JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let quadratic curve passing through the point \((-1,0)\) and touching the line \(y=x\) at \((1,1)\) be \(y=\) \(f(x)\). Then the \(x\)-intercept of the normal to the curve at the point \((\alpha, \alpha+1)\) in the first quadrant is \(..........\).
- A \(10\)
- B \(12\)
- C \(11\)
- D \(13\)
Answer & Solution
Correct Answer
(C) \(11\)
Step-by-step Solution
Detailed explanation
\(f(x)=(x+1)(a x+b)\) \(1=2 a+2 b \quad(1)\) \(f(x)=(a x+b)+a(x+1)\) \(1=(3 a+b) \quad(2)\) \(\Rightarrow b=1 / 4, a=1 / 4\) \(f(x)=\frac{(x+1)^2}{4}\) \(f^{\prime}(x)=\frac{x}{2}+\frac{1}{2} \quad \alpha+1=\frac{(\alpha+1)^2}{4}, \alpha > -1\) \(\alpha+1=4\) \(\alpha=3\) normal…
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