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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_2⁡x))=0,〗 〗 then the values of x is
•  32
•  125
•  625
•  125
Solution
We have, log_5⁡〖(log_5⁡(log_2⁡x ) )=0〗 ⇒log_5⁡〖(log_2⁡x)〗=5^0⇒log_2⁡〖x=5⇒x=2^5 〗

Q2.Q∑_(n=0)^∞▒(log_e⁡x )^n/(n !) is equal
•  log_e⁡x
•  x
•  log_xe
•  None of these
Solution
Required sum =∑_(n=0)^∞▒(log_e⁡x )^n/(n !)=e^log_e⁡x =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
•   2
•  3
•  3/2
•  6
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〗 〗

Q4. The number of solutions of log_2⁡〖(x-1)=2 log_2⁡〖(x-3)〗 〗 is
•  2
•  1
•  6
•  7
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
•  0
•  1
•  2
•  3
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

Q6. If 0<ϕ<π/2,x=∑_(n=0)^∞▒cos^2n⁡ϕ ,y=∑_(n=0)^∞▒sin^2n⁡ϕ and z=∑_(n=0)^∞▒〖cos^2n⁡ϕ sin^2n⁡ϕ 〗, then
•  xyz=xz+y
•  xyz=xy+z
•  xyz=x+y+z
•  xyz=yz+x
Solution
Since, x=∑_(n=0)^∞▒cos^2n⁡ϕ =1+cos^2⁡ϕ+cos^4⁡ϕ+... =1/(1-cos^2⁡ϕ )=1/sin^2⁡ϕ [∵|cos⁡x |<1 and="" cos="" similarly="" span="" x.1="" xy-1="" xy="" xyz="xy+z" y="" z="1/(1-sin^2⁡ϕ">

Q7.log_e⁡3-log_e⁡9/2^2 +log_e⁡27/3^2 -log_e⁡81/4^2 +... is
•  (log_e⁡3 )(log_e⁡2)
•  (log_e⁡3 )
•  (log_e⁡2)
•  (log_e5/log_e⁡3)
Solution
log_e⁡3-log_e⁡9/2^2 +log_e⁡27/3^2 -log_e⁡81/4^2 +... =(log_e⁡3 )(1-1/2+1/3-1/4...) =(log_e⁡3 ) log_e⁡2

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
•  First terms
•  Sum of first and last terms
•  Last terms
•  Half of the sum of the series
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.

Q9.The coefficients of x^n in the expansion of log_a⁡〖(1+x)〗 is
•  (-1)^(n-1)/n
•  (-1)^(n-1)/n log_a⁡e
•   (-1)^(n-1)/n log_e⁡a
•   (-1)^n/n log_a⁡e
Solution
log_a⁡〖(1+x)〗=log_e⁡〖(1+x)〗 log_a⁡e=(log_a⁡e )[∑_(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_a⁡e

Q10. 0.14189189189… can be expressed as a rational number
•  7/3000
•  7/50
•  525/111
•  21/148
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 ## Want to know more

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