JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the curve \(y=a x^{2}+b x+c, x \in R,\) passes through the point \((1,2)\) and the tangent line to this curve at origin is \(y = x\), then the possible values of \(a , b , c\) are :
- A \(a =\frac{1}{2}, b =\frac{1}{2}, c =1\)
- B \(a =1, b =0, c =1\)
- C \(a =1, b =1, c =0\)
- D \(a =-1, b =1, c =1\)
Answer & Solution
Correct Answer
(C) \(a =1, b =1, c =0\)
Step-by-step Solution
Detailed explanation
\(a+b+c=2\) and \(\left.\frac{ dy }{ dx }\right|_{(0,0)}=1\) \(2 a x+\left.b\right|_{(0,0)}=1\) \(b =1\) Curve passes through origin So, \(c=0\) and \(a=1\)
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