JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 7)
The slope of the tangent to the curve $${\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2}$$ at the point $$(1, 3)$$ is
Answer
8
Explanation
Given curve
$$ \begin{aligned} & \left(y-x^5\right)^2=x\left(1+x^2\right)^2 \\\\ & \Rightarrow 2\left(y-x^5\right)\left(\frac{d y}{d x}-5 x^4\right)=\left(1+x^2\right)^2+2 x\left(1+x^2\right) \cdot 2 x \end{aligned} $$
Now putting $(1,3)$ in it, we get
$$ 2(3-1)\left(\frac{d y}{d x}-5\right)=1\{2(2) 2\}+(1+1)^2 $$
$\Rightarrow 4\left(\frac{d y}{d x}-5\right)=8+4 \Rightarrow \frac{d y}{d x}=8$
Thus, the slope at $(1,3)$ is 8 .
$$ \begin{aligned} & \left(y-x^5\right)^2=x\left(1+x^2\right)^2 \\\\ & \Rightarrow 2\left(y-x^5\right)\left(\frac{d y}{d x}-5 x^4\right)=\left(1+x^2\right)^2+2 x\left(1+x^2\right) \cdot 2 x \end{aligned} $$
Now putting $(1,3)$ in it, we get
$$ 2(3-1)\left(\frac{d y}{d x}-5\right)=1\{2(2) 2\}+(1+1)^2 $$
$\Rightarrow 4\left(\frac{d y}{d x}-5\right)=8+4 \Rightarrow \frac{d y}{d x}=8$
Thus, the slope at $(1,3)$ is 8 .
Comments (0)
