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 logx/(a-b)=logy/(b-c)=logz/(c-a), then xyz is equal to

Solution

We have, logx/(a-b)=logy/(b-c)=logz/(c-a)=Î»(say) ⇒log〖x=Î»(a-b),〗 log〖y=Î»(b-c)〗,log〖z=Î»〗 (c-a) ⇒log〖x+log〖y+logz 〗 〗=0⇒log〖xyz=0⇒xyz=1〗

We have, logx/(a-b)=logy/(b-c)=logz/(c-a)=Î»(say) ⇒log〖x=Î»(a-b),〗 log〖y=Î»(b-c)〗,log〖z=Î»〗 (c-a) ⇒log〖x+log〖y+logz 〗 〗=0⇒log〖xyz=0⇒xyz=1〗

**Q2.**If x≠0, then the sum of series 1+x/(2 !)+(2 x^2)/(3 !)+(3 x^3)/(4 !)+⋯to ∞, is

Solution

We have, 1+x/(2 !)+(2 x^2)/(3 !)+(3 x^3)/(4 !)+⋯∞ =1+∑_(n=1)^∞▒((n-1) x^(n-1))/(n !) =1+∑_(n=1)^∞▒x^(n-1)/((n-1) !)-∑_(n=1)^∞▒x^(n-1)/(n !) =1+∑_(n=1)^∞▒x^(n-1)/((n-1) !)-1/x ∑_(n=1)^∞▒x^n/(n !) =1+e^x-1/x (e^x-1)=(e^x (x-1)+x+1)/x

We have, 1+x/(2 !)+(2 x^2)/(3 !)+(3 x^3)/(4 !)+⋯∞ =1+∑_(n=1)^∞▒((n-1) x^(n-1))/(n !) =1+∑_(n=1)^∞▒x^(n-1)/((n-1) !)-∑_(n=1)^∞▒x^(n-1)/(n !) =1+∑_(n=1)^∞▒x^(n-1)/((n-1) !)-1/x ∑_(n=1)^∞▒x^n/(n !) =1+e^x-1/x (e^x-1)=(e^x (x-1)+x+1)/x

**Q3.**If the sum of the roots of the equation ax^2+bx+c=0 be equal to the sum of the reciprocals of their squares, then bc^2,ca^2,ab^2 will be in

Solution

(a) Given equation is ax^2+bx+c=0 and let the roots are Î±,Î². So Î±+Î²=-b/a and Î±Î²=c/a. Now, Î±^2+Î²^2=(Î±+Î²)^2-2Î± Î² =b^2/a^2 -2c/a Now, 1/Î±^2 +1/Î²^2 =(Î±^2 〖+Î²〗^2)/(Î±^2 Î²^2 )=(b^2/a^2 -2c/a)/(c^2/a^2 )=(b^2-2ac)/c^2 According to given condition, Î±+Î²=1/Î±^2 +1/Î²^2 ⇒ -b/a=(b^2-2ac)/c^2 ⇒ -bc^2=ab^2-2a^2 c Hence, 2a^2 c=ab^2+bc^2 ⇒ ab^2,ca^2,bc^2 or bc^2,ca^2 be in AP.

(a) Given equation is ax^2+bx+c=0 and let the roots are Î±,Î². So Î±+Î²=-b/a and Î±Î²=c/a. Now, Î±^2+Î²^2=(Î±+Î²)^2-2Î± Î² =b^2/a^2 -2c/a Now, 1/Î±^2 +1/Î²^2 =(Î±^2 〖+Î²〗^2)/(Î±^2 Î²^2 )=(b^2/a^2 -2c/a)/(c^2/a^2 )=(b^2-2ac)/c^2 According to given condition, Î±+Î²=1/Î±^2 +1/Î²^2 ⇒ -b/a=(b^2-2ac)/c^2 ⇒ -bc^2=ab^2-2a^2 c Hence, 2a^2 c=ab^2+bc^2 ⇒ ab^2,ca^2,bc^2 or bc^2,ca^2 be in AP.

**Q4.**Jairam purchased a house in Rs 15000 and paid Rs 5000 at once. Rest money he promised to pay in annual installment of Rs 1000 with 10% per annum interest. How much money is to be paid by Jairam?

