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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
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Verified Answer
The correct answer is:
$(-2,8)$
$$
\begin{aligned}
& \text { } \frac{d y}{d x}=3 x^2-5 \\
& \Rightarrow d y=\left(3 x^2-5\right) d x \\
& \Rightarrow \quad f(x)=y=\int\left(3 x^2-5\right) d x=x^3-5 x+C \\
& \quad f(1)=2 \Rightarrow 2=1-5+C \Rightarrow C=6 \\
& \therefore \quad f(x)=x^3-5 x+6 \\
& \left.\frac{d y}{d x}\right|_{(1,2)}=-2
\end{aligned}
$$
Equation of tangent at $(1,2)$
$$
\begin{aligned}
& y-2=-2(x-1) \Rightarrow y-2=-2 x+2 \\
\Rightarrow & 2 x+y=4 \\
& y=4-2 x
\end{aligned}
$$
From (1) and (2)
$$
\begin{aligned}
& 4-2 x=x^3-5 x+6 \Rightarrow x^3-3 x+2=0 \\
\therefore \quad & x=1,-2
\end{aligned}
$$
For $x=1 \Rightarrow y=4-2 x=2$
For $x=-2 \Rightarrow y=4-2(-2)=8$
$\therefore \quad(-2,8)$ is solution.
\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 y}{d x}\right|_{(1,2)}=-2
\end{aligned}
$$
Equation of tangent at $(1,2)$
$$
\begin{aligned}
& y-2=-2(x-1) \Rightarrow y-2=-2 x+2 \\
\Rightarrow & 2 x+y=4 \\
& y=4-2 x
\end{aligned}
$$
From (1) and (2)
$$
\begin{aligned}
& 4-2 x=x^3-5 x+6 \Rightarrow x^3-3 x+2=0 \\
\therefore \quad & x=1,-2
\end{aligned}
$$
For $x=1 \Rightarrow y=4-2 x=2$
For $x=-2 \Rightarrow y=4-2(-2)=8$
$\therefore \quad(-2,8)$ is solution.
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