A) \[x+y-1=0,x-y-2=0\]
B) \[x-y-1=0,\text{ }x-y=0\]
C) \[x-y+2=0,x-y-1=0\]
D) \[x+y=0,\text{ }x-y=0\]
Correct Answer: A
Solution :
Curve \[y={{x}^{2}}-3x+2\] ...(i) Cuts\[x-\]axis. So putting\[y=0,\] \[{{x}^{2}}-3x+2=0\] \[(x-1)(x-2)=0\] \[\Rightarrow \] \[x=1,2\] \[\therefore \]Intersection points are (1,0) and (2,0) On differentiating Eq. (i) with respect to\[x,\] \[\frac{dy}{dx}=2x-3,\] \[{{\left( \frac{dy}{dx} \right)}_{(1,0)}}=2-3=-1\] \[{{\left( \frac{dy}{dx} \right)}_{(2,0)}}=4-3=1\] \[\therefore \]Equation of tangent at the point (1, 0). \[y-0=-1(x-1)\] \[\Rightarrow \] \[y=-x+1\] \[\Rightarrow \] \[y+x-1=0\] and equation of tangent at the point (2, 0) \[y-0=1(x-2)\] \[y=x-2\] \[\Rightarrow \] \[y-x+2=0\] \[\Rightarrow \] \[x-y-2=0\]
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