JEE PYQ: Application Of Derivatives Question 2
Question 2 - 2021 (16 Mar Shift 1)
If the normal to the curve $y(x) = \int_0^x (2t^2 - 15t + 10),dt$ at a point $(a, b)$ is parallel to the line $x + 3y = -5$, $a > 1$, then the value of $|a + 6b|$ is equal to ______.
Show Answer
Answer: 406
Solution
$y’(x) = 2x^2 - 15x + 10$. Slope of normal $= -\frac{1}{3}$, so $y’(a) = 3$. Solving $2a^2 - 15a + 10 = 3$ gives $a = 7$ (rejecting $a = \frac{1}{2}$). Then $b = \int_0^7 (2t^2 - 15t + 10),dt = \frac{2 \cdot 343}{3} - \frac{15 \cdot 49}{2} + 70$. So $6b = 4 \times 7^3 - 45 \times 49 + 60 \times 7$ and $|a + 6b| = 406$.
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