Question

For a twice continuously differentiable function $g : R \to R$ , define
$u_{g} (x, y) = \frac{1}{y} \int_{- y}^{y} g (x + t) d t for (x, y) \in R^{2}, y > 0.$
Which one of the following holds for all such g?

$\frac{\partial^{2} u_{g}}{\partial x^{2}} = \frac{2}{y} \frac{\partial u_{g}}{\partial y} + \frac{\partial^{2} u_{g}}{\partial y^{2}}$
$\frac{\partial^{2} u_{g}}{\partial x^{2}} = \frac{1}{y} \frac{\partial u_{g}}{\partial y} + \frac{\partial^{2} u_{g}}{\partial y^{2}}$
$\frac{\partial^{2} u_{g}}{\partial x^{2}} = \frac{2}{y} \frac{\partial u_{g}}{\partial y} - \frac{\partial^{2} u_{g}}{\partial y^{2}}$
$\frac{\partial^{2} u_{g}}{\partial x^{2}} = \frac{1}{y} \frac{\partial u_{g}}{\partial y} - \frac{\partial^{2} u_{g}}{\partial y^{2}}$

Answer: $\frac{\partial^{2} u_{g}}{\partial x^{2}} = \frac{2}{y} \frac{\partial u_{g}}{\partial y} + \frac{\partial^{2} u_{g}}{\partial y^{2}}$

Solution :-

We have, u_{g} = \frac{1}{y} \int_{- y}^{y} g (x + t) d t

First partial derivative:

First partial derivative with respect to $x$

\begin{array}{rcl} \frac{\partial u_{g}}{\partial x} & = & \frac{\partial}{\partial x} (\frac{1}{y} \int_{- y}^{y} g (x + t) d t) \\ = & \frac{1}{y} \cdot \frac{\partial}{\partial x} (\int_{- y}^{y} g (x + t) d t) \\ = & \frac{1}{y} \cdot [\int_{- y}^{y} \frac{\partial (g (x + t))}{\partial x} \cdot d t + g (x + y) \frac{\partial (y)}{\partial x} - g (x - y) \frac{\partial (- y)}{\partial x}] \\ = & \frac{1}{y} \cdot [\int_{- y}^{y} g^{'} (x + t) \cdot d t + g (x + y) \cdot 0 - g (x - y) \cdot 0] \\ = & \frac{1}{y} \cdot [\int_{- y}^{y} g^{'} (x + t) \cdot d t + 0 - 0] \\ = & \frac{1}{y} \cdot [\int_{- y}^{y} g^{'} (x + t) \cdot d t] . \end{array}

First partial derivative with respect to $y$

\begin{array}{rcl} \frac{\partial u_{g}}{\partial y} & = & \frac{\partial}{\partial y} (\frac{1}{y} \cdot \int_{- y}^{y} g (x + t) d t) \\ = & \frac{\partial (\frac{1}{y})}{\partial y} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} \cdot \frac{\partial}{\partial y} (\int_{- y}^{y} g (x + t) d t) \\ = & - \frac{1}{y^{2}} (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} \cdot (\int_{- y}^{y} \frac{\partial (g (x + t))}{\partial y} \cdot d t + g (x + y) \frac{\partial (y)}{\partial y} - g (x - y) \frac{\partial (- y)}{\partial y}) \\ = & - \frac{1}{y^{2}} (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} \cdot (\int_{- y}^{y} 0 \cdot d t + g (x + y) + g (x - y)) \\ = & - \frac{1}{y^{2}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} \cdot [0 + g (x + y) + g (x - y)] \\ = & - \frac{1}{y^{2}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} \cdot [g (x + y) + g (x - y)] . \end{array}

Second partial derivative:

Second partial derivative with respect to $x$

\begin{array}{rcl} \frac{\partial^{2} u_{g}}{\partial x^{2}} & = & \frac{\partial}{\partial x} (\frac{1}{y} \int_{- y}^{y} g^{'} (x + t) d t) \\ = & \frac{1}{y} \cdot \frac{\partial}{\partial x} (\int_{- y}^{y} g^{'} (x + t) d t) \\ = & \frac{1}{y} \cdot [\int_{- y}^{y} \frac{\partial (g^{'} (x + t))}{\partial x} \cdot d t + g^{'} (x + y) \frac{\partial (y)}{\partial x} - g^{'} (x - y) \frac{\partial (- y)}{\partial x}] \\ = & \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t + g^{'} (x + y) \cdot 0 - g^{'} (x - y) \cdot 0] \\ = & \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t + 0 - 0] \\ = & \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] . \end{array}

Second partial derivative with respect to $y$

\begin{array}{rcl} \frac{\partial^{2} u_{g}}{\partial y^{2}} & = & \frac{\partial}{\partial y} (- \frac{1}{y^{2}} (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} [g (x + y) + g (x - y)]) \\ = & \frac{\partial}{\partial y} (- \frac{1}{y^{2}} (\int_{- y}^{y} g (x + t) d t)) + \frac{\partial}{\partial y} (\frac{1}{y} [g (x + y) + g (x - y)]) \\ = & \frac{\partial (- \frac{1}{y^{2}})}{\partial y} \cdot (\int_{- y}^{y} g (x + t) d t) + (- \frac{1}{y^{2}}) \cdot \frac{\partial}{\partial y} (\int_{- y}^{y} g (x + t) d t) + \\ \frac{\partial (\frac{1}{y})}{\partial y} \cdot (g (x + y) + g (x - y)) + \frac{1}{y} \cdot \frac{\partial}{\partial y} (g (x + y) + g (x - y)) \\ = & \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + (- \frac{1}{y^{2}}) \cdot (\int_{- y}^{y} \frac{\partial (g (x + t))}{\partial y} \cdot d t + g (x + y) \frac{\partial (y)}{\partial y} - g (x - y) \frac{\partial (- y)}{\partial y}) + \\ (- \frac{1}{y^{2}}) \cdot (g (x + y) + g (x - y)) + \frac{1}{y} \cdot (\frac{\partial (g (x + y))}{\partial y} + \frac{\partial (g (x - y))}{\partial y}) \\ = & \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) - \frac{1}{y^{2}} \cdot (g (x + y) + g (x - y)) - \frac{1}{y^{2}} \cdot (g (x + y) + g (x - y)) + \\ \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ = & \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) - \frac{2}{y^{2}} \cdot (g (x + y) + g (x - y)) + \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) . \end{array}

So, up till now, we get

\begin{array}{llllllllll} \frac{\partial u_{g}}{\partial x} & = & \frac{1}{y} \cdot [\int_{- y}^{y} g^{'} (x + t) \cdot d t] \\ \frac{\partial u_{g}}{\partial y} & = & - \frac{1}{y^{2}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} \cdot [g (x + y) + g (x - y)] \\ \frac{\partial^{2} u_{g}}{\partial x^{2}} & = & \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] \\ \frac{\partial^{2} u_{g}}{\partial y^{2}} & = & \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) - \frac{2}{y^{2}} \cdot (g (x + y) + g (x - y)) + \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \end{array}

Now, from options

Option 1. $\frac{\partial^{2} u_{g}}{\partial x^{2}} = \frac{2}{y} \frac{\partial u_{g}}{\partial y} + \frac{\partial^{2} u_{g}}{\partial y^{2}}$

\begin{array}{llllllllll} \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & \frac{2}{y} \cdot [- \frac{1}{y^{2}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} \cdot [g (x + y) + g (x - y)]] + \\ \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) - \frac{2}{y^{2}} \cdot (g (x + y) + g (x - y)) + \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & - \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{2}{y^{2}} \cdot [g (x + y) + g (x - y)] + \\ \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) - \frac{2}{y^{2}} \cdot (g (x + y) + g (x - y)) + \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \int_{- y}^{y} g^{′′} (x + t) \cdot d t & = & g^{'} (x + y) - g^{'} (x - y) . \end{array}

Here, L.H.S. is equal to R.H.S. So, yes this option is right.

