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Let `f: Rvec(0,1)`be a continuous function. Then, which of the following function (s) has(have) the value zero at some point in the interval (0,1)?`e^x-int_0^xf(t)sintdt`(b) `f(x)+int_0^(pi/2)f(t)sintdt``x-int_0^(pi/2-x)f(t)costdt`(d)`x^9-f(x)`A. `e^(x)-int _(0)^(x)f(t) sin t dt `B. `f(x)+int_(0)^((pi)/(2))f(t)sint dt `C. `x-int_(0)^((pi)/(2)-x)f(t) cos t dt `D. `x^(9) - f(x)` |
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Answer» Correct Answer - C::D (a) `because e^(x) in (1,e) " in " (0,1) and int_(0)^(x)f(t) sin t dt in (0, 1) " in " (0, 1)` `therefore e^(x) - int_(0)^(x) f(t) sin t dt ` cannot be zero. So, option (a) is incorrect. (b) `f(x) + int_(0)^((pi)/(2)) f(t) sin t dt ` always positive `therefore` Option (b) is incorrect. (c ) Let `h(x)=x-int_(0)^((pi)/(2)-x) f(t) cos t dt`, `h(0)=-int _(0)^((pi)/(2))f(t) cost dt lt 0 rArr h(1) =1 int_(0)^((pi)/(2)-1)f(t)cost dt gt 0` `therefore` Option (c) is correct. (d) Let `g(x) =x^(9)-f(x) rArr g(0) = -f(0) lt 0` g(1) =1-f(1) gt 0` ` therefore ` Option (d) is correct. |
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