A) \[\left( \frac{1}{2},\frac{5}{24} \right)\] and \[\left( -1,-\frac{1}{6} \right)\]
B) \[\left( \frac{1}{2},\frac{4}{9} \right)\] and \[(-1,0)\]
C) \[\left( \frac{1}{3},\frac{1}{7} \right)\] and \[\left( -3,\frac{1}{2} \right)\]
D) \[\left( \frac{1}{3},\frac{4}{47} \right)\] and \[\left( -1,-\frac{1}{3} \right)\]
Correct Answer: A
Solution :
[a] \[y=\frac{2}{3}{{x}^{3}}+\frac{1}{2}{{x}^{2}}\therefore \frac{dy}{dx}=\frac{2}{3}3{{x}^{2}}+\frac{1}{2}2x=2{{x}^{2}}+x\] Since the tangent makes equal angles with the axes. \[\therefore \frac{dy}{dx}=\pm 1or2{{x}^{2}}+x=\pm 1\] or \[2{{x}^{2}}+x-1=0(2{{x}^{2}}+x+1=0\] has no real roots) or \[(2x-1)(x+1)=0\] i.e., \[x=\frac{1}{2}\] or \[x=-1\].
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