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Question

If the normal to the curve $$y=f(x)$$ at the point $$(3,4) $$ makes an angle $$3\pi /4 $$ with the positive x-axis, then $$f'(3)=$$


A
1
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B
0
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C
1
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D
3
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Solution

The correct option is C $$1$$
Tangent being perpendicular to given line of slope 2, will have its slope as -$$\displaystyle \dfrac{1}{2}$$.
Slope of tangent=$$\displaystyle -\dfrac{fx}{fy}=-\dfrac{6x+1}{2(y+1)}=-\dfrac{1}{2}\therefore y=6x$$.
Sloping with the given curve, we have $$\displaystyle 3x^{2}+36x^{2}+x+12x=0 $$
or $$13x(3x+1)=0\therefore x=0, -1/3 \therefore y=0, -2$$
 Hence the two points are $$\displaystyle (0,0),\left ( -\dfrac{1}{3},-2 \right )$$ $$\displaystyle \therefore y=-\dfrac{1}{2}x$$ and $$y+2=-\dfrac{1}{2}\left ( x+\dfrac{1}{3} \right ) $$ or $$ 2y+x=0$$ and $$\displaystyle 2y+x+\dfrac{13}{3}=0$$

Ans: D

Mathematics

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