The tangent to the curve y = x^2 - 5x + 5, parallel to the… — JEE Main 2019 Mathematics PYQ

JEE Main 2019 · 12 Jan · Shift II

The tangent to the curve

y = x^2 - 5x + 5

, parallel to the line

2y = 4x + 1

, also passes through the point

A.
$\left(\dfrac{1}{4}, 2\right)$
B.
$\left(\dfrac{7}{2}, \dfrac{1}{4}\right)$
C.
$\left(-\dfrac{1}{8}, 7\right)$
D.
$\left(\dfrac{1}{8}, -7\right)$

Show answer and solution

Correct option: D

Solution

The line 2y = 4x + 1 has slope 2. Setting y' = 2x - 5 = 2 gives x = 7/2, and y = (7/2)² - 5(7/2) + 5 = 49/4 - 35/2 + 5 = -1/4. So the tangent line passes through (7/2, -1/4) with slope 2: y + 1/4 = 2(x - 7/2), i.e., y = 2x - 15/4. Checking (1/8, -7): 2(1/8) - 15/4 = 1/4 - 15/4 = -14/4 = -7/2 ≠ -7... Let me recheck: y = (7/2)² - 5(7/2) + 5 = 49/4 - 70/4 + 20/4 = -1/4. Tangent: y = 2x - 7 + (-1/4) = 2(1/8) - 7 - 1/4. Equation: y - (-1/4) = 2(x - 7/2) → y = 2x - 7 - 1/4 = 2x - 29/4. At x = 1/8: y = 2/8 - 29/4 = 1/4 - 29/4 = -28/4 = -7. ✓ So the tangent line y = 2x - 29/4 passes through (1/8, -7), confir

[JEE Main 2019 · 12 Jan · Shift II]

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