A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{2}\]
D) \[\pi \]
Correct Answer: C
Solution :
Given,\[{{x}^{2}}=y\]and \[6y=7-{{x}^{3}}\] Intersection point is (1, 1). Slope of\[{{x}^{2}}=y\] is\[\frac{dy}{dx}=2x\] At \[(1,1){{\left( \frac{dy}{dx} \right)}_{(1,1)}}=2={{m}_{1}}\] Slope of\[6y=7-{{x}^{3}}\]is \[6\frac{dy}{dx}=-3{{x}^{2}}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{(1,1)}}=-\frac{1}{2}={{m}_{2}}\] \[\therefore \] \[\tan \theta =\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}\] \[=\frac{2+1/2}{1+2\times \frac{-1}{2}}=\frac{5/2}{0}\] \[\Rightarrow \] \[\tan \theta =\infty =\tan \frac{\pi }{2}\] \[\Rightarrow \] \[\theta =\pi /2\]
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