Verified Solution • 2025–26 Session moderate QJEEM24Mab628: JEE Main 2024 Mathematics - Class 12 JEEM24Mab628 Solution: If the function defined by is one - one and onto, then the… If the function f:(−∞,−1]→(a,b]f:(-\infty,-1] \rightarrow(a, b]f:(−∞,−1]→(a,b] defined by f(x)=ex3−3x+1f(x)=e^{x^3-3 x+1}f(x)=ex3−3x+1 is one - one and onto, then the distance of the point P(2b+4,a+2)P(2 b+4, a+2)P(2b+4,a+2) from the line x+e−3y=4x+e^{-3} y=4x+e−3y=4 is : Multiple Choice Options A 21+e62 \sqrt{1+e^6}21+e6 B 1+e6\sqrt{1+e^6}1+e6 C 31+e63 \sqrt{1+e^6}31+e6 D 41+e64 \sqrt{1+e^6}41+e6 Step-by-Step Verified Solution amp;f(x)=ex3−3x+1amp;f′(x)=ex3−3x+1⋅(3x2−3)amp;=ex3−3x+1⋅3(x−1)(x+1)\begin{aligned}& f(x)=e^{x^3-3 x+1} \\& f^{\prime}(x)=e^{x^3-3 x+1} \cdot\left(3 x^2-3\right) \\& =e^{x^3-3 x+1} \cdot 3(x-1)(x+1)\end{aligned}amp;f(x)=ex3−3x+1amp;f′(x)=ex3−3x+1⋅(3x2−3)amp;=ex3−3x+1⋅3(x−1)(x+1) For f′(x)≥0\mathrm{f}^{\prime}(\mathrm{x}) \geq 0f′(x)≥0 ∴f(x)\therefore \mathrm{f}(\mathrm{x})∴f(x) is increasing function amp;∴a=e−∞=0=f(−∞)amp;b=e−1+3+1=e3=f(−1)amp;P(2 b+4,a+2)amp;∴P(2e3+4,2)\begin{aligned}& \therefore \mathrm{a}=\mathrm{e}^{-\infty}=0=\mathrm{f}(-\infty) \\& \mathrm{b}=\mathrm{e}^{-1+3+1}=\mathrm{e}^3=\mathrm{f}(-1) \\& \mathrm{P}(2 \mathrm{~b}+4, \mathrm{a}+2) \\& \therefore \mathrm{P}\left(2 \mathrm{e}^3+4,2\right)\end{aligned}amp;∴a=e−∞=0=f(−∞)amp;b=e−1+3+1=e3=f(−1)amp;P(2 b+4,a+2)amp;∴P(2e3+4,2) d=(2e3+4)+2e−3−41+e−6=21+e6\mathrm{d}=\frac{\left(2 \mathrm{e}^3+4\right)+2 \mathrm{e}^{-3}-4}{\sqrt{1+\mathrm{e}^{-6}}}=2 \sqrt{1+\mathrm{e}^6}d=1+e−6(2e3+4)+2e−3−4=21+e6 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring
