JEE Mains · Maths · STD 12 - 6. Application of derivatives
If the lines \(x+y=a\) and \(x-y=b\) touch the curve \(y = x ^{2}-3 x +2\) at the points where the curve intersects the \(x\)-axis, then \(\frac{ a }{ b }\) is equal to
- A \(1.50\)
- B \(2.00\)
- C \(1.00\)
- D \(0.5\)
Answer & Solution
Correct Answer
(D) \(0.5\)
Step-by-step Solution
Detailed explanation
\(y=x^{0}-3 x+2\) At \(x\) -axis \(y=0=x^{2}-3 x+2\) \(x=1,2\) \(\frac{d y}{d x}=2 x-3\) \(A(1,0) B(2,0)\) \(\left(\frac{ dy }{ dx }\right)_{ x =1}=-1\) and \(\left(\frac{ dy }{ dx }\right)_{ x =2}=1\) \(x+y=a \Rightarrow \frac{d y}{d x}=-1\) So \(A(1,0)\) lies on it…
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