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The value of `f(0),`so that the function `f(x)=((27-2x)^(1/3)-3)/(9-3(243+5x)^(1//5))(x!=0)`is continuous, is given by(a)`2/3`(b) 6(c) 2 (d) 4 |
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Answer» `f(x) = ((27- 2x)^(1/3) - 3)/(9- 3(243 + 5x)^(1/5))` `= (3(1- 2x/27)^(1/3) - 3)/(9- 3*3*(1+ (5x)/(243))^(1/5))` `= (3(1-(2x)/27 xx 1/3) -3)/(9- 9(1 + (5x)/243) xx 1/5)` `= (3 - (2x)/27 - 3)/(9 - 9 -9 xx (5x)/243 xx 1/5)` `= ((-2x)/27)/((-x)/27)` `= 2` option c is correct |
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