Let `f:[a,b]to[1,oo)` be a continuous function and let `g:RtoR` be defined as
`g(x)={(0,"if",xlta),(int_(a)^(x)f(t)dt,"if",alexleb),(int_(a)^(b)f(t)dt,"if",xgtb):}` Then

Options

A. g(x) is continuous but not differentiable at x=a
B. g(x) is differentiable on R
C. g(x) is continuous but not differentiable at x=b
D. g(x) is continuous and differentiable at either x=a or x=b but not both

Answer

A, C

Explanation

Continuity of g(x): `("LHL at x=a")=underset(x to a^(-))lim g(x)=underset(x to a^(-))lim 0=0`
`("RHL at x=a")=underset(x to a^(+))lim g(x)=underset(x to a^(+))lim underset(a)overset(x)int f(t)dt=0`
and , ` g(a) = overset(a) underset a int f (t) dt = 0 `
` therefore lim_(x to a ^( - ) ) g (x) = lim_( x to a ^( + )) g (x) = g (a) `
So, g(x) is continuous at x =a.
(LHL at x = b ) = `lim_( x to b ^( - )) g (x) =lim_( x - b ^( - )) overset x underset a int f ( t) dt = oversetb underset b f (t) dt `
( RHL at x = b ) = `lim_(x to b^(+ )) g(x) =lim_( x to b^(+ )) overset b underseta int f(t) dt = overset b underset a int f (t) dt `
and, ` g(b) = oversetb underset a int f (t)dt `
and, ` g (b) = oversetb underseta f(t) dt `
` therefore lim_(x to b^(- )) g (x) = lim_( x to b ^( + )) g (x) = g (b) `
So, g (x) is continuous at x = b.
Thus, g (x) is continuous at x = a and x = b .
Differentiability of g (x) :
(LHD of g (x) at x = a) = `lim_( x to a ^( - ) ) (g (x) - g ( a))/( x - a )`
` " " = lim_( x to a^( - ))(0 - overseta underseta int f (t) dt )/(x - a ) = 0 `
( RHD of g(x) at x = a) = `lim_( x to a ^( + )) (g (x) - g(a))/(x - a )`
` = lim_( x to a ^(+ ))(overset x underset a int f ( t) dt - overset a underset a int f ( t) dt ) /( x- a )`
` = lim _( x to a ^( + )) ( overset x underset a f ( t) dt - 0 ) /( x - a )`
` =lim_(x to a ^( + ))( overset x underset a f ( t) dt) /( x - a )`
` = lim_( x to a ^( + )) (f (x))/(1 )" "`[By L' Hospital's rule ]
` = lim_( x to a^(+)) f (x) ge 1" "[because " Codomain " f (x) = [1, oo)]`
` therefore ` ( LHD of g (x) at x = a) ` ne ` ( RHD of g (x) at x = a )
So, g (x) is not differentiable at x = a.
(LHD of g(x) at x = b )
` = lim_( x to b ^( - )) (g (x) - g(b))/( x - b )`
` = lim_( x to b ^( - )) ( oversetx underset aint f (t) dt - overset b underset a int f ( t) dt)/( x - b )`
` = lim_( x to b^(-)) (f (x) - 0 )/( 1 - 0 ) = lim_( x to b ^(-)) f (x) ge 1 " " [because " Range " f (x) sube [1, oo)]`
( RHD of g (x) at x = b ) = ` lim_(x to b ^( + ) ) (g (x) - g (b))/(x - b ) `
` = lim_( x to b ^( + ) ) ( overset b underset a int f (t) dt - overset b ) underset a int f (t)dt)/(x - b )`
` = lim_( x to b ^(+ )) (0)/(x - b ) = 0 `
` therefore ` (LHD of g (x) at x = b ) ` ne ` ( RHD of g (x) at x = b )
So, g (x) is not differentiable at ` x = b `.