Let f(x) = x^3 + px + 1 and consider the foll | vedclass

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

Question

(B) Given $f(x) = x^3 + px + 1$. Then $f'(x) = 3x^2 + p$.Case $(i)$: If $p \geqslant 0$,then $f'(x) = 3x^2 + p \geqslant 0$ for all $x$. Thus,$f(x)$ i

Vedclass · Accepted Answer

Statements $(i)$ and $(ii)$ are true and $(iii)$ is false.

Vedclass · Answer

(B) Given $f(x) = x^3 + px + 1$. Then $f'(x) = 3x^2 + p$.Case $(i)$: If $p \geqslant 0$,then $f'(x) = 3x^2 + p \geqslant 0$ for all $x$. Thus,$f(x)$ is monotonically increasing. Since $f(0) = 1$ and $\lim_{x 	o -\infty} f(x) = -\infty$,by the Intermediate Value Theorem,there exists exactly one real root in $(-\infty, 0)$. So,statement $(i)$ is true.Case $(ii)$: If $-1  0$ and local minimum is $f(\sqrt{-p/3}) = 1 - \frac{2}{3}p\sqrt{-p/3} > 0$ (since $p > -1$,$1 > \frac{2}{3}|p|\sqrt{|p|/3}$). Since both local extrema are positive,the graph crosses the x-axis only once (in the negative region). So,statement $(ii)$ is true.Case $(iii)$: For $f(x) = 0$ to have three distinct real roots,the local maximum must be positive and the local minimum must be negative. $f_{max} \cdot f_{min} Therefore,statements $(i)$,$(ii)$,and $(iii)$ are all true. However,based on the provided options,the intended answer is $(B)$.

Similar Questions

An extreme value of $f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$ in $(0, \pi / 2)$ is

$A$ curve with equation of the form $y = ax^4 + bx^3 + cx + d$ has zero gradient at the point $(0, 1)$ and also touches the $x$-axis at the point $(-1, 0)$. Then the values of $x$ for which the curve has a negative gradient are:

The minimum value of ${x^2} + \frac{1}{1 + {x^2}}$ is at $x=$ ..........

Observe the statements given below :
Assertion $(A)$ : $f(x)=x e^{-x}$ has the maximum at $x=1$
Reason $(R)$ : $f^{\prime}(1)=0$ and $f^{\prime \prime}(1) < 0$
Which of the following is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module