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.**If log_5〖(log_5〖(log_2x))=0,〗 〗 then the values of x is

Solution

We have, log_5〖(log_5(log_2x ) )=0〗 ⇒log_5〖(log_2x)〗=5^0⇒log_2〖x=5⇒x=2^5 〗

We have, log_5〖(log_5(log_2x ) )=0〗 ⇒log_5〖(log_2x)〗=5^0⇒log_2〖x=5⇒x=2^5 〗

**Q2.**Q∑_(n=0)^∞▒(log_ex )^n/(n !) is equal

Solution

Required sum =∑_(n=0)^∞▒(log_ex )^n/(n !)=e^log_ex =x

Required sum =∑_(n=0)^∞▒(log_ex )^n/(n !)=e^log_ex =x

**Q3.**If ∑_(r=1)^n▒〖a_r=1/6 n(n+1)(n+2)〗 for all n≥1, then lim┬(n→∞)∑_(r=1)^n▒1/a^r is

Solution

For n≥1, we have a_n=∑_(r=1)^n▒〖a_r-∑_(r=1)^(n-1)▒〖a_r=1/6 n(n+1)(n+2)-1/6 (n-1)(n)(n+1) 〗〗 ⇒a_n=(n(n+1))/2 ∴∑_(r=1)^n▒〖1/a_r =2∑_(r=1)^n▒〖1/r(r+1) =2∑_(r=1)^n▒(1/r-1/(r+1)) 〗〗 ⇒∑_(r=1)^n▒〖1/a_r =2(1-1/(n+1)) 〗 ⇒lim┬(n→∞)∑_(r=1)^n▒〖1/a_r =〖lim┬(n→∞) 2〗〖(1+1/(n+1))=2〗 〗

For n≥1, we have a_n=∑_(r=1)^n▒〖a_r-∑_(r=1)^(n-1)▒〖a_r=1/6 n(n+1)(n+2)-1/6 (n-1)(n)(n+1) 〗〗 ⇒a_n=(n(n+1))/2 ∴∑_(r=1)^n▒〖1/a_r =2∑_(r=1)^n▒〖1/r(r+1) =2∑_(r=1)^n▒(1/r-1/(r+1)) 〗〗 ⇒∑_(r=1)^n▒〖1/a_r =2(1-1/(n+1)) 〗 ⇒lim┬(n→∞)∑_(r=1)^n▒〖1/a_r =〖lim┬(n→∞) 2〗〖(1+1/(n+1))=2〗 〗

**Q4.**The number of solutions of log_2〖(x-1)=2 log_2〖(x-3)〗 〗 is

Solution

We have, log_2〖(x-1)〗=2 log_2〖(x-3)〗 ⇒x-1=(x-3)^2⇒x^2-7x+10=0⇒x=2,5 But, log_2〖(x-1)〗 and log_2〖(x-3)〗 are defined for x>3 Hence, x=5 is the only solution

We have, log_2〖(x-1)〗=2 log_2〖(x-3)〗 ⇒x-1=(x-3)^2⇒x^2-7x+10=0⇒x=2,5 But, log_2〖(x-1)〗 and log_2〖(x-3)〗 are defined for x>3 Hence, x=5 is the only solution

**Q5.**If a^x=b,b^y=c,c^z=a, then value of xyz is

Solution

We have, a^x=b,b^y=c and c^z=a ⇒b=(c^z )^x [∵b=a^x and a=c^z] ⇒b=c^zx ⇒b=(b^y )^zx [∵c=b^y] ⇒b=b^xyz⇒xyz=1

We have, a^x=b,b^y=c and c^z=a ⇒b=(c^z )^x [∵b=a^x and a=c^z] ⇒b=c^zx ⇒b=(b^y )^zx [∵c=b^y] ⇒b=b^xyz⇒xyz=1

**Q6.**If 0<Ï•<Ï€/2,x=∑_(n=0)^∞▒cos^2nÏ• ,y=∑_(n=0)^∞▒sin^2nÏ• and z=∑_(n=0)^∞▒〖cos^2nÏ• sin^2nÏ• 〗, then

Solution

Since, x=∑_(n=0)^∞▒cos^2nÏ• =1+cos^2Ï•+cos^4Ï•+... =1/(1-cos^2Ï• )=1/sin^2Ï• [∵|cosx |<1 and="" cos="" similarly="" span="" x.1="" xy-1="" xy="" xyz="xy+z" y="" z="1/(1-sin^2Ï•">

Since, x=∑_(n=0)^∞▒cos^2nÏ• =1+cos^2Ï•+cos^4Ï•+... =1/(1-cos^2Ï• )=1/sin^2Ï• [∵|cosx |<1 and="" cos="" similarly="" span="" x.1="" xy-1="" xy="" xyz="xy+z" y="" z="1/(1-sin^2Ï•">

**Q7.**log_e3-log_e9/2^2 +log_e27/3^2 -log_e81/4^2 +... is

Solution

log_e3-log_e9/2^2 +log_e27/3^2 -log_e81/4^2 +... =(log_e3 )(1-1/2+1/3-1/4...) =(log_e3 ) log_e2

log_e3-log_e9/2^2 +log_e27/3^2 -log_e81/4^2 +... =(log_e3 )(1-1/2+1/3-1/4...) =(log_e3 ) log_e2

**Q8.**In an AP the sum of any two terms, such that the distance of one of them from the beginning is same as that of the other from the end, is

Solution

In an AP, the distance of one of them from the beginning is same as that of the other from the end is equal to the sum of first and last terms.

In an AP, the distance of one of them from the beginning is same as that of the other from the end is equal to the sum of first and last terms.

**Q9.**The coefficients of x^n in the expansion of log_a〖(1+x)〗 is

Solution

log_a〖(1+x)〗=log_e〖(1+x)〗 log_ae=(log_ae )[∑_(n=1)^∞▒〖(-1)^(n-1) x^n/n〗] So, the coefficient of x^n in log_a〖(1+x)〗 is (-1)^(n-1)/n log_ae

log_a〖(1+x)〗=log_e〖(1+x)〗 log_ae=(log_ae )[∑_(n=1)^∞▒〖(-1)^(n-1) x^n/n〗] So, the coefficient of x^n in log_a〖(1+x)〗 is (-1)^(n-1)/n log_ae

**Q10.**0.14189189189… can be expressed as a rational number

Solution

0.14189189189… =0.14+0.00189+0.00000189+⋯ =14/100+189[1/10^5 +1/10^8 +...∞] =7/50+189/(999×100) =7/50+7/3700=21/148

0.14189189189… =0.14+0.00189+0.00000189+⋯ =14/100+189[1/10^5 +1/10^8 +...∞] =7/50+189/(999×100) =7/50+7/3700=21/148