SATHEE: Theory of Equations 5 Question 11

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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Answer:

Correct Answer: 11. (c)

Solution:

Given, $f (x) = 4 x^{3} + 3 x^{2} + 2 x + 1$

$\begin{aligned} f^{'} (x) & = 2 (6 x^{2} + 3 x + 1) \\ \Rightarrow D & = 9 - 24 < 0 \end{aligned}$

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 $- \frac{3}{4}, - \frac{1}{2}$ .

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Theory of Equations 5 Question 11

11. Find all real values of x which satisfy x2−3x+2>0 and x2−2x−4≤0.

Integer Answer Type Question

Answer:

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11. Find all real values of $x$ which satisfy $x^{2} - 3 x + 2 > 0$ and $x^{2} - 2 x - 4 \leq 0$ .