Q.
Find the range of the function f:[0,1]→R,f(x)=x3−x2+4x+2sin−1x ?
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J & K CETJ & K CET 2014Inverse Trigonometric Functions
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Solution:
We have f:[0,1]→Rf(x)=x3−x2+4x+2sin−1x Now f′(x)=3x2−2x+4+1−x22 For x[0,1],f′(x)>0 Hence, it is a increasing function at x=0,f(0)=0−0+4(0)+2sin−1(0)=0 at x=1,f(1)=1−1+4(1)+2sin−1(1) =4+2(2π)=4+π ∴ Range of f(x)[0,4+π]