TS EAMCET · Maths · Application of Derivatives
If the slope of the tangent drawn at any point \((x, y)\) to the curve \(\mathrm{y}=f(\mathrm{x})\) is \(3 \mathrm{x}^2-5\) and \(f(1)=2\), then the tangent at \((1,2)\) to the curve \(\mathrm{y}=f(\mathrm{x})\) intersects the curve at the point
- A \((2,0)\)
- B \((-2,8)\)
- C \((3,-2)\)
- D \((-1,6)\)
Answer & Solution
Correct Answer
(B) \((-2,8)\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } \frac{d y}{d x}=3 x^2-5 \\ & \Rightarrow d y=\left(3 x^2-5\right) d x \\ & \Rightarrow \quad f(x)=y=\int\left(3 x^2-5\right) d x=x^3-5 x+C \\ & \quad f(1)=2 \Rightarrow 2=1-5+C \Rightarrow C=6 \\ & \therefore \quad f(x)=x^3-5 x+6 \\ & \left.\frac{d…
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