Solution :
The equation of the curve is y = 2x3–3
Differentiating with respect to x, we get
dydx = 6x2
Now, m1 = (Slope of the tangent at x = 2) = (dydx)x=2 = 6×(2)2 = 24
and, m2 = (Slope of the tangent at x = -2) = (dydx)x=−2 = 6×(−2)2 = 24
Clearly m1 = m2.
Thus, the tangents to the given curve at the points where x = 2 and x = -2 are parallel.
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