If the tangent to the curve, y = x^{3} + ax – b at the point (1, -5) is perpendicular to the line,

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JEE Mains Previous Paper 1 (Held On: 09 Apr 2019 Shift 1)

Option 4 : (2, -2)

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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Given curve, y = x^{3 }+ ax – b ----(i)

Since, the point (1, -5) lies on the curve.

Put x = 1, y = -5 in equation (i), we get

⇒ (1)^{3} + a(1) – b = -5

⇒1 + a – b = - 5

⇒ a – b = - 6 …. (ii)

Now, \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = 3{{\rm{x}}^2} + {\rm{a}}\)

\(\Rightarrow {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)_{{\rm{\;at\;x}} = 1}} = 3{\left( 1 \right)^2} + {\rm{a}} = 3 + {\rm{a}}\)

Since, required line is perpendicular to y = x – 4,

Then, slope of tangent at the point P(1, -5) = -1

∴ 3 + a = -1

⇒ a = - 4

Put a = - 4 in equation (ii), we get

- 4 – b = - 6

⇒ b = 2

∴ The equation of the curve is y = x^{3 }– 4x – 2

From option (a), put x = 2 in the above equation,

y = 2^{3} – 4(2) – 2

= 8 – 8 – 2 = -2

⇒ (2, -2) lies on the curve.