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→0}(e

^{ax}-e

^{bx})/x is equal to

Solution

lim

lim

_{x→0}(e^{ax}-e^{bx})/x =lim_{x→0}((1+ax/1!+(ax)^{2}/2!+...)-(1+bx/1!+(bx)^{2}/2!+...))/x =a-b Alternate lim_{x→0}(e^{ax}-e^{bx})/x=lim_{x→0}(ae_{ax}-be_{bx})/1=a-b**Q2.**If z

_{r}=cos rÎ±/n

^{2}+i sin rÎ±/n

^{2}, where r=1,2,3,…,n, then lim

_{n→∞}z

_{1}z

_{2}…z

_{n}is equal to

Solution

lim

lim

_{n→∞}z_{1}z_{2}…z_{n}=lim_{n→∞}(cos Î±/n^{2}+i sin Î±/n^{2})×(cos 2Î±/n^{2}+i sin 2Î±/n^{2})…(cos nÎ±/n^{2}+i sin nÎ±/n^{2}) =lim_{n→∞}[cos{Î±/n^{2}(1+2+3+...+n)}+i sin{Î±/n^{2}(1+2+3+...+n)} ] =lim_{n→∞}[cos (Î±n(n+1))/(2n^{2})+i sin (Î±n(n+1))/(2n^{2 ) ] =cos Î±/2+i sin Î±/2=e(iÎ±/2) }**Q3.**The value of lim

_{x→∞}(1.∑

^{n}

_{r=1}r+2.∑

^{n-1}

_{r=1}r+3.∑

^{n-2}

_{r=1}r+⋯n.1)/n

^{4}is

Solution

We have, 1∙∑

We have, 1∙∑

^{n}_{r=1}r+2∙∑^{n-1}_{r=1}r+3∙∑^{n-2}_{r=1}+⋯+n .1 =∑^{n}_{k=1}{k ∑_{r=1}^{n-k+1}r} =∑^{n}_{k=1}{k ((n-k+1)(n-k+2))/2} =1/2 ∑^{n}_{k=1}k {(n+1)-k}{(n+2)-k} =1/2 ∑^{n}_{k=1}{(n+1)(n+2)k-(2n+3) k^{2}+k^{3}} =1/2 [(n+1)(n+2) (n(n+1))/2-(2n+3) n(n+1)(2n+1)/6+{n(n+1)/2}^{2}] =1/2 [(n(n+1)^{2}(n+2))/2-n(n+1)(2n+1)(2n+3)/6+(n^{2}(n+1)^{2})/4] =(n(n+1))/24 [6(n+1)(n+2)-2(2n+1)(2n+3)+3n(n+1)] =(n(n+1))/24 [6n^{2}+18n+12-8n^{2}-16n-6+3n^{2}+3n] =n(n+1)/24 (n^{2}+5n+6) =(n(n+1)(n+2)(n+3))/24 ∴ Required limit =lim_{n→∞}(n(n+1)(n+2)(n+3))/(24n^{4}) =1/24 lim_{n→∞}(1+1/n) (1+2/n)(1+3/n)=1/24

**Q4.**lim

_{x→0}d/dx ∫ (1-cosx)/x

^{2}dx is equal to

Solution

lim

lim

_{x→0}d/dx ∫ ((1-cosx)/x^{2}) dx=lim_{x→0}(1-cosx)/x^{2}=lim_{x→0}(2 sin^{2}x/2)/(4.x^{2}/4) =1/2 lim_{x→0}(sin x⁄2/(x⁄2))^{2}=1/2**Q5.**The value of lim

