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If`f(x)={asinpi/2(x+1), ,xlt=0(tanx-sinx)/(x^3),x >0`is continuous at `x=0`, then a equal(a)`1/2`(b) `1/3`(c) `1/4`(d) `1/6` |
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Answer» `f(x) = { a sin pi/2 ( x+1) x<= 0 ` `= { (tanx - sin x)/x^3 x>0` `f(0^-) = (tanx- sin x)/x^3` `= lim_(x->0) (tan x - sinx)/x^3` `= lim_(x->0) (sinx[1/cosx - 1])/x^3` `= lim_(x-0) (1-cos x)/(x^2 cos x)` `= lim_(x->0) (1-(1- 2sin^2 (x/2)))/(x^2 cos x)` `= lim_(x->0) (2sin^2 (x/2))/(4 (x/2)^2 cosx)` `= lim_(x->0) 1/(2cos x) = 1/2` `a=1/2` option a is correct |
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