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If $y=5x^2+3$, then the tangent at $x=0$, $y=3$

  1. passes through $x=0,y=0$
  2. has a slope of $+1$
  3. is parallel to the $x$-axis
  4. has a slope of $-1$
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Given, $y=5x^2+3$
Slope $dy/dx = 10x$
Given point $x=0,y=3: (0,3)$
Now, at $(0,3)$ we have the slope as $dy/dx = 10 \times 0=0,$ which means no matter how much ever we change 'x',change in y is zero or y is a constant (y=3,here) i.e line is parallel to $X$- axis.
Options B and D eliminated as they have slope as '+1' and '-1'.
Option A is eliminated as given the tangent is at point $(0,3)$ and we found $slope=0$, which means it is parallel to $X$ axis. The equation of line is nothing but $y=3$, which never passes through $(0,0).$ Hence A too eliminated.
Answer: Option $C)$
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Option (c) 

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