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If f(x)={:{(3x^2+12"," x-1),(37-x ","2 lt x le 3):} then |
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Answer» f(X) INCREASING on [-1,2] So f(x) is CONTINUOUS on [-1,3] THUS options (b) and (c ) are correct . ALSO `(LHD at x =2) ={d/(dx)(37 -x)}_(atx=2)`=24 Clearly f(x) is not differentiableat x=2 So f(2) does not exist. Now `f(x)={(6x+12","-1 lt xlt 2 ),(-1","2 ltx lt 3):}` Cleary `f' (x)lt 0 for all x in [-1, 2] ` So f(x) in increasing on [-1,2] Thus option (a) is correct . |
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