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प्रश्न
The equations of tangent at those points where the curve y = x2 − 3x + 2 meets x-axis are _______________ .
पर्याय
x − y + 2 = 0 = x − y − 1
x + y − 1 = 0 = x − y − 2
x − y − 1 = 0 = x − y
x − y = 0 = x + y
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उत्तर
`x + y − 1 = 0 = x − y − 2`
Let the tangent meet the x-axis at point (x, 0).
Now,
\[y = x^2 - 3x + 2\]
\[ \Rightarrow \frac{dy}{dx} = 2x - 3\]
\[\text { The tangent passes through point (x, 0) }.\]
\[ \therefore 0 = x^2 - 3x + 2\]
\[ \Rightarrow \left( x - 2 \right)\left( x - 1 \right) = 0\]
\[ \Rightarrow x = 2 \ or \ x = 1\]
\[\text { Case 1: When } x=2:\]
\[\text { Slope of the tangent },m= \left( \frac{dy}{dx} \right)_\left( 2, 0 \right) =4-3=1\]
\[ \therefore \left( x_1 , y_1 \right) = \left( 2, 0 \right)\]
\[\text { Equation of the tangent }:\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 0 = 1 \left( x - 2 \right)\]
\[ \Rightarrow x - y - 2 = 0\]
\[\text { Case 2: When } x=1:\]
\[\text { Slope of the tangent },m= \left( \frac{dy}{dx} \right)_\left( 2, 0 \right) =2-3=-1\]
\[ \therefore \left( x_1 , y_1 \right) = \left( 1, 0 \right)\]
\[\text { Equation of the tangent }:\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 0 = - 1 \left( x - 1 \right)\]
\[ \Rightarrow x + y - 1 = 0\]
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