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Introduction to linear inequalities

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Question

$If f (x) = a x^{2} + b x + c and f (- 1) \geq - 4, f (1) \leq 0 and f (3) \geq 5 then the least value of a is$

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Solution

$\begin{array}{l} f (- 1) \geq - 4 \\ \Rightarrow a - b + c \geq - 4 - - - - - (i) \\ f (1) \leq 0 \\ \Rightarrow a + b + c \leq 0 \\ \Rightarrow - a - b - c \geq 0 - - - - (i i) \\ f (3) \geq 5 \\ and 9 a + 3 b + c \geq 5 - - - - (i i i) \\ From (i) + (ii) \\ \Rightarrow - 2 b \geq - 4 - - - - (i v) \\ From (ii) + (iii) + (iv) \\ \Rightarrow 8 a \geq 1 \\ \Rightarrow a \geq 1 / 8 \end{array}$

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The solution of $x - 1 = (x - [x]) (x - {x})$ (where $[x]$ and ${x}$ are the integral and fractional part of $x$ ) is :

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