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Question

# If x is real, then the maximum and minimum values of the expression x2−3x+4x2+3x+4 will be

A
2,1
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B
5,15
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C
7,17
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D
None of these
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Solution

## The correct option is C 7,17Let y=x2−3x+4x2+3x+4 ⇒(y−1)x2+3(y+1)x+4(y−1)=0 For x is real D≥0 ⇒9(y+1)2−16(y−1)2≥0 ⇒−7y2+50y−7≥0⇒7y2−50y+7≤0 ⇒(y−7)(7y−1)≤0 Now, the product of two factors is negative if one in -ve and one in +ve. Case I : (y−7)≥0 and (7y−1)≤0 ⇒y≥7 and y≤17. But it is impossible Case II : (y−7)≤0 and (7y−1)≥0 ⇒y≤7 and y≥17⇒17≤y≤7 Hence maximum value is 7 and minimum value is 17

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