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The set of values of $x$ for which the inequalities $x^2-3 x-10 < 0, \quad 10 x-x^2-16>0 \quad$ hold simultaneously, is
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The correct answer is:
$(2,5)$
Given inequalities are
$\begin{array}{lrl}x^2-3 x-10 & \\ \Rightarrow & x^2-5 x+2 x-10 & < 0 \\ \Rightarrow & & x \in(-2,5) \\ \Rightarrow & & (x+2)(x-5) < 0 \\ \text { and } & & 10 x-x^2-16>0 \\ \Rightarrow & x^2-10 x+16 < 0 \\ \Rightarrow & x^2-8 x-2 x+16 < 0 \\ \Rightarrow & & (x-2)(x-8) < 0 \\ \Rightarrow & x \in(-2,5) \cap(2,8) \\ \therefore & x \in(2,5)\end{array}$
$\begin{array}{lrl}x^2-3 x-10 & \\ \Rightarrow & x^2-5 x+2 x-10 & < 0 \\ \Rightarrow & & x \in(-2,5) \\ \Rightarrow & & (x+2)(x-5) < 0 \\ \text { and } & & 10 x-x^2-16>0 \\ \Rightarrow & x^2-10 x+16 < 0 \\ \Rightarrow & x^2-8 x-2 x+16 < 0 \\ \Rightarrow & & (x-2)(x-8) < 0 \\ \Rightarrow & x \in(-2,5) \cap(2,8) \\ \therefore & x \in(2,5)\end{array}$
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