Theory of Equations 5 Question 11
11. Find all real values of $x$ which satisfy $x^{2}-3 x+2>0$ and $x^{2}-2 x-4 \leq 0$.
$(1983,2 M)$
Integer Answer Type Question
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Solution:
- Given, $f(x)=4 x^{3}+3 x^{2}+2 x+1$
$$ \begin{aligned} f^{\prime}(x) & =2\left(6 x^{2}+3 x+1\right) \\ \Rightarrow \quad D & =9-24<0 \end{aligned} $$
Hence, $f(x)=0$ has only one real root.
$$ \begin{aligned} f-\frac{1}{2} & =1-1+\frac{3}{4}-\frac{4}{8}>0 \\ f-\frac{3}{4} & =1-\frac{6}{4}+\frac{27}{16}-\frac{108}{64} \\ & =\frac{64-96+108-108}{64}<0 \end{aligned} $$
$f(x)$ changes its sign in $-\frac{3}{4},-\frac{1}{2}$
Hence, $f(x)=0$ has a root in $-\frac{3}{4},-\frac{1}{2}$.
