Question

If A is the area in the first quadrant enclosed by the curve \(C: 2x^2 – y + 1 = 0\), the tangent to C at the point (1, 3) and the line x + y = 1, then the value of 60A is ________

Answer 1

Correct Answer: 16

To solve the problem of finding the area \(A\) enclosed by the curve \(C: 2x^2-y+1=0\), the tangent to \(C\) at the point \((1,3)\), and the line \(x+y=1\), follow these steps:
1. **Equation of the Tangent Line:**
- Differentiate \(2x^2-y+1=0\) implicitly to find the slope:\(
\frac{d}{dx}(2x^2) - \frac{d}{dx}(y) = 0 \implies 4x-\frac{dy}{dx}=0 \implies \frac{dy}{dx}=4x\).
- At \((1,3)\), the slope is \(4 \times 1 = 4\). Thus, the tangent line at \((1,3)\) is \(y-3=4(x-1) \Rightarrow y=4x-1\).
2. **Find Intersection Points:**
- Intersection of \(2x^2-y+1=0\) and \(x+y=1:\)
Substituting \(y=1-x\) into \(2x^2-y+1=0:\)
\(2x^2-(1-x)+1=0 \implies 2x^2+x=0 \implies x(x+\frac{1}{2})=0\).
- Solutions are \(x=0\) and \(x=-\frac{1}{2}\) but as we are in the first quadrant, we only consider \(x=0\).
- Intersection of \(x+y=1\) and \(y=4x-1:\)
Equating, \(1-x=4x-1 \Rightarrow 5x=2 \Rightarrow x=\frac{2}{5}\), \(y=1-\frac{2}{5}=\frac{3}{5}\). Thus, the intersection point is \(\left(\frac{2}{5},\frac{3}{5}\right)\).
3. **Calculate Area:**
- Integral from \(x=0\) to \(x=\frac{2}{5}\) gives the area between the curves \(4x-1\) and \(1-x\):
\(\int_0^{\frac{2}{5}}[(1-x)-(4x-1)]\,dx=\int_0^{\frac{2}{5}}-5x+2\,dx\).
- Evaluate: \(-\frac{5}{2}x^2+2x \big|_0^{\frac{2}{5}}=\left[-\frac{5}{2}\left(\frac{2}{5}\right)^2+2\left(\frac{2}{5}\right)\right]\).
- Compute: \(-\frac{5}{2}\times\frac{4}{25}+\frac{4}{5}=-\frac{10}{25}+\frac{20}{25}=\frac{10}{25}=\frac{2}{5}\).

Thus, \(A=\frac{2}{5}\). Therefore, \(60A=60\times\frac{2}{5}=24\).
Since the provided range is 16 and 16, there might have been a misinterpretation or omission in the problem data. Please verify the given conditions. Assuming all steps are correct, \(60A=24\) does not match the supposed range.