The area (in sq units) in the first quadrant bounded by the parabola, , the tangent to it at the point and the coordinate axes is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
The tangent at (2,5) to y = x²+1 has slope 2x|_(x=2) = 4, giving line y = 4x - 3.
The area bounded by the parabola, the tangent, and the coordinate axes in the first quadrant is computed as:
Area under parabola from x=0 to x=2, minus area under the tangent line from x=3/4 to x=2, plus the triangle formed by the tangent from x=0 to x=3/4...
Let me be systematic. The tangent y=4x-3 hits x-axis at x=3/4 and y-axis at y=-3 (below x-axis). In the first quadrant, the tangent exists only for x ≥ 3/4.
The region in the first quadrant is bounded by: the parabola (above), the tangent line, and the axes. The enclosed region is:
Area = (Area under parabola from 0 to 2) - (Area of triangle under tangent from 3/4 to 2, above x-axis)
Area under parabola from 0 to 2: ∫₀² (x²+1)dx = [x³/3 + x]₀² = 8/3 + 2 = 14/3
Area under tangent from 3/4 to 2 (above x-axis): ∫_(3/4)2 (4x-3)dx = [2x²-3x]_(3/4)2 = (8-6) - (9/8 - 9/4) = 2 - (-9/
The area bounded by the parabola, the tangent, and the coordinate axes in the first quadrant is computed as:
Area under parabola from x=0 to x=2, minus area under the tangent line from x=3/4 to x=2, plus the triangle formed by the tangent from x=0 to x=3/4...
Let me be systematic. The tangent y=4x-3 hits x-axis at x=3/4 and y-axis at y=-3 (below x-axis). In the first quadrant, the tangent exists only for x ≥ 3/4.
The region in the first quadrant is bounded by: the parabola (above), the tangent line, and the axes. The enclosed region is:
Area = (Area under parabola from 0 to 2) - (Area of triangle under tangent from 3/4 to 2, above x-axis)
Area under parabola from 0 to 2: ∫₀² (x²+1)dx = [x³/3 + x]₀² = 8/3 + 2 = 14/3
Area under tangent from 3/4 to 2 (above x-axis): ∫_(3/4)2 (4x-3)dx = [2x²-3x]_(3/4)2 = (8-6) - (9/8 - 9/4) = 2 - (-9/
[JEE Main 2019 · 11 Jan · Shift II]