Question

Let G(x, t) =

x(t - 1),whenx \leq t \\ t(x - 1),whent < xandtiscontinuousfunctionofxin\lbrack 0,1\rbrack\end{matrix} \right.$$ If g(x) = $\int_{0}^{1}{f(t)}$G(x, t)dt, then which is incorrect

• A. g(0) + g(1) = 0
• B. g(0) = 0
• C. g(1) = 1
• D. g ¢¢ (x) = f(x)

Answer 1

Answer: (C)

g(1) = 1

Answer 2

G(0, t) = 0 for t £ 0 so g(0) = $\int_{0}^{1}{f(t)}$ . 0 dt = 0

G(1, t) = t . (1 – t) = 0 for t < 1.

Hence g(1) = $\int_{0}^{1}{f(t)}$. 0 dt = 0

Also, g(x) = $\int_{0}^{x}{f(t)}$t(x – 1) dt + $\int_{x}^{1}{f(t)}$x (t – 1) dt

= (x – 1) $\int_{x}^{x}{tf(t)}$ dt + x $\int_{x}^{1}{f(t)}$ (t – 1) dt

Henc, g¢(x) = (x – 1) x f(x) + $\int_{0}^{x}{tf(t)}$dt + $\int_{x}^{1}{f(t)}$ (t – 1) dt – xf (x) (x – 1)

= $\int_{x}^{1}{tf(t)}$dt – $\int_{x}^{1}{f(t)}$dt

Thus g¢¢(x) = f(x).