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The point on the curve $y=x^2+7$ nearest to point (3, 7) is: |
(1, 8) (3, 16) (-1, 8) (4, 23) |
(1, 8) |
The correct answer is Option (1) → (1, 8) $y = x^2 + 7$ $D^2 = (x-3)^2 + (x^2+7-7)^2 = (x-3)^2 + x^4$ $\frac{d}{dx}(D^2) = 2(x-3) + 4x^3$ $= 2x - 6 + 4x^3 = 0$ $2x^3 + x - 3 = 0$ $x=1 \Rightarrow 2(1)^3+1-3=0$ $y = 1^2 + 7 = 8$ $\text{Nearest point} = (1,8)$ |