Solution

It will take 10yr for Jairam to pay off Rs 10000 in 10 yearly installments. ∴ He pays 10% annual interest on remaining amount. ∴ Money given in the first year =1000+(10000×10)/100=1000+1000 =Rs 2000 Money given in second year =1000+interest of (10000-1000) =1000+(9000×10)/100=100+900= Rs 1900 Similarly, money paid in third year = Rs 1800 etc. So, money given by Jairam in 10 yr will be Rs 2000, Rs 1900, Rs 1800, Rs 1700 … Which is in arithmetic progression, whose first term a=2000 and d=-100 Total money given in 10 yr =10/2 [2(2000)+(10-1)(-100)]= Rs 15500 Therefore, total money given by Jairam =5000+15500= Rs 20500

It will take 10yr for Jairam to pay off Rs 10000 in 10 yearly installments. ∴ He pays 10% annual interest on remaining amount. ∴ Money given in the first year =1000+(10000×10)/100=1000+1000 =Rs 2000 Money given in second year =1000+interest of (10000-1000) =1000+(9000×10)/100=100+900= Rs 1900 Similarly, money paid in third year = Rs 1800 etc. So, money given by Jairam in 10 yr will be Rs 2000, Rs 1900, Rs 1800, Rs 1700 … Which is in arithmetic progression, whose first term a=2000 and d=-100 Total money given in 10 yr =10/2 [2(2000)+(10-1)(-100)]= Rs 15500 Therefore, total money given by Jairam =5000+15500= Rs 20500

**Q5.**Let f(x) be a polynomial function of second degree. If f(1)=f(-1) and a,b,c are in AP, then f^' (a),f^' (b) and f^' (c) are in

Solution

Let f(x)=Ax^2+Bx+C ∴ f(1)=A+B+C and f(-1)=A-B+C ∵f(1)=f(-1) ⇒A+B+C=A-B+C ⇒B=0 ∴f(x)=Ax^2+C ⇒f^' (x)=2Ax ⇒f^' (a)=2Aa,f^' (b)=2Ab and f^' (c)=2Ac Also, a,b,c are in AP ∴2Aa,2Ab,2Ac are in AP ⇒ f^' (a),f^' (b),f^' (c) are in AP

Let f(x)=Ax^2+Bx+C ∴ f(1)=A+B+C and f(-1)=A-B+C ∵f(1)=f(-1) ⇒A+B+C=A-B+C ⇒B=0 ∴f(x)=Ax^2+C ⇒f^' (x)=2Ax ⇒f^' (a)=2Aa,f^' (b)=2Ab and f^' (c)=2Ac Also, a,b,c are in AP ∴2Aa,2Ab,2Ac are in AP ⇒ f^' (a),f^' (b),f^' (c) are in AP

**Q6.**If log_a〖x,〗 log_b〖x,〗 log_cx are in A.P., where x≠1, then c^2=

Solution

It is given that log_a〖x,log_b〖x,log_cx 〗 〗 are in A.P. ⇒2 log_b〖x=log_a〖x+log_cx 〗 〗 ⇒(2 logx)/logb =logx/loga +logx/logc ⇒(2 log〖a logc 〗)/logb =log〖a+logc 〗 ⇒(2 log〖a logc 〗)/logb =log〖(ac)〗 ⇒2 log〖c log_ba 〗=logac ⇒log〖c^(2 log_ba )=logac 〗 ⇒c^(2 log_ba )=ac⇒(c^2 )=(ac)^log_ab

It is given that log_a〖x,log_b〖x,log_cx 〗 〗 are in A.P. ⇒2 log_b〖x=log_a〖x+log_cx 〗 〗 ⇒(2 logx)/logb =logx/loga +logx/logc ⇒(2 log〖a logc 〗)/logb =log〖a+logc 〗 ⇒(2 log〖a logc 〗)/logb =log〖(ac)〗 ⇒2 log〖c log_ba 〗=logac ⇒log〖c^(2 log_ba )=logac 〗 ⇒c^(2 log_ba )=ac⇒(c^2 )=(ac)^log_ab

