Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then
Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then
- A
$f(x) = 0$ has no root in $(0,1)$
- B
$f(x) = 0$ has at least one root in $(0,1)$
- C
$f' (c)$ vanishes for some $c\, \in \,(0,1)$
- D
Both $(B)$ and $(C)$
Similar Questions
Suppose that $f (0) = - 3$ and $f ' (x) \le 5$ for all values of $x$. Then the largest value which $f (2)$ can attain is
Suppose that $f (0) = - 3$ and $f ' (x) \le 5$ for all values of $x$. Then the largest value which $f (2)$ can attain is
Which of the following function can satisfy Rolle's theorem ?
Which of the following function can satisfy Rolle's theorem ?
Examine the applicability of Mean Value Theorem:
$(i)$ $f(x)=[x]$ for $x \in[5,9]$
$(ii)$ $f(x)=[x]$ for $x \in[-2,2]$
$(iii)$ $f(x)=x^{2}-1$ for $x \in[1,2]$
Examine the applicability of Mean Value Theorem:
$(i)$ $f(x)=[x]$ for $x \in[5,9]$
$(ii)$ $f(x)=[x]$ for $x \in[-2,2]$
$(iii)$ $f(x)=x^{2}-1$ for $x \in[1,2]$
Let $f(x)=2+\cos x$ for all real $x$.
$STATEMENT -1$ : For each real $\mathrm{t}$, there exists a point $\mathrm{c}$ in $[\mathrm{t}, \mathrm{t}+\pi]$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$. because
$STATEMENT -2$: $f(t)=f(t+2 \pi)$ for each real $t$.
Let $f(x)=2+\cos x$ for all real $x$.
$STATEMENT -1$ : For each real $\mathrm{t}$, there exists a point $\mathrm{c}$ in $[\mathrm{t}, \mathrm{t}+\pi]$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$. because
$STATEMENT -2$: $f(t)=f(t+2 \pi)$ for each real $t$.
- [IIT 2007]
If $c$ is a point at which Rolle's theorem holds for the function, $f(\mathrm{x})=\log _{\mathrm{e}}\left(\frac{\mathrm{x}^{2}+\alpha}{7 \mathrm{x}}\right)$ in the interval $[3,4],$ where $\alpha \in \mathrm{R},$ then $f^{\prime \prime}(\mathrm{c})$ is equal to
If $c$ is a point at which Rolle's theorem holds for the function, $f(\mathrm{x})=\log _{\mathrm{e}}\left(\frac{\mathrm{x}^{2}+\alpha}{7 \mathrm{x}}\right)$ in the interval $[3,4],$ where $\alpha \in \mathrm{R},$ then $f^{\prime \prime}(\mathrm{c})$ is equal to
- [JEE MAIN 2020]