Now, Option 2. $\frac{\partial^{2} u_{g}}{\partial x^{2}} = \frac{1}{y} \frac{\partial u_{g}}{\partial y} + \frac{\partial^{2} u_{g}}{\partial y^{2}}$

\begin{array}{llllllllll} \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & \frac{1}{y} \cdot [- \frac{1}{y^{2}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} \cdot [g (x + y) + g (x - y)]] + \\ \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) - \frac{2}{y^{2}} \cdot (g (x + y) + g (x - y)) + \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & - \frac{1}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y^{2}} \cdot [g (x + y) + g (x - y)] + \\ \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) - \frac{2}{y^{2}} \cdot (g (x + y) + g (x - y)) + \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & \frac{1}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) - \frac{1}{y^{2}} \cdot [g (x + y) + g (x - y)] + \\ \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y^{3}} \cdot [\int_{- y}^{y} g (x + t) \cdot d t] & = & \frac{1}{y^{2}} \cdot (g (x + y) + g (x - y)) (\begin{array}{llllllllll} from option 1, the left side term and \\ the third term of the right side are equal \end{array}) \\ \int_{- y}^{y} g (x + t) \cdot d t & = & y \cdot [g (x + y) + g (x - y)] . \end{array}

Here, L.H.S. is not equal to R.H.S. because we don’t know what is integretion of the L.H.S. so we don’t say L.H.S. is equal to R.H.S.
So, this option is wrong.

Now, Option 3. $\frac{\partial^{2} u_{g}}{\partial x^{2}} = \frac{2}{y} \frac{\partial u_{g}}{\partial y} - \frac{\partial^{2} u_{g}}{\partial y^{2}}$

\begin{array}{llllllllll} \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & \frac{2}{y} \cdot [- \frac{1}{y^{2}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} \cdot [g (x + y) + g (x - y)]] - \\ \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{2}{y^{2}} \cdot (g (x + y) + g (x - y)) - \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & - \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{2}{y^{2}} \cdot [g (x + y) + g (x - y)] - \\ \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{2}{y^{2}} \cdot (g (x + y) + g (x - y)) - \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & - \frac{4}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{4}{y^{2}} \cdot [g (x + y) + g (x - y)] - \\ \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & - \frac{4}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{4}{y^{3}} \cdot y [g (x + y) + g (x - y)] - \\ \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{2}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & - \frac{4}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{4}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) (from option 1 and 2) \\ \frac{2}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & 0 \\ \int_{- y}^{y} g^{′′} (x + t) \cdot d t & = & 0. \end{array}

Which is wrong from option 1. So, this option is also wrong.

Now, Option 4. $\frac{\partial^{2} u_{g}}{\partial x^{2}} = \frac{1}{y} \frac{\partial u_{g}}{\partial y} - \frac{\partial^{2} u_{g}}{\partial y^{2}}$

\begin{array}{llllllllll} \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & \frac{1}{y} \cdot [- \frac{1}{y^{2}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y} \cdot [g (x + y) + g (x - y)]] - \\ \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{2}{y^{2}} \cdot (g (x + y) + g (x - y)) - \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & - \frac{1}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{1}{y^{2}} \cdot [g (x + y) + g (x - y)] - \\ \frac{2}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{2}{y^{2}} \cdot (g (x + y) + g (x - y)) - \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & - \frac{3}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{3}{y^{2}} \cdot [g (x + y) + g (x - y)] - \\ \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{1}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & - \frac{3}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{3}{y^{3}} \cdot y [g (x + y) + g (x - y)] - \\ \frac{1}{y} \cdot (g^{'} (x + y) - g^{'} (x - y)) \\ \frac{2}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & - \frac{3}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) + \frac{3}{y^{3}} \cdot (\int_{- y}^{y} g (x + t) d t) (from option 1 and 2) \\ \frac{2}{y} \cdot [\int_{- y}^{y} g^{′′} (x + t) \cdot d t] & = & 0 \\ \int_{- y}^{y} g^{′′} (x + t) \cdot d t & = & 0. \end{array}

Which is wrong from option 1. So, this option is also wrong.

So, correct answer is option-1, $\to \frac{\partial^{2} u_{g}}{\partial x^{2}} = \frac{2}{y} \frac{\partial u_{g}}{\partial y} + \frac{\partial^{2} u_{g}}{\partial y^{2}}$

#Integral Calculus #IIT JAM PYQ 2024 #Mathematics #IIT JAM PYQs

Question

For a twice continuously differentiable function g⁡:R→R , define ug⁡(x,y)=1y∫−yyg⁡(x+t)⁢d⁢tfor(x,y)∈R2,y>0. Which one of the following holds for all such g?

Solution :-

First partial derivative:

Second partial derivative:

For a twice continuously differentiable function $g : R \to R$ , define
$u_{g} (x, y) = \frac{1}{y} \int_{- y}^{y} g (x + t) d t for (x, y) \in R^{2}, y > 0.$
Which one of the following holds for all such g?