At the point P(a, a^n) on the graph of y = x^ | vedclass

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(C) Given the curve $y = x^n$.The slope of the tangent at point $P(a, a^n)$ is $\frac{dy}{dx} = n x^{n-1} = n a^{n-1}$.The slope of the normal at poin

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$2$

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(C) Given the curve $y = x^n$.The slope of the tangent at point $P(a, a^n)$ is $\frac{dy}{dx} = n x^{n-1} = n a^{n-1}$.The slope of the normal at point $P$ is $m_n = -\frac{1}{n a^{n-1}}$.The equation of the normal line passing through $(a, a^n)$ is $y - a^n = -\frac{1}{n a^{n-1}}(x - a)$.To find the $y-$ intercept $b$,we set $x = 0$ in the equation:$b - a^n = -\frac{1}{n a^{n-1}}(0 - a)$$b - a^n = \frac{a}{n a^{n-1}} = \frac{1}{n a^{n-2}}$$b = a^n + \frac{1}{n a^{n-2}}$.Now,we evaluate the limit $\mathop {Lim}\limits_{a 	o 0} b = \mathop {Lim}\limits_{a 	o 0} (a^n + \frac{1}{n a^{n-2}})$.If $n=2$,then $b = a^2 + \frac{1}{2 a^0} = a^2 + \frac{1}{2}$.As $a 	o 0$,$b 	o 0 + \frac{1}{2} = \frac{1}{2}$.Thus,$n = 2$ satisfies the given condition.

At the point $P(a, a^n)$ on the graph of $y = x^n$ $(n \in N)$ in the first quadrant,a normal is drawn. The normal intersects the $y-$ axis at the point $(0, b)$. If $\mathop {Lim}\limits_{a \to 0} b = \frac{1}{2}$,then $n$ equals

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