KVPY SX SB Exam 2013 Question Paper

Question : 12 of 120

Marks: +1, -0

For real x with

-10 ≤x≤ 1 0

define

f ( x ) = ∫↖{x}↙{-10}2^[t] dt,

where for a real number r we denote by [r] the largest integer less than or equal to r. The number of points of discontinuity of

f

in the interval (-10,10) is

Solution:

Let r be an integer in (-10, 10)
Now, LHL

= \lim↙{x→r^{-}}∫↖{x}↙{-10}2^{[\t]}dt

=\lim↙{h→0^{+}}[∫↖{-9}↙{-10}2^{[\t]}dt

+∫↖{-8}↙{-9}2^{[\t]}dt

+...+∫↖{r-h}↙{r-1}2^{[\t]}dt]

=\lim↙{h→0}[2^{-10}+2^{-9}

+...+2^{r-1}(1-h)]

= 2 ^{-10} + 2 ^{-9} + ... + 2 ^{ r-1}

... (1)

\lim↙{x→r^{+}}∫↖{x}↙{-10}2^{[t]}dt

= 2 ^{-10} + 2 ^{-9}

+ ..... + 2 ^{r-1}

... (2)

\f(r)=∫↖{r}↙{-10}2^{[t]}dt

= 2^{-10} + 2^{-9} + ... + 2^{r-1}

... (3)
From (1), (2) & (3)
f(x) is continous at all integers.

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