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)=(2x

^{2}+7)/(x

^{3}+3x

^{2}-x-3) is discontinuous for

Solution

Given, f(x)=(2x

Given, f(x)=(2x

^{2}+7)/((x^{2}-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}/logcos

^{m}(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