If the equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x = 0$,where $a_1 \neq 0$ and $n \ge 2$,has a positive root $x = \alpha$,then the equation $n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$ has a positive root,which is:

  • A
    Greater than or equal to $\alpha$
  • B
    Equal to $\alpha$
  • C
    Greater than $\alpha$
  • D
    Smaller than $\alpha$

Explore More

Similar Questions

Given $f(x) = 4 - (\frac{1}{2} - x)^{2/3}$,$g(x) = \begin{cases} \frac{\tan([x])}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$,$h(x) = \{x\}$,and $k(x) = 5^{\log_2(x + 3)}$. Then,in the interval $[0, 1]$,Lagrange's Mean Value Theorem is $NOT$ applicable to:

If $f:[-5,5] \rightarrow R$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere,then prove that $f(-5) \neq f(5).$

If Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1]$ for the point $c = \frac{1}{2}$,then the value of $2a + b$ is

If $a, b, c \in \mathbb{R}$ and satisfy $3a + 5b + 15c = 0$,then the equation $ax^4 + bx^2 + c = 0$ has:

If $f(x)$ is a differentiable function,$f^{\prime}(x) \geq 5$ for all $x \in [2, 6]$,$f(2) = 4$ and $f(3) = 15$,then a possible value of $f(6)$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo