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प्रश्न
Find the equation of tangent to the curve y = 2x3 – x2 + 2 at `(1/2, 2)`.
उत्तर
Given, y = 2x3 – x2 + 2
∴ `(dy)/(dx)` = 6x2 – 2x
∴ `((dy)/(dx))_((1/2"," 2)) = 6(1/2)^2 - 2 xx 1/2`
= `3/2 - 1 = 1/2`
∴ Slope of tangent is `1/2`
∴ Equation of tangent at `(1/2, 2)` is `(y - 2) = 1/2(x - 1/2)`
⇒ `2y - 4 = x - 1/2`
⇒ 2(2y – 4) = 2x – 1
⇒ 4y – 8 = 2x –1
⇒ 2x – 1 – 4y + 8 = 0
⇒ 2x – 4y + 7 = 0
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