Suppose a is a positive real number such that | vedclass

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

Question

(c)

Given, $a^5-a^3+a=2$

$\Rightarrow \quad a^5-a^3+a-2=0$

Let $f(a)=a^5-a^3+a-2$

$f^{\prime}(a)=5 a^4-3 a^2+1$

$f^{\prime}(a)>0, \f

Vedclass · Accepted Answer

$3 < a^6 < 4$

Vedclass · Answer

(c) Given, $a^5-a^3+a=2$ $\Rightarrow \quad a^5-a^3+a-2=0$ Let $f(a)=a^5-a^3+a-2$ $f^{\prime}(a)=5 a^4-3 a^2+1$ $f^{\prime}(a)>0, \forall a \in R$ $ herefore a^5-a^3+a-2=0$ has only one roots. for $a^6=3 \Rightarrow a=(3)^{1 / 6}=12$ [by calculation] $f\left(4^{1 / 6} ight) > 0$ and at $a^6=4, a=(4)^{1 / 6}$ So one root lies in $(3,4)$. $ herefore 3 < a^6 < 4$

Suppose $a$ is a positive real number such that $a^5-a^3+a=2$. Then,

Similar Questions

If the graph of $y = ax^3 + bx^2 + cx + d$ is symmetric about the line $x = k$ then

For a real number $x$, let $[x]$ denote the largest integer less than or equal to $x$, and let $\{x\}=x-[x]$. The number of solutions $x$ to the equation $[x]\{x\}=5$ with $0 \leq x \leq 2015$ is

If $\alpha , \beta$ and $\gamma$ are the roots of ${x^3} + 8 = 0$, then the equation whose roots are ${\alpha ^2},{\beta ^2}$ and ${\gamma ^2}$ is

If $S$ is a set of $P(x)$ is polynomial of degree $ \le 2$ such that $P(0) = 0,$$P(1) = 1$,$P'(x) > 0{\rm{ }}\forall x \in (0,\,1)$, then

Let $\alpha$ and $\beta$ be two real numbers such that $\alpha+\beta=1$ and $\alpha \beta=-1 .$ Let $p _{ n }=(\alpha)^{ n }+(\beta)^{ n },p _{ n -1}=11$ and $p _{ n +1}=29$ for some integer $n \geq 1 .$ Then, the value of $p _{ n }^{2}$ is .... .