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4-2.Quadratic Equations and Inequations
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मान लें कि $a$ एक धनात्मक वास्तविक संख्या इस प्रकार है कि $a^5-a^3+a=2$. तब
A
$a^6 < 2$
B
$2 < a^6 < 3$
C
$3 < a^6 < 4$
D
$4 \leq a^6$
(KVPY-2016)
Solution
(c)
Given, $a^5-a^3+a=2$
$\Rightarrow \quad a^5-a^3+a-2=0$
Let $f(a)=a^5-a^3+a-2$
$f^{\prime}(a)=5 a^4-3 a^2+1$
$f^{\prime}(a)>0, \forall a \in R$
$\therefore a^5-a^3+a-2=0$ 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 $(3,4)$.
$\therefore 3 < a^6 < 4$
Standard 11
Mathematics
