/2 - x} f(t) \cos t \, dt$. Then $m(0) = 0 - \int_0^{\pi/2} f(t) \cos t \, dt  0$. By $IVT$,there exists $c \in (0,1)$ such that $m(c) = 0$.

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 $(0,1)$?

Similar Questions

Let the function $f(x) = x^2 + x + \sin x - \cos x + \log(1 + |x|)$ be defined over the interval $[0, 1]$. The odd extension of $f(x)$ to the interval $[-1, 1]$ is:

Let $f(x) = a^x$ $(a > 0)$ be written as $f(x) = f_1(x) + f_2(x)$,where $f_1(x)$ is an even function and $f_2(x)$ is an odd function. Then $f_1(x + y) + f_1(x - y)$ equals

If the function $f:(-\infty,-1] \rightarrow(a, b]$ defined by $f(x)=e^{x^3-3 x+1}$ is one-one and onto,then the distance of the point $P(2 b+4, a+2)$ from the line $x+e^{-3} y=4$ is:

Which of the following statements is false?

Which of the four statements given below is different from the others?

Vedclass Test Series

Exam Paper Generator

Online Exam Module