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If the slope of the tangent of the curve $y=a x^3+b x+4$ at $(2,14)$ is 21 , then the values of $a$ and $b$ respectively
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Verified Answer
The correct answer is:
$2,-3$
The curve $y=a x^3+b x+4$ is passes through (2, 14).
$$
\begin{array}{rlrlrl}
& \therefore & & 14 & =a(2)^3+b(2)+4 \\
\Rightarrow & & 14 & =8 a+2 b+4 \\
\Rightarrow & & 5 & =4 a+b
\end{array}
$$
Slope of tangent to the curve $y=a x^3+b x+4$
i.e. $\frac{d y}{d x}=3 a x^2+b$
$$
\begin{array}{rlrl}
\Rightarrow & \left(\frac{d y}{d x}\right)_{(2,14)} & =3 a(2)^2+b \\
\Rightarrow & & 21 & =12 a+b
\end{array}
$$
$$
\left[\because \frac{d y}{d x}=21\right]
$$
Solving Eqs. (i) and (ii), we get
$$
a=2, b=-3
$$
$$
\begin{array}{rlrlrl}
& \therefore & & 14 & =a(2)^3+b(2)+4 \\
\Rightarrow & & 14 & =8 a+2 b+4 \\
\Rightarrow & & 5 & =4 a+b
\end{array}
$$
Slope of tangent to the curve $y=a x^3+b x+4$
i.e. $\frac{d y}{d x}=3 a x^2+b$
$$
\begin{array}{rlrl}
\Rightarrow & \left(\frac{d y}{d x}\right)_{(2,14)} & =3 a(2)^2+b \\
\Rightarrow & & 21 & =12 a+b
\end{array}
$$
$$
\left[\because \frac{d y}{d x}=21\right]
$$
Solving Eqs. (i) and (ii), we get
$$
a=2, b=-3
$$
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