The minimum possible value of the expression 58−42x+149−1401−x2,−1≤x≤1 equals k then the sum of the digits of k2 equals

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The minimum possible value of the expression $\sqrt{58 - 42 x} + \sqrt{149 - 140 \sqrt{1 - x^{2}}}, - 1 \leq x \leq 1$ equals $k$ then the sum of the digits of $k^{2}$ equals

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answer is 10.

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Detailed Solution

Put $x = \cos θ$ , see that two cosine rules for two triangles. Max value $= \sqrt{109} \Rightarrow s u m = 10$

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