What is the slope of normal to the curve y = 2x3 - 5x2 + x - 2 at the point (1, -1)?

Question

Question

This question was previously asked in

AAI ATC Junior Executive Official Paper (Held On: 27 Dec, 2023 Shift 2)

Answer 1

Concept:

The slope of the tangent to a curve y = f(x) is m = \(\rm dy\over dx\)

The slope of the normal = \(\rm -{1\over m}\) = \(\rm -{1\over {dy\over dx}}\)

Calculation:

Given curve y = 2x³ - 5x² + x - 2

Differentiating the equation wrt x

\(\rm dy\over dx\) = 6x² - 10x + 1

Slope at the point (1, -1)

\(\rm dy\over dx\) at x = 1 = 6(1)² - 10(1) + 1

\(\rm dy\over dx\) = 6 - 10 + 1

\(\rm dy\over dx\) = -3

The slope of the normal (m') = \(\rm -{1\over {dy\over dx}}\)

m' = \(\boldsymbol{\rm -{1 \over -3}}\) = \(\boldsymbol{\rm {1 \over 3}}\)

The slope of the normal to the curve at the point (1, -1) is 1/3.

∴ Option 2 is correct