As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.
Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
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Q1. If f(x)=x sin(1/x),x≠0, then the value of function at x=0, so that the function is continuous at x=0 is
Q2. The number of points at which the function f(x)=(|x-1|+|x-2|+cosx) where x∈[0,4] is not continuous, is
Solution
Given, f(x)=|x-1|+|x-2|+cosx Since, |x-1|,|x-2| and cosx are continuous in [0, 4] ∴ f(x) being sum of continuous functions is also continuous
Given, f(x)=|x-1|+|x-2|+cosx Since, |x-1|,|x-2| and cosx are continuous in [0, 4] ∴ f(x) being sum of continuous functions is also continuous
Q3. The function f(x)=(2x2+7)/(x3+3x2-x-3) is discontinuous for
Solution
Given, f(x)=(2x2+7)/((x2-1)(x+3)) Since, at x=1,-1,-3,f(x)=∞ Hence, function is discontinuous
Given, f(x)=(2x2+7)/((x2-1)(x+3)) Since, at x=1,-1,-3,f(x)=∞ Hence, function is discontinuous
Q6. Let f(x) be an even function. Then f'(x)
Solution
Since f(x) is an even function ∴f(-x)=f(x) for all x ⇒-f' (-x)=f'(x) for all x ⇒f' (-x)=-f'(x) for all x ⇒f'(x) is an odd function
Since f(x) is an even function ∴f(-x)=f(x) for all x ⇒-f' (-x)=f'(x) for all x ⇒f' (-x)=-f'(x) for all x ⇒f'(x) is an odd function
Q10. Let g(x)=(x-1)n/logcosm(x-1) ;0 < x < 2, m and n are integers. m=0, n > 0, and let p be the left hand derivative of |x-1| at x=1. If lim(x→1)+ g(x)=p, then