PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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Solution for The Equations of Tangent at Those Points Where the Curve Y = X2 − 3x + 2 Meets X-axis Are (A) X − Y + 2 = 0 = X − Y − 1 (B) X + Y − 1 = 0 = X − Y − 2 (C) X − Y − 1 = 0 = X − Y (D) X − Y = 0 = X + Y - PUC Karnataka Science Class 12 - Mathematics

Question

The equations of tangent at those points where the curve y = x2 − 3x + 2 meets x-axis are
(a) x − y + 2 = 0 = x − y − 1
(b) x + y − 1 = 0 = x − y − 2
(c) x − y − 1 = 0 = x − y
(d) x − y = 0 = x + y

Solution

(b) x + y − 1 = 0 = x − y − 2
Let the tangent meet the x-axis at point (x, 0).
Now,

$y = x^2 - 3x + 2$

$\Rightarrow \frac{dy}{dx} = 2x - 3$

$\text { The tangent passes through point (x, 0) }.$

$\therefore 0 = x^2 - 3x + 2$

$\Rightarrow \left( x - 2 \right)\left( x - 1 \right) = 0$

$\Rightarrow x = 2 \ or \ x = 1$

$\text { Case 1: When } x=2:$

$\text { Slope of the tangent },m= \left( \frac{dy}{dx} \right)_\left( 2, 0 \right) =4-3=1$

$\therefore \left( x_1 , y_1 \right) = \left( 2, 0 \right)$

$\text { Equation of the tangent }:$

$y - y_1 = m \left( x - x_1 \right)$

$\Rightarrow y - 0 = 1 \left( x - 2 \right)$

$\Rightarrow x - y - 2 = 0$

$\text { Case 2: When } x=1:$

$\text { Slope of the tangent },m= \left( \frac{dy}{dx} \right)_\left( 2, 0 \right) =2-3=-1$

$\therefore \left( x_1 , y_1 \right) = \left( 1, 0 \right)$

$\text { Equation of the tangent }:$

$y - y_1 = m \left( x - x_1 \right)$

$\Rightarrow y - 0 = - 1 \left( x - 1 \right)$

$\Rightarrow x + y - 1 = 0$

Is there an error in this question or solution?
Solution The Equations of Tangent at Those Points Where the Curve Y = X2 − 3x + 2 Meets X-axis Are (A) X − Y + 2 = 0 = X − Y − 1 (B) X + Y − 1 = 0 = X − Y − 2 (C) X − Y − 1 = 0 = X − Y (D) X − Y = 0 = X + Y Concept: Tangents and Normals.
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