JEE Mains · Maths · STD 12 - 8. Application and integration
If \(A\) is the area in the first quadrant enclosed by the curve \(C: 2 x^2-y+1=0\), the tangent to \(C\) at the point \((1,3)\) and the line \(x+y=1\), then the value of \(60 A\) is
- A \(16\)
- B \(14\)
- C \(12\)
- D \(10\)
Answer & Solution
Correct Answer
(A) \(16\)
Step-by-step Solution
Detailed explanation
\(y=2 x^2+1\) Tangent at \((1,3)\) \(y =4 x -1\) \(A =\int \limits_0^1\left(2 x ^2+1\right) dx \text {-area of }(\Delta QOT )-\text { area of }\) \((\Delta PQR )+\text { area of }(\Delta QRS )\) \(A =\left(\frac{2}{3}+1\right)-\frac{1}{2}-\frac{9}{8}+\frac{9}{40}=\frac{16}{60}\)
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