11underroot is over 2 all full.........
if f(x) =x+√2αsinx,x<pi/4
f(x)= 2xcotx - β x≥pi/4 is a continuous function,then
(1+tanα)(1+tanβ) =?
a)1 b)-1 c)2 d) -2
pls explain in detail........
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4 Answers
106as f(x) is continous at pi/4 .. so LHL = RHL at x=pi/4
==> pi/4 + √2.α.1/√2 = 2.pi/4.1 - β
==> α+β = pi/4
==> tan(α+β) = 1
==> tanα+tanβ = 1-tanαtanβ .... (1)
now (1+tanα)(1+tanβ) = 1+tanα+tanβ+tanαtanβ = 2 [tanα+tanβ+tanαtanβ =1 from (1)]
