TS EAMCET · Maths · Application of Derivatives
The curve \(y=x^3-2 x^2+3 x-4\) intersects the horizontal line \(y=-2\) at the point \(P(h, k)\). If the tangent drawn to this curve at P meets the \(X\)-axis at \(\left(x_1, y_1\right)\), then \(x_1=\)
- A 1
- B 2
- C 3
- D -3
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & x^3-2 x^2+3 x-4=-2 \\ & \Rightarrow x^3-2 x^2+3 x-2=0 \end{aligned}\) By hit and trial method, \(x=1\) Point of intersection is \((1,-2)\) \(\frac{d y}{d x}=3 x^2-4 x+3 ;\left.\frac{d y}{d x}\right|_{(1,-2)}=3-4+3=2\) Equation of tangent is \(y+2=2(x-1)\)…
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