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.**The value of lim

_{x→∞}((x+3)/(x+1))

^{(x+2)}is

Solution

lim

lim

_{x→∞} ((x+3)/(x+1))^{(x+2)}=lim_{x→∞}[(1+2/(x+1))^{((x+2)/2) }]^{(2/(x+2)×(x+2)) }=lim_{x→∞}(1+2/(x+1))^{(x+2)/2}[(2(x+2))/(x+1)] =e^{(limx→∞}((2+4/x))/((1+1/x))) =e^{2}**Q2.**The value of lim

_{x-0} 1/x

^{3}∫

^{x}

_{0}(t log(1+t))/(t

^{4}+4) dt is

Solution

Given limit =lim

Given limit =lim

_{x→0}(∫^{x}_{0}(t log(1+t))/(t^{4}+4) dt)/x^{3}Using L’ Hospital’s rule, =lim_{x→0}((x log(1+x))/(x^{4}+4))/3x^{2}=lim_{x→0}log(1+x)/3x.1/(x^{4}+4) =1/3.1/4=1/12**Q3.**lim

_{x→1} (x

^{8}-2x+1)/(x

^{4}-2x+1) equals

Solution

lim

lim

_{x→1} (x^{8}-2x+1)/(x^{4}-2x+1)=lim_{x→1} (8x^{7}-2)/(4x^{3}-2) =(8-2)/(4-2)=3 [using L’ Hospital’s rule]

**Q4.**lim

_{x→0}((1+x)

^{8}-1)/((1+x)

^{2}-1) is equal to

Solution

lim

lim

_{x→0} ((1+x)^{8}-1)/((1+x)^{2}-1) =lim_{x→0}([(1+x)^{4}+1][(1+x)^{2}+1][(1+x)^{2}-1])/((1+x)^{2}-1) =2×2=4 Alternate lim_{x→0}((1+x)^{8}-1)/((1+x)^{2}-1) (0/0 form) lim_{x→0}(8(1+x)^{7})/(2(1+x)) (by L’ Hospital’s rule) =4**Q5.**lim

_{x→0}(x/(√(1+x)-√(1-x)))is equal to

Solution

lim

lim

_{x→0} (x/(√(1+x)-√(1-x))) =lim_{x→0}((x(√(1+x)+√(1-x)))/2x) =1**Q6.**The value of lim

_{x→∞}x

^{(3/2)}(√(x

^{3}+1)-√(x

^{3}-1)), is

Solution

We have, lim

We have, lim

_{x→∞} x^{(3/2)}(√(x^{3}+1)-√(x^{3}-1)) =lim_{x→∞} (2x^{(3/2)})/(√(x^{3}+1)+√(x^{3}-1)) =lim_{x→∞}(2x^{(3/2)})/(√(1+1/x^{3})+√(1-1/x^{3}))=2/(1+1)=1**Q7.**lim

_{x→0} (2 sin

^{2}3x)/x

^{2}is equal to

Solution

lim

lim

_{x→0} (2 sin^{2}3x)/x^{2}=lim_{x→0}2(sin3x/3x)^{2}×9/1=18**Q8.**lim

_{n→∞}(1

^{2}/(1-n

^{3})+2

^{2}/(1-n

^{3})+...+n

^{2}/(1-n

^{3})) is equal to

Solution

lim

lim

_{n→∞}1/(1-n^{3}) ∑^{n}_{(r=1)}r^{2}= lim_{n→∞}1/(n^{3}(1/n^{3}-1) ) (n(n+1)(2n+1))/6 =lim_{n→∞}(n^{3}(1+1/n)(2+1/n))/(n^{3}(1/n^{3}-1)6)=-1/3**Q9.**If l

_{1}=lim

_{x→2}

^{+}) (x+[x]), l

_{2}=lim

_{x→2}

^{-}) (2x-[x]) and l

_{3}=lim

_{x→Ï€/2} cosx/((x-Ï€/2)), then

Solution

l

l

_{1}=lim_{x→2}^{+}(x+[x] ) =lim_{h→0}2+h+[2+h]=4 l_{2}=lim_{x→2}^{-})(2x-[x]) =lim_{h→0}{2(2-h)-[2-h]} =lim_{h→0}{2(2-h)-1}=3 l_{3}=lim_{x→Ï€/2} cosx/(x-Ï€/2)=lim_{x→Ï€/2}-sinx=-1 [by L’Hospital’s rule] Thus, l_{3}<l_{2}<l_{1}**Q10.**lim

_{h→0}(sin (a+3h)-3 sin(a+2h)+3 sin(a+h)-sina)/h

^{3}is equal to

Solution

We have, lim

We have, lim

_{h→0}(sin (a+3h) -3 sin(a+2h)+3 sin(a+h)-sina)/h^{3}=lim_{h→0}({sin (a+3h)-sina }-3{sin (a+2h)-sin(a+h) })/h^{3}=lim_{h→0}(2 sin 3h/2 cos (a+3h/2)-6 cos(a+3h/2) sin h/2 )/h^{3}=lim_{h→0}(2 cos(a+3h/2)(sin 3h/2 -3 sin h/2 ))/h^{3}=-8lim_{h→0}cos(a+3h/2) sin^{3} h/2 /h^{3}=-lim_{h→0}cos(a+3h/2) {sin h/2 /(h/2)}^{3}=-cosa