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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Mathematics
Chapter
Limits

JEE Advanced 2025 · Paper 2 · Mathematics · Limits

Let x_0 be the real number such that e^ x_0 +x_0=0 . For a given...

Let x0x_0 be the real number such that ex0+x0=0e^{x_0}+x_0=0. For a given real number α\alpha, define g(x)=3xex+3x−αex−αx3(ex+1)g(x)=\frac{3 x e^x+3 x-\alpha e^x-\alpha x}{3\left(e^x+1\right)} for all real numbers xx. Then which one of the following statements is TRUE?Let x_0 be the real number such that e^ x_0 +x_0=0 . For a given real number α , define g(x)=(3 x e^x+3 x-α e^x-αLet x_0 be the real number such that e^ x_0 +x_0=0 . For a given real number α , define g(x)=(3 x e^x+3 x-α e^x-αLet x_0 be the real number such that e^ x_0 +x_0=0 . For a given real number α , define g(x)=(3 x e^x+3 x-α e^x-αLet x_0 be the real number such that e^ x_0 +x_0=0 . For a given real number α , define g(x)=(3 x e^x+3 x-α e^x-α

Options

  1. ADetermine the average velocity of a projectile between the instants it crosses half the maximum height. It is projected
  2. BDetermine the average velocity of a projectile between the instants it crosses half the maximum height. It is projected
  3. CDetermine the average velocity of a projectile between the instants it crosses half the maximum height. It is projected
  4. DDetermine the average velocity of a projectile between the instants it crosses half the maximum height. It is projected

Correct answer

C. Determine the average velocity of a projectile between the instants it crosses half the maximum height. It is projected

Step-by-step solution

ex0+x0=0g(x)=3xex+3x−αex−αx3(ex+1)g(x)=x−α(ex+x)3(ex+1),g(x)+ex0=x+ex0−α3(ex+xex+1)g(x)+ex0=(x−x0)−α3(ex+xex+1)⇒ As very clear g(x0)+x0⇒0\begin{aligned} & e^{x_0}+x_0=0 \\ & g(x)=\frac{3 x e^x+3 x-\alpha e^x-\alpha x}{3\left(e^x+1\right)} \\ & g(x)=x-\frac{\alpha\left(e^x+x\right)}{3\left(e^x+1\right)}, g(x)+e^{x_0}=x+e^{x_0}-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) \\ & g(x)+e^{x_0}=\left(x-x_0\right)-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) \\ & \Rightarrow \text{ As very clear } g\left(x_0\right)+x_0 \Rightarrow 0\end{aligned}lim⁡x→x0∣g(x)+ex0x−x0∣0↑0↑=lim⁡x→x0∣g′(x)1∣=∣g′(x0)∣g′(x)=1−α3((ex+1)(ex+1)−(ex+x)ex(ex+1)2)g′(x0)=1−α3((ex0+1)2(ex0+1)2)=1−α3⇒lim⁡x→x0∣g(x)+ex0x−x0∣=∣g′(x0)∣=∣1−α3∣\begin{aligned} & \lim _{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right| \frac{0 \uparrow}{0 \uparrow}=\lim _{x \rightarrow x_0}\left|\frac{g^{\prime}(x)}{1}\right|=\left|g^{\prime}\left(x_0\right)\right| \\ & g^{\prime}(x)=1-\frac{\alpha}{3}\left(\frac{\left(e^x+1\right)\left(e^x+1\right)-\left(e^x+x\right) e^x}{\left(e^x+1\right)^2}\right) \\ & g^{\prime}\left(x_0\right)=1-\frac{\alpha}{3}\left(\frac{\left(e^{x_0}+1\right)^2}{\left(e^{x_0}+1\right)^2}\right)=1-\frac{\alpha}{3} \\ & \Rightarrow \lim _{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=\left|g^{\prime}\left(x_0\right)\right|=\left|^{1}-\frac{\alpha}{3}\right|\end{aligned}$
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