InterviewSolution
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InterviewSolution
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The value of `a`for which the function `f(x)=(4a-3)(x+log5)+2(a-7)cotx/2sin^2x/2`does not possess critical points is`(-oo,-4/3)`(b) `(-oo,-1)``[1,oo)`(d) `(2,oo)`A. `(-,oo,-4//3)`B. `(-oo,-1)`C. `[1,oo)`D. `(2,oo)` |
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Answer» Correct Answer - 1,4 we have `f(x)=(4a-3)(x+log5)+2(a-7)cot (x)/(2)sin^(2)(x)/(2)` `=(4a-3)(x+log5)+(a-7)sinx)` or `f(X)=(4a-3)+(a-7)cosx` If f(X) does ot have critical ponts then f(x) =0 does not have any solution in R Now `f(x)=0 cos x =(4a-3)/(7-a)` or `|(4a-3)/(7-a)|le1` or `-1le(4a-3)/(7-a)le1` or `a-7le4a-3le7-a` Thus f(x) =0m has solution in R if `-4//3leale2` `ain(-oo,-4//3)cup(2,oo)` |
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