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The Equations of Tangent at Those Points Where the Curve Y = X2 − 3x + 2 Meets X-axis Are

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प्रश्न

The equations of tangent at those points where the curve y = x2 − 3x + 2 meets x-axis are _______________ .

विकल्प

  • 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

MCQ
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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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 15: Tangents and Normals - Exercise 16.5 [पृष्ठ ४३]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 15 Tangents and Normals
Exercise 16.5 | Q 11 | पृष्ठ ४३
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