A) 2
B) 0
C) 1
D) Infinite
Correct Answer: D
Solution :
[d] Here, \[{{f}_{1}}(x)={{x}^{2}}-x+1\] and \[{{f}_{2}}(x)={{x}^{3}}-{{x}^{2}}-2x+1\] or \[{{f}_{1}}'({{x}_{1}})=2{{x}_{1}}-1\] and \[{{f}_{2}}'({{x}_{2}})=3x_{2}^{2}-2{{x}_{2}}-2\] Let the tangents drawn to the curves \[y={{f}_{1}}(x)\] and \[y={{f}_{2}}(x)\] at \[({{x}_{1}},{{f}_{1}}({{x}_{1}}))\] and \[({{x}_{2}},{{f}_{2}}({{x}_{2}}))\] be parallel. Then \[2{{x}_{1}}-1=3x_{2}^{2}-2{{x}_{2}}-2\] or \[2{{x}_{1}}=(3x_{2}^{2}-2{{x}_{2}}-1)\] So, which is possible for infinite numbers of ordered pairs. So, there are infinite solutions.
You need to login to perform this action.
You will be redirected in
3 sec
