Let $T > 0$ be a fixed number. Suppose $f$ is a continuous function such that for all $x \in R,\,f(x + T) = f(x)$. If $I = \int_{\,0}^{\,T} {f(x)\,dx} $ then the value of $\int_{\,3}^{\,3 + 3T} {f(2x)\,dx,} $ is
- A
$\frac{3}{2}I$
- B
$2I$
- ✓
$3I$
- D
$6I$
✓
Answer
c
(c) Put $2x = z.$ Then, $\int_{\,3}^{\,3 + 3T} {f(2x)dx} = \frac{1}{2}\int_{\,6}^{\,6 + 6T} {f(z)\,dz} $
$ = \frac{1}{2}\left[ {\int_{\,6}^{\,0} {f(x)\,dx} + \int_{\,0}^{\,T} {f(x)\,dx} + \int_{\,T}^{\,6 + 6T} {f(x)\,dx} } \right]$
$ = \frac{1}{2}\left[ { - \int_{\,0}^{\,6} {f(x)\,dx} + I + \int_{\,T}^{\,6 + 5T} {f(T + z)\,dz} } \right]$
$ = \frac{1}{2}\left[ { - \int_{\,0}^{\,6} {f(x)\,dx} + I + \int_{\,0}^{\,T} {f(z)\,dz} + \int_{\,T}^{\,6 + 5T} {f(T + z)\,dz} } \right]$
$ = \frac{1}{2}\left[ { - \int_{\,0}^{\,6} {f(x)\,dx} + I + I + \int_{\,T}^{\,6 + 5T} {f(x)\,dz} } \right]$
$ = \frac{1}{2}\left[ { - \int_{\,0}^{\,6} {f(x)\,dx} + I + 5I + \int_{\,T}^{\,6 + T} {f(x)\,dz} } \right]$
$ = \frac{1}{2}\left[ { - \int_{\,0}^{\,6} {f(x)\,dx} + 6I + \int_{\,0}^{\,6} {f(z + T)\,dz} } \right]$
$ = \frac{1}{2}\left[ { - \int_{\,0}^{\,6} {f(x)\,dx} + 6I + \int_{\,0}^{\,6} {f(z)\,dz} } \right] = \frac{1}{2} \times 6I = 3I.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.