Ans.

(i)

Sol. lim at – 2

|– 2 – h + 1|

= | – h – 1|

= 1

2(–2 + h) +3

= – 4 + h + 3

= – 1

function is discontinuous at point x = – 2 lim at 0

(0 + h) 2 + 3 = 3

2(0 – h) + 3 = 3

continuous at x = 0

lim at x = 3 (3 + h) 3 – 15

= 27 – 15 = 12

(3 – h) 2 + 3

= 9 + 3 = 12

continuous at x = 3

(ii)

Sol. Define the greatest integer between the integer.

So that [x] =

Now f(x) =

Now apply the conditions for continuity as previous problem.

(iii)

Sol. g (x) =

For R.H.L. at x = 1

g(x) =

= =

 → 0

For L.H.L. at x = 1

g(x) =

(1 – h) m → 0

g(x) =

For g(1) =

= (1 ∞ )

=

= e 2 g(1) = R.H.L. = L.H.L. at x = 1

h(1) = 6e

2 – 1 and f(1) = 2e 2

Now

2g(1) + 2f(1) – h(1) = 2e 2 + 4e 2 – 6e 2 + 1

= 1