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प्रश्न
Find the equation of normal to the curve y = 2x3 – x2 + 2 at `(1/2, 2)`
उत्तर
y = 2x3 – x2 + 2
∴ `("d"y)/("d"x)` = 6x2 – 2x
∴ `(("d"y)/("d"x))_((1/2"," 2)) = 6(1/2)^2 - 2(1/2)`
= `3/2 - 1`
= `2 1/2`
Slope of the normal at `(1/2, 2)` is `(-1)/(("d"y)/("d"x))_((1/2","2))`
= `(-1)/(1/2)`
= – 2
∴ Equation of the normal at `(1/2, 2)` is
y – 2 = `-2(x - 1/2)`
∴ y – 2 = – 2x + 1
∴ 2x + y – 3 = 0
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