Question

Let the equation of a curve passing through the point (0, 1) be given by y = $\int_{}^{}{x^{2}.e^{x^{3}}}$dx. If the equation of the curve is written in the form x = ƒ(y) then ƒ(y) is –

• A. $\sqrt{\ln\left( \frac{3y - 2}{3} \right)}$
• B. $3\sqrt{\ln\left( \frac{2 - 3y}{3} \right)}$
• C. $\sqrt[3]{\ln\left( \frac{3y - 2}{3} \right)}$
• D. None of these

Answer 1

Answer: (C)

$\sqrt[3]{\ln\left( \frac{3y - 2}{3} \right)}$

Answer 2

Q y = $\int_{}^{}{x^{2}.e^{x^{3}}}$dx

= $\frac { 1 } { 3 }$ $\int_{}^{}{3x^{2}.e^{x^{3}}}$dx

y = $\frac{1}{3}$ . $e^{x^{3}}$ + c

it is passing through (0, 1) then

1 = $\frac{1}{3}$ + c Ž c = $\frac{2}{3}$

Then y = $\frac { 1 } { 3 }$ $e^{x^{3}}$ + $\frac{2}{3}$

Ž = $\frac{(3y - 2)}{3}$

\ x³ = ln $\left( \frac{3y - 2}{3} \right)$ or x = $\sqrt[3]{\ln\left( \frac{3y - 2}{3} \right)}$