InterviewSolution
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InterviewSolution
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Let `f(x)=4cossqrt(x^2-pi^2/9` Then the range of `f(x)` isA. `[-1, 1] `B. `[-4, 4]`C. `[0, 1 ]`D. none of these |
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Answer» Correct Answer - B Clearly, f(x) is defined for `x^(2) - (pi^(2))/(9) ge 0` `rArr x le -(pi)/(3) or x ge (pi)/(3)` `rArr x in (-oo, -pi//3]uu[pi//3, oo)` Let `y = 4 cos sqrtx^(2) - (pi)/(2))` `rArr cos^(-1) (y)/(4) = sqrt(x^(2) - (pi^(2))/(9))` `rArr x^(2) - (pi)/(9) =(cos^(-1) .(y)/(4))^(2) rArr x = pm sqrt((pi^(2))/(9) + (cos^(-1).(y)/(4))^(2))` clearly, x is real all y for which `cos^(-1).(y)/(4)`is meaningful. Now, `cos^(-1)((y)/(4))` is defined, if `-1 le (y)/(4) lt 1 rArr - 4 lt y lt 4rArr y in [-4,4]` Hence, range of `f =[-4,4]` |
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