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Here, you will find year-wise important questions and their detailed solutions for IIT JEE Advanced. Our goal is to provide you with a comprehensive collection of high-quality problems and their conceptual solution to strengthen your preparation and boost your confidence for the exam.
IIT JEE Advanced Paper-1, 2025 Questions and Solutions
Question-1
IIT JEE Advanced 2025 Moderate Level Question Based on Leibniz Integral Formula and L’Hopital’s Rule
IIT JEE Advanced 2025, Paper-1, Marks (+4, 0); Limits Question on Leibniz Rule
Let \alpha and \beta be the real numbers such that ,
{\lim\limits_{x\to 0}{\frac{1}{x^3}(\frac{\alpha}{2}\int_{0}^{x}\frac{1}{1-t^2}dt+\beta x \cos x)}=2}
then the value of \alpha + \beta is__________.
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IIT JEE Advanced Paper-2, 2025 Questions and Solutions
Question-1
IIT JEE Advanced 2025 Paper-2 a Good Question Based on L’Hopital’s Rule
IIT JEE Advanced 2025, Paper-2, (+3, –1); Limits Question
Let x_0 be the real number such that e^{x_0}+x_0=0 . For a given real number \alpha , define
g(x)=\frac{3xe^x+3x-\alpha e^x-\alpha x}{3(e^x +1)}
for all real numbers x .
Then which one of the following statements is TRUE?
(A) For \alpha = 2 , \lim_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=0
(B) For \alpha = 2 , \lim_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=1
(C) For \alpha = 3 , \lim_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=0
(D) For \alpha = 3 , \lim_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=\frac{2}{3}