If m is slope of common tangents y = $x^{2}$ - x + 1, y = $x^{2}$ - 3x + 1 then m is

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-2

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Solution

The correct option is D -2
Let y = mx + c is common tangent
$∴ m x + c = x^{2} - x + 1 \Rightarrow x^{2} - (1 + 4) x + 1 - c = 0 m x + c = x^{2} - 3 x + 1 x^{2} - (3 + m) x + 1 - c = 0$
Clearly $Δ = 0$
$\Rightarrow (1 + 4)^{2} - 4 (1 - c) = 0 \Rightarrow m^{2} + 2 m - 3 + 4 c = 0 \Rightarrow (3 + m)^{2} - 4 (1 - c) = 0 \Rightarrow m^{2} + 6 m + 5 + 4 c = 0$
On equality $^{'} m^{2} + 4 c^{'}$ value on both the equations
2m-3=6m+5
$\Rightarrow m = - 2$

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