A) \[4\]
B) \[3\]
C) \[2\]
D) \[1\]
Correct Answer: D
Solution :
[d] Equation of normal at \[P\left( \frac{3}{4}{{y}_{1}}^{3},{{y}_{1}} \right)\] is \[y-{{y}_{1}}=\frac{-9{{y}_{1}}^{2}}{4}\left( x-\frac{3}{4}{{y}_{1}}^{3} \right)\] If it passes from \[(0,1),\]then \[27{{y}_{1}}^{5}+16{{y}_{1}}-16=0\] ...(1) Now, consider the function \[f({{y}_{1}})=27{{y}_{1}}^{5}+16{{y}_{1}}-16.\] Then \[f'({{y}_{1}})=135{{y}_{1}}^{4}+16>0\] So, \[f({{y}_{1}})\] is an increasing function. Hence, equation (1) has only one root. Therefore, there is only one normal through the point \[(0,1)\].
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