Question

Consider two twice differentiable functions $f(x)$ and $\phi(x)$ satisfying $f''(x) + f'(x)\phi'(x) = \phi(x)\phi'(x), x > 0, f'(1) = f(1) = 0, \phi(1) = 0$. Match the entires of List-I with the correct entries of List-II. (where [.] denotes the greatest integer function)

• A. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• B. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• C. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• D. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• E. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• F. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• G. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• H. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• I. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• J. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• K. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• L. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• M. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• N. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• O. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• P. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• Q. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• R. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• S. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• T. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• U. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• V. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• W. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• X. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• Y. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• Z. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• [. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• \. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• ]. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• ^. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• _. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• `. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• a. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• b. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• c. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• d. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• e. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• f. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• g. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• h. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• i. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• j. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• k. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• l. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• m. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• n. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• o. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• p. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• q. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• r. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• s. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• t. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• u. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• v. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• w. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• x. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• y. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• z. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• {. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• |. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• }. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• ~. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• . If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• €. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• . If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• ‚. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• ƒ. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• „. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• †. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• ‡. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• ˆ. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• ‰. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• Š. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• ‹. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• Œ. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• . If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• Ž. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• . If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• . If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• ‘. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• ’. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• “. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• ”. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• •. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• –. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• —. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• ˜. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• ™. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• š. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• ›. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• œ. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• . If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• ž. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• Ÿ. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• ¡. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• ¢. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• £. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• ¤. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to
• ¥. If $\phi(x) = \ln x$, then $[f(e)]$ is equal to
• ¦. If $\phi(x) = \ln x$, then $[f(\frac{1}{e})]$ is equal to
• §. If $\phi(x) = x - 1$, then $[f(2)]$ is equal to
• ¨. If $\phi(x) = x - 1$, then $[f(\frac{1}{2})]$ is equal to

Answer 1

Answer: (¥), (¦), (§), (¨)

(P) $\rightarrow$ (3), (Q) $\rightarrow$ (2), (R) $\rightarrow$ (3), (S) $\rightarrow$ (2)

Answer 2

The given differential equation is $f''(x) + f'(x)\phi'(x) = \phi(x)\phi'(x)$. This can be rewritten as $\frac{d}{dx}(f'(x)e^{\phi(x)}) = \phi(x)\phi'(x)e^{\phi(x)}$. Integrating both sides with respect to $x$: $f'(x)e^{\phi(x)} = \int \phi(x)\phi'(x)e^{\phi(x)} dx$ Let $u = \phi(x)$, then $du = \phi'(x) dx$. The integral becomes $\int u e^u du$. Using integration by parts: $\int u e^u du = u e^u - \int e^u du = u e^u - e^u = e^u(u-1)$. So, $f'(x)e^{\phi(x)} = e^{\phi(x)}(\phi(x) - 1) + C_1$. Dividing by $e^{\phi(x)}$, we get $f'(x) = \phi(x) - 1 + C_1 e^{-\phi(x)}$.

Using the initial condition $f'(1) = 0$ and $\phi(1) = 0$: $0 = \phi(1) - 1 + C_1 e^{-\phi(1)}$ $0 = 0 - 1 + C_1 e^0$ $0 = -1 + C_1 \implies C_1 = 1$. Thus, $f'(x) = \phi(x) - 1 + e^{-\phi(x)}$.

Integrating $f'(x)$ to find $f(x)$: Since $f(1) = 0$, we have $f(x) = \int_1^x (\phi(t) - 1 + e^{-\phi(t)}) dt$.

Case (P): $\phi(x) = \ln x$ $\phi'(x) = 1/x$. $f'(x) = \ln x - 1 + e^{-\ln x} = \ln x - 1 + 1/x$. $f(x) = \int_1^x (\ln t - 1 + 1/t) dt = [t \ln t - t - t + \ln t]_1^x$ $f(x) = [t \ln t - 2t + \ln t]_1^x = (x \ln x - 2x + \ln x) - (1 \ln 1 - 2(1) + \ln 1)$ $f(x) = x \ln x - 2x + \ln x + 2$. $f(e) = e \ln e - 2e + \ln e + 2 = e - 2e + 1 + 2 = 3 - e$. $[f(e)] = [3 - e] \approx [3 - 2.718] = [0.282] = 0$. So, (P) matches with (3).

Case (Q): $\phi(x) = \ln x$ Using $f(x) = x \ln x - 2x + \ln x + 2$: $f(1/e) = (1/e) \ln(1/e) - 2(1/e) + \ln(1/e) + 2 = (1/e)(-1) - 2/e - 1 + 2 = -1/e - 2/e + 1 = 1 - 3/e$. $[f(1/e)] = [1 - 3/e] \approx [1 - 3/2.718] = [1 - 1.104] = [-0.104] = -1$. So, (Q) matches with (2).

Case (R): $\phi(x) = x - 1$ $\phi'(x) = 1$. $f'(x) = (x - 1) - 1 + e^{-(x - 1)} = x - 2 + e^{1 - x}$. $f(x) = \int_1^x (t - 2 + e^{1 - t}) dt = [\frac{t^2}{2} - 2t - e^{1 - t}]_1^x$ $f(x) = (\frac{x^2}{2} - 2x - e^{1 - x}) - (\frac{1}{2} - 2 - e^0) = \frac{x^2}{2} - 2x - e^{1 - x} - (\frac{1}{2} - 2 - 1)$ $f(x) = \frac{x^2}{2} - 2x - e^{1 - x} + \frac{5}{2}$. $f(2) = \frac{2^2}{2} - 2(2) - e^{1 - 2} + \frac{5}{2} = 2 - 4 - e^{-1} + \frac{5}{2} = -2 - \frac{1}{e} + \frac{5}{2} = \frac{1}{2} - \frac{1}{e}$. $[f(2)] = [\frac{1}{2} - \frac{1}{e}] \approx [0.5 - 0.368] = [0.132] = 0$. So, (R) matches with (3).

Case (S): $\phi(x) = x - 1$ Using $f(x) = \frac{x^2}{2} - 2x - e^{1 - x} + \frac{5}{2}$: $f(1/2) = \frac{(1/2)^2}{2} - 2(1/2) - e^{1 - 1/2} + \frac{5}{2} = \frac{1}{8} - 1 - e^{1/2} + \frac{5}{2} = \frac{1}{8} - \frac{8}{8} - \sqrt{e} + \frac{20}{8} = \frac{13}{8} - \sqrt{e}$. $[f(1/2)] = [\frac{13}{8} - \sqrt{e}] \approx [1.625 - 1.649] = [-0.024] = -1$. So, (S) matches with (2).

The correct matches are: (P) $\rightarrow$ (3) (Q) $\rightarrow$ (2) (R) $\rightarrow$ (3) (S) $\rightarrow$ (2)