Let `f: RvecR` be a continuous function which satisfies `f(x)=` `int_0^xf(t)dtdot` Then the value of `f(1n5)` is______

Options

A. `e^(x)-int_(0)^(x)f(t)sintdt`
B. `x^(9)-f(x)`
C. `f(x)+int_(0)^((pi)/2)f(t)sin tdt`
D. `x-int_(0)^((pi)/2-x)f(t)costdt`

Answer

B, D

Explanation

For option 1
Let `g(x)=e^(x)-int_(0)^(x)f(t)sin t dt`
`:.g'(x)=e^(x)-(f(x).sinx)gt0AAxepsilon(0,1)`
`[ :' f(x) epsilon(0,1)` and `e^(x)gt1` for `x epsilon(0,1)]`
So `g(x)` is strictly increasing function.
Also `g(0)=1`
`impliesg(x)gt 1AAxepsilon(0,1)`
Thus, option 1 is not possible
For option 2
Let `h(x)=x^(9)-f(x)`
Now `h(0)=-f(0)lt0( :' f epsilon(0,1))`
Also `h(1)=1-f(1)gt0 ( :' fepsilon(0,1))`
`impliesh(0).h(1)lt0`
So option 2 is correct (using intermediate value theorem)
for option 3
Let `G(x)=f(x)+int_(0)^((pi)/2) f(t). sin t dt`
`impliesG(x)gt0AAxepsilon(0,1) ( :' f epsilon(0,1))`
So option 3 is not possible.
For option 4
Let `H(x)=x-int_(0)^((pi)/2-x)f(t)cost dt`
`:. H(0)=0-int_(0)^((pi)/2) f(t).costdtlt0`
Also `H(1)=1-int_(0)^((pi)/2-1) f(t).costdtgt0`
`impliesH(0).H(1)lt0`
So option 4 is correct (using intermediate value theorem)