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Solution for If the Tangent to the Curve Y = X3 + Ax + B at (1, − 6) is Parallel to the Line X − Y + 5 = 0, Find a and B ? - CBSE (Commerce) Class 12 - Mathematics

Question

If the tangent to the curve y = x3 + ax + b at (1, − 6) is parallel to the line x − y + 5 = 0, find a and b ?

Solution

$\text { Given }:$

$x - y + 5 = 0$

$\Rightarrow y = x + 5$

$\Rightarrow \frac{dy}{dx} = 1$

$\text { Now,}$

$y = x^3 + ax + b . . . \left( 1 \right)$

$\Rightarrow \frac{dy}{dx} = 3 x^2 + a$

$\text { Slope of the tangent at }\left( 1, - 6 \right)= \text { Slope of the given line }$

$\Rightarrow \left( \frac{dy}{dx} \right)_\left( 1, - 6 \right) = 1$

$\Rightarrow 3 + a = 1$

$\Rightarrow a = - 2$

$\text { On substitutinga }= - 2, x=1 \text { and}y=-6 \text { in eq.} (1), \text { we get}$

$- 6 = 1 - 2 + b$

$\Rightarrow b = - 5$

$\therefore a = - 2 \text { and} \ b = - 5$

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Solution for question: If the Tangent to the Curve Y = X3 + Ax + B at (1, − 6) is Parallel to the Line X − Y + 5 = 0, Find a and B ? concept: Tangents and Normals. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)
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