_{x→2}5/(√2-√x) is

Solution

LHL=lim

LHL=lim

_{x→2}^{-5/(√2-√x)}=lim_{h→0}5/((√2-√(2-h)))×((√2+√(2-h)))/((√2+√(2-h))) =lim_{h→0}(5(√2+√(2-h)))/(2-2+h)=∞ RHL=lim_{x→2+ ) 5/(√2-√x)}=lim_{h→0}5/((√2-√(2+h))×((√2+√(2+h))/((√2+√(2+h)) =lim_{h→0}(5(√2+√(2+h))/(2-2-h)=-∞ ∴ LHL≠RHL Hence, limit does not exist.**Q6.**If g(x) is a polynomial satisfying g(x)g(y)=g(x)+g(y)+g(xy)-2 for all real x and y and g(2)=5, then lim

_{x→3}g(x) is

Solution

Since, g(x)g(y)=g(x)+g(y)+g(xy)-2 ...(i) Now, at x=0,y=2, we get g(0)g(2)=g(0)+g(2)+g(0)-2 ⇒ 5g(0)=5+2g(0)-2 [∵ g(2)=5] ⇒ g(0)=1 g(x)is given in a polynomial and by the relation given g(x) cannot be linear. Let g(x)=x

Since, g(x)g(y)=g(x)+g(y)+g(xy)-2 ...(i) Now, at x=0,y=2, we get g(0)g(2)=g(0)+g(2)+g(0)-2 ⇒ 5g(0)=5+2g(0)-2 [∵ g(2)=5] ⇒ g(0)=1 g(x)is given in a polynomial and by the relation given g(x) cannot be linear. Let g(x)=x

^{2}+k ⇒ g(x)=x^{2}+1 [∵ g(0)=1] ∴ g(x) is satisfied in Eq. (i) ∴ lim_{x→3})g(x)=g(3)=3^{2}+1=10**Q7.**If lim

_{x→0}log(3+x)-log(3-x) /x=k, the value of k is

Solution

We have, lim

We have, lim

_{x→0}log_{e}(3+x)-log_{e}(3-x) /x=k ⇒lim_{x→0}log_{e}(1+x/3)-log_{e}(1-x/3) /x=k ⇒lim_{x→0}1/3×log_{e}(1+x/3)/(x/3)+1/3+lim_{x→0}log(1-x/3)/((-x/3) )=k ⇒1/3+1/3=k⇒k=2/3**Q8.**Let Î± and Î² be the roots of a x

^{2}+b x+c=0, then lim

_{x→a}(1-cos (a x

^{2}+bx+c))/(x-Î±)

^{2}is equal to

Solution

We have, lim

We have, lim

_{x→a}(1-cos (ax^{2}+bx+c))/(x-Î±)^{2}=2lim_{x→Î±}sin^{2}{((ax^{2}+bx+c))/2}/(x-Î±)^{2}=2lim_{x→Î±}sin^{2}{(a(x-Î±)(x-Î²))/2}/(x-Î±)^{2}[(∵Î±,Î² are roots of ax^{2}+bx+c=0 ∴ax^{2}+bx+c =a(x-Î±)(x-Î²))] =2lim_{x→Î±}[sin{(a(x-Î±)(x-Î²))/2}/((a(x-Î±)(x-Î²))/2)]^{2}×a^{2}/4 (x-Î²)^{2}=2(1)^{2}×a^{2}/4 (Î±-Î²)^{2}=a^{2}/2 (Î±-Î²)^{2}**Q9.**lim

_{x→0}(3x+|x|)/(7x-|x|)

Solution

We have, lim

We have, lim

_{x→0-} (3x+|x|)/(7x-|x|)=lim_{x→0} (3x-x)/(7x+5x)=1/6 and, lim_{x→0+ }(3x+|x|)/(7x-5|x|)=lim_{x→0} (3x+x)/(7x-5x)=2 So, lim_{x→0} (3x+|x|)/(7x-5|x|) does not exist**Q10.**The derivative of function f(x) istan

^{4}x. If f(x)=0, then lim

_{x→0} (f(x))/x is equal to

Solution

lim

lim

_{x→0}(f(x))/x=lim_{x→0}(f'(x))/1 =lim_{x→0}tan^{4}x/1=0