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The slope of the normal to the curve `y=2x^2+3`sin x at `x = 0`is(A) 3 (B) `1/3` (C)`-3` (D) `-1/3` |
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Answer» Slope of normal to a curve is given as, `m = -1/(dy/dx)` Here, equation of the curve is, `y = 2x^2 + 3sinx` `=>dy/dx = 4x + 3cosx` `:. dy/dx|_(x = 0) = 4(0)+3cos 0 = 3` `:.` Slope of the normal ` = -1/3.` |
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