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. .

**Q1.**lim

_{x→0}(x cosx-log(1+x))/x

^{2}equals

Solution

Using expressions of cosx and log (1+x), the given limit is equal to lim

Using expressions of cosx and log (1+x), the given limit is equal to lim

_{x→0}(x{1-x^{2}/2!+x^{4}/4!-x^{6}/6!…}-{x-x^{2}/2+x^{3}/3-x^{4}/4…})/x^{2}=lim_{x→0}(1/2-x/2!-x/3+⋯)=1/2**Q2.**If f(x) is differentiable and strictly increasing function, then the value of lim

_{x→0} (f(x

^{2})-f(x))/(f(x)-f(0)) is

Solution

Here, lim

Here, lim

_{x→0} (f(x^{2})-f(x))/(f(x)-f(0)) =lim_{x→0})(f^{'}(x^{2}).2x-f^{'}(x))/(f^{'}(x) ) =(-f^{'}(0))/(f^{'}(0))=-1**Q3.**The value of lim

_{n→∞}cos (x/2) cos (x/4) cos (x/8)…cos(x/2

^{n}) is

Solution

We know that cos A cos 2A cos4A …cos 2

We know that cos A cos 2A cos4A …cos 2

^{n-1}A=sin 2^{n}A/(2^{n}sinA ) lim_{n→∞}cos (x/2) cos (x/4)…cos (x/2^{n-1}) cos(x/2^{n}) =lim_{n→∞}sinx/(2^{n}sin (x/2^{n}) ) [put A=x/2^{n}] =lim_{n→∞}sinx/x.((x/2^{n}))/sin (x/2^{n}) =sinx/x

**Q4.**The value of lim

_{x→1} (log

_{2}2x )

^{logx5}is

Solution

We have, lim

We have, lim

_{x→1} (log_{2}2x )^{logx5}=lim_{x→1} (log_{2}2+log_{2}x )^{logx5}=lim_{x→1} (1+log_{2}x )^{1/log5x }=e^{(limx→1log2x .1/log5x )}=e^{limx→1log25}=e^{log25}**Q5.**The value of lim

_{x→7} (2-√(x-3))/(x

^{2}-49) is

Solution

lim

lim

_{x→7} (2-√(x-3))/(x^{2}-49)=lim_{x→7} (-1/(2√(x-3)))/2x =-1/4.7.2=-1/56**Q6.**lim

_{x→∞} (∫

^{2x}

_{0}xe

^{(x2 }dx)/e

^{4x2 }equals

Solution

lim

lim

_{x→∞} (∫^{2x}_{0}xe^{x}^{2}dx)/e^{4x2 }=lim_{x→∞} (∫^{2x}_{0}e^{x}^{2}dx^{2})/(2e^{4x2}) ) =lim_{x→∞} ([e^{(x2}) ]^{2x}_{0}/(2e^{4x2 }) =lim_{x→∞} (e^{4x2 }-1)/(2e^{4x2}) =lim_{x→∞} (1/2-1/e^{4x2})=1/2**Q7.**If f(x)=sin(e

^{x-2}-1)/log(x-1) , then lim

_{x→2} f(x) is given by

Solution

We have, ⇒lim

We have, ⇒lim

_{x→2>} f(x)=lim_{x→2} sin e^{x-2}-1)/log (x-1) ⇒lim_{x→2} f(x)=lim_{x→2}{sin(e^{x-2}-1)/(e^{x-2}-1).(e^{x-2}-1)/(x-2).(x-2)/log(1+(x-2) ) } ⇒lim_{x→2} f(x)=1×1×1=1**Q8.**The value of lim

_{x→∞}√(a

^{2}x

^{2}+ax+1)-√(a

^{2}x

^{2}+1), is

Solution

We have, lim

We have, lim

_{x→∞}√(a^{2}x^{2}+ax+1)-√(a^{2}x^{2}+1) =lim_{x→∞} ax/(√(a^{2}x^{2}+ax+1)+√(a^{2}x^{2}+1)) =lim_{x→∞} a/(√(a^{2}+a/x+1/x^{2})+√(a^{2}+1/x^{2}))=a/(2 a)=1/2**Q9.**The value of lim

_{x→1}(log

_{5}5x )

^{logx5}is

Solution

We have, lim

We have, lim

_{x→1}(log_{4}5x )^{logx5}=lim_{x→1}(log_{5}5+log_{5}x )^{logx5}=lim_{x→1} (1+log_{5}x )^{(1/log5x )}=e^{limx→1 log5x.1/logx5 }=e^{1}=e**Q10.**The value of lim

_{x→0} (e

^{x}+log (1+x)-(1-x)

^{-2})/x

^{2}is equal to

Solution

lim

lim

_{x→0} (e^{x}+log (1+x)-(1-x)^{-2})/x^{2}[0/0 form] =lim_{x→0} (e^{x}+(1+x)^{-1} -2(1-x)^{-3})/2x [by L’ Hospital’s rule] =lim_{x→0}(e^{x}-(1+x)^{-2}-6(1-x)^{-4})/2 [by L’ Hospital’s rule] =(e^{0}-1-6)/2=-3