JEE Class main Answered
the slope of the normal at the point with absicissa x=-2 of graph of f(x)=|x^2-|x|| is
Asked by neerajmeena034356 | 22 Nov, 2023, 11:06: PM
Expert Answer
At x=2 , f(x) = | x2 - |x|| = ( x2 - x )
at x=2 , tangent of f(x) is df/dx = 2x-1= 2(2)-1 = 3
Hence at x=2 , slope of normal to the function f(x) = | x2 - |x|| is -1/3 .
Answered by Thiyagarajan K | 23 Nov, 2023, 11:02: AM
JEE main - Maths
Asked by neerajmeena034356 | 22 Nov, 2023, 11:06: PM
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