MCQ
Suppose $a$ is a positive real number such that $a^5-a^3+a=2$. Then,
- A$a^6 < 2$
- B$2 < a^6 < 3$
- ✓$3 < a^6 < 4$
- D$4 \leq a^6$
Given, $a^5-a^3+a=2$
$\Rightarrow \quad a^5-a^3+a-2=0$
Let $f(a)=a^5-a^3+a-2$
$f^{\prime}(a)=5 a^4-3 a^2+1$
$f^{\prime}(a)>0, \forall a \in R$
$\therefore a^5-a^3+a-2=0$ has only one roots.
for $a^6=3 \Rightarrow a=(3)^{1 / 6}=12$ [by calculation]
$f\left(4^{1 / 6}\right) > 0$ and at $a^6=4, a=(4)^{1 / 6}$
So one root lies in $(3,4)$.
$\therefore 3 < a^6 < 4$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ The ratio of the areas of the triangles $P Q S$ and $P Q R$ is
$(A)$ $1: \sqrt{2}$ $(B)$ $1: 2$ $(C)$ $1: 4$ $(D)$ $1: 8$
$2.$ The radius of the circumcircle of the triangle PRS is
$(A)$ $5$ $(B)$ $3 \sqrt{3}$ $(C)$ $3 \sqrt{2}$ $(D)$ $2 \sqrt{3}$
$3.$ The radius of the incircle of the triangle $P Q R$ is
$(A)$ $4$ $(B)$ $3$ $(C)$ $8 / 3$ $(D)$ $2$
Give the answer question $1,2$ and $3.$