**Q7.**If logx/(a^2+ab+b^2 )=logy/(b^2+bc+c^2 )=logz/(c^2+ca+a^2 ), then x^(a-b).y^(b-c).z^(c-a)=

Solution

We have, logx/(a^2+ab+b^2 )=logy/(b^2+bc+c^2 )=logz/(c^2+ca+a^2 )=Î»(say) ⇒((a-b) logx)/(a^3-b^3 )=((b-c) logy)/(b^3-c^3 )=((c-a) logz)/(c^3-a^3 )=Î» ⇒log〖x^(a-b) 〗/(a^3-b^3 )=log〖y^(b-c) 〗/(b^3-c^3 )=log〖z^(c-a) 〗/(c^3-a^3 )=Î» ⇒log〖x^(a-b) 〗=Î»(a^3-b^3 ),log〖y^(b-c) 〗=Î»(b^3-c^3 ),log〖z^(c-a) 〗=Î»(c^3-a^3) ⇒log〖x^(a-b) 〗+log〖y^(b-c) 〗+log〖z^(c-a)=0〗 ⇒log〖(x^(a-b) y^(b-c) z^(c-a) )=0〗 ⇒x^(a-b) y^(b-c) z^(c-a)=1

We have, logx/(a^2+ab+b^2 )=logy/(b^2+bc+c^2 )=logz/(c^2+ca+a^2 )=Î»(say) ⇒((a-b) logx)/(a^3-b^3 )=((b-c) logy)/(b^3-c^3 )=((c-a) logz)/(c^3-a^3 )=Î» ⇒log〖x^(a-b) 〗/(a^3-b^3 )=log〖y^(b-c) 〗/(b^3-c^3 )=log〖z^(c-a) 〗/(c^3-a^3 )=Î» ⇒log〖x^(a-b) 〗=Î»(a^3-b^3 ),log〖y^(b-c) 〗=Î»(b^3-c^3 ),log〖z^(c-a) 〗=Î»(c^3-a^3) ⇒log〖x^(a-b) 〗+log〖y^(b-c) 〗+log〖z^(c-a)=0〗 ⇒log〖(x^(a-b) y^(b-c) z^(c-a) )=0〗 ⇒x^(a-b) y^(b-c) z^(c-a)=1

**Q8.**The sum of the series 1+(1+2)/(2 !)+(1+2+2^2)/(3 !)+(1+2+2^2+2^3)/(4 !)+⋯, is

Solution

**Q9.**If S_n=1^3+2^3+...+n^3 and T_n=1+2+...+n, then

Solution

Given, S_n=1^3+2^3+...+n^3=Î£n^3 and T_n=1+2+...+n=Î£n ∵S_n=Î£n^3=[(n(n+1))/2]⇒S_n={Î£n}^2=T_n^2

Given, S_n=1^3+2^3+...+n^3=Î£n^3 and T_n=1+2+...+n=Î£n ∵S_n=Î£n^3=[(n(n+1))/2]⇒S_n={Î£n}^2=T_n^2

**Q10.**If log_x〖(4x^log_5x +5)〗=2 log_5〖x,〗 then x equals to

Solution

We have, log_x〖(4.x^log_5x +5)=2 log_5x 〗 ⇒log_x〖(4.x^log_5x +5)=log_5〖x^2 〗 〗 ⇒4.x^log_5x +5=x^log_5〖x^2 〗 ⇒4.x^log_5x +5=x^(2 log_5x ) ⇒4y+5=y^2, where y=x^log_5x ⇒y^2-4 y-5=0 ⇒y=5,-1 ⇒x^log_5x =5 [∵y≠-1] ⇒log_5x=log_x5 ⇒(log_5x )^2=1⇒log_5〖x=±1⇒x=〖5,5〗^(-1) 〗

We have, log_x〖(4.x^log_5x +5)=2 log_5x 〗 ⇒log_x〖(4.x^log_5x +5)=log_5〖x^2 〗 〗 ⇒4.x^log_5x +5=x^log_5〖x^2 〗 ⇒4.x^log_5x +5=x^(2 log_5x ) ⇒4y+5=y^2, where y=x^log_5x ⇒y^2-4 y-5=0 ⇒y=5,-1 ⇒x^log_5x =5 [∵y≠-1] ⇒log_5x=log_x5 ⇒(log_5x )^2=1⇒log_5〖x=±1⇒x=〖5,5〗^(-1) 〗