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.**Let L=lim

_{x→0}(a-√(a

^{2}-x

^{2})-x

^{2}/4)/x

^{4},a>0. If L is finite, then

Solution

We have, L=lim

We have, L=lim

_{x→0}(a-√(a^{2}-x^{2})-x^{2}/4)/x^{4}⇒L=lim_{x→0} (a-a(1-x^{2}/a^{2})^{1/2})-x^{2}/4)/x^{4}⇒L=lim_{x→0} (a-a{1-1/2∙x^{2}/a^{2}-1/8∙x^{4}/a^{4}+1/16∙x^{6}/a^{6}…}-x^{2}/4)/x^{4}⇒L=lim_{x→0} ((1/2∙x^{2}/a+1/8∙x^{4}/a^{3}-1/16∙x^{6}/a^{5}…)-x^{2}/4)/x^{4}⇒L=lim_{x→0} x^{2}/2 (1/a-1/2)+1/(8a^{3})-1/16∙x^{2}/a^{5}+⋯ ⇒1/a-1/2=0 and in that case L=1/(8a^{3}) [∵L is finite] ⇒a=2 and L=1/64**Q2.**lim

_{x→0} x log

_{e} (sinx) is equal to

Solution

lim

lim

_{x→0}log_{e} (sinx )^{x}=log_{e}[lim_{x→0} (sinx )^{x}] =log_{e}[lim_{x→0} (1+sinx-1)^{(x(sin x-1)/(sin x-1)}] =log_{e}[e^{(limx→0 x(sin x-1))}] =log_{e}1**Q3.**If f:R→R is defined by f(x)=[x-3]+[x-4] for x∈R, then lim

_{x→3-} f(x) is equal to

Solution

∴ lim

∴ lim

_{x→3-} f(x)=lim_{x→3-} ([x-3]+|x-4|) =lim_{h→0} ([3-h-3]+|3-h-4|) =lim_{h→0}([-h]+1+h) =-1+1+0=0

**Q4.**lim

_{n→∞} (3.2

^{n+1}- 4.5

^{n+1})/( 5.2

^{n}+ 7.5

^{n}) is equal to

Solution

lim

lim

_{n→∞} ( 3.2^{n+1}- 4.5^{n+1})/( 5.2^{n}+ 7.5^{n}) =lim_{n→∞}(5^{n}(6.(2/5)^{n}-20))/(5^{n}(5.(2/5)^{n}+7) )=-20/7**Q5.**lim

_{x→0} (e

^{x2 }-cosx)/x

^{2}is equal to

Solution

lim

lim

_{x→0} (e^{x2}-cosx)/x^{2}(0/0 from) =lim_{x→0} ( 2xe^{x2 })-sinx)/2x (0/0 from) =lim_{x→0} ( 2e^{(x2}+4x^{2}e^{x2})+cosx)/2 =(2+0+1)/2=3/2**Q6.**The value oflim

_{x→2}{[(x

^{3}-4x)/(x

^{3}-8))

^{-1}-((x+√2x)/(x-2)-√2/(√x-√2))

^{-1}} is

Solution

We have, lim

We have, lim

_{x→2}{((x^{3}-4x)/(x^{3}-8))^{-1}-((x+√2x)/(x-2)-√2/(√x-√2))^{-1}} =lim_{x→2}{(x^{2}+2x+4)/(x(x+2))-((√x (x-2)-√2(x-2))/((x-2)(√x-√2)))^{-1}} =lim_{x→2}{(x^{2}+2x+4)/(x(x+2))-(((x-2)(√x-√2))/((x-2)(√x-√2)))^{-1}} =lim_{x→2}{(x^{2}+2x+4)/(x(x+2))}=12/8-1=1/2**Q7.**lim

_{x→0} (x tan2x-2x tanx)/(1-cos2x )

^{2}, is

Solution

1/2

1/2

**Q8.**The value of lim

_{x→2} (e

^{(3x-6)}-1)/sin(2-x) is

Solution

lim

lim

_{x→2} (e^{(3x-6)}-1)/sin (2-x) =lim_{x→2} (e^{(3x-6)}(3))/ -cos (2-x) [using L’ Hospital’s rule] =-(3e^{0})/cos0 =-3**Q9.**The value of lim

_{x→∞} {(a

_{1}

^{1/x}+a

_{2}

^{1/x}+⋯+a

_{n}

^{1/x})/n}

^{1}, is

Solution

Let x=1/y. Then, lim

Let x=1/y. Then, lim

_{x→∞}{(a_{1}^{1/x})+a_{2}^{1/x}+⋯+a_{n}^{1/x})/n}^{nx}=lim_{y→0}{(a_{1}^{y}+a_{2}^{y}+⋯+a_{n}^{y})/n}^{n/y}=lim_{y→0} {(1+a_{1}^{y}+a_{2}^{y}+⋯+a_{n}^{y}-n)/n}^{n/y}=e^{limy→0 {(a1y-1)/y+(a2y-1)/y+⋯+any-1}n/y }=e^{(log a1+ log a2 +⋯+log an )}=e^{log(a1 a2……an )}=a_{1}a_{2}a_{3}…a_{n}**Q10.**lim

_{x→-3} (3x

^{2}+ax+a-7)/(x

^{2}+2x-3) exists, then a is equal to

Solution

Here, lim

Here, lim

_{x→-3}x^{2}+2x-3=0 ∴lim_{x→-3}3x^{2}+ax+a-7must be zero, in order to limit exist. ⇒ 3(-3)^{2}+a(-3)+a-7=0 ⇒ 27-2a-7=0 ⇒ 2a=20 ⇒ a=10