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प्रश्न
Solve the following : Determine the area of the triangle formed by the tangent to the graph of the function y = 3 – x2 drawn at the point (1, 2) and the coordinate axes.
उत्तर
y = 3 – x2
∴ `dy/dx = d/dx(3 - x^2) = 0 - 2x = - 2x`
∴ `(dy/dx)_(at (1 , 2)) = -2(1) = -2`
= slope of the tangent at (1 2)
∴ equation of the tangent at (1, 2) is
y – 2 = 2(x – 1)
∴ y – 2 = – 2x + 2
∴ 2x + y = 4
Let this tangent cuts the coordinate axes at A (a, 0) and B (0, b).
∴ 2a + 0 = 4 and 2(0) + b = 4
∴ a = 2 and b = 4
area of required triangle
= `(1)/(2) xx l("OA") xx l("OB")`
= `(1)/(2)ab`
= `(1)/(2)(2)(4)`
= 4 sq units.
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