ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 6. Application of derivatives

If the normal to the curve \(y(x)=\int_{0}^{x}\left(2 t^{2}-15 t+10\right) d t\) at a point \((a, b)\) is parallel to the line \(x+3 y=-5, a>1,\) then the value of \(|a +6 b|\) is equal to..........

  1. A \(324\)
  2. B \(406\)
  3. C \(512\)
  4. D \(376\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(406\)

Step-by-step Solution

Detailed explanation

\(y(x)=\int_{0}^{x}\left(2 t^{2}-15 t+10\right) d t\) \(\left.y^{\prime}(x)\right]_{x=a}=\left[2 x^{2}-15 x+10\right]_{a}=2 a^{2}-15 a+10\) Slope of normal \(=-\frac{1}{3}\) \(\Rightarrow \quad 2 a^{2}-15 a+10=3 \Rightarrow a=7\) \(a=\frac{1}{2}(\) rejected \()\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app