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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. The value of limx→0⁡(eax-ebx)/x is equal to
•  a+b
•  a-b
•  eab
•  1
Solution
limx→0⁡(eax-ebx)/x =limx→0⁡((1+ax/1!+(ax)2/2!+...)-(1+bx/1!+(bx)2/2!+...))/x =a-b Alternate limx→0⁡(eax-ebx)/x=limx→0⁡(aeax-bebx)/1=a-b

Q2.If zr=cos⁡ rα/n2 +i sin⁡ rα/n2 , where r=1,2,3,…,n, then limn→∞⁡z1 z2…zn is equal to
•  cos⁡ α+i sin⁡α
•  cos⁡(α/2)-i sin⁡(α/2)
•  e(iα/2)
•  ∛(e )
Solution
limn→∞⁡z1 z2…zn =limn→∞ (cos⁡ α/n2 +i sin⁡ α/n2 )×(cos⁡ 2α/n2 +i sin⁡ 2α/n2 )…(cos⁡ nα/n2 +i sin⁡ nα/n2 ) =limn→∞ [cos⁡{α/n2 (1+2+3+...+n)}+i sin⁡{α/n2 (1+2+3+...+n)} ] =limn→∞ [cos⁡ (αn(n+1))/(2n2 )+i sin⁡ (αn(n+1))/(2n2 ) ] =cos⁡ α/2+i sin⁡ α/2=e(iα/2)

Q3. The value of limx→∞⁡(1.∑nr=1r+2.∑n-1r=1 r+3.∑n-2r=1 r+⋯n.1)/n4 is
•   1/24
•  1/12
•  1/6
•  None of these
Solution
We have, 1∙∑nr=1 r+2∙∑n-1r=1 r+3∙∑n-2r=1 +⋯+n .1 =∑nk=1 {k ∑r=1n-k+1 r} =∑nk=1 {k ((n-k+1)(n-k+2))/2} =1/2 ∑nk=1 k {(n+1)-k}{(n+2)-k} =1/2 ∑nk=1 {(n+1)(n+2)k-(2n+3) k2+k3 } =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+(n2 (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 [6n2+18n+12-8n2-16n-6+3n2+3n] =n(n+1)/24 (n2+5n+6) =(n(n+1)(n+2)(n+3))/24 ∴ Required limit =limn→∞⁡(n(n+1)(n+2)(n+3))/(24n4 ) =1/24 limn→∞⁡(1+1/n) (1+2/n)(1+3/n)=1/24

Q4. limx→0⁡d/dx ∫ (1-cos⁡x)/x2 dx is equal to
•  1
•  0
•  1/2
•  None of these
Solution
limx→0⁡d/dx ∫ ((1-cos⁡x)/x2 ) dx=limx→0⁡(1-cos⁡x)/x2 =limx→0⁡(2 sin2⁡x/2)/(4.x2/4) =1/2 limx→0⁡(sin⁡ x⁄2/(x⁄2))2=1/2

Q5.The value of limx→2⁡5/(√2-√x) is
•  10√2
•  +∞
•  -∞
•  Does not exist
Solution
LHL=limx→2-⁡5/(√2-√x) =limh→0⁡5/((√2-√(2-h)))×((√2+√(2-h)))/((√2+√(2-h))) =limh→0 (5(√2+√(2-h)))/(2-2+h)=∞ RHL=limx→2+ )⁡ 5/(√2-√x) =limh→0⁡5/((√2-√(2+h))×((√2+√(2+h))/((√2+√(2+h)) =limh→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 limx→3⁡g(x) is
•  9
•  10
• 25
•  20
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)=x2+k ⇒ g(x)=x2+1 [∵ g(0)=1] ∴ g(x) is satisfied in Eq. (i) ∴ limx→3)g(x)=g(3)=32+1=10

Q7.If limx→0⁡log⁡(3+x)-log⁡(3-x) /x=k, the value of k is
•  -2/3
•  0
•  -1/3
•  2/3
Solution
We have, limx→0⁡loge⁡(3+x)-loge⁡(3-x) /x=k ⇒limx→0⁡loge⁡(1+x/3)-loge⁡(1-x/3) /x=k ⇒limx→0⁡1/3×loge⁡(1+x/3)/(x/3)+1/3+limx→0⁡log⁡(1-x/3)/((-x/3) )=k ⇒1/3+1/3=k⇒k=2/3

Q8. Let α and β be the roots of a x2+b x+c=0, then limx→a⁡(1-cos⁡ (a x2+bx+c))/(x-α)2 is equal to
•  0
•  1/2 (α-β)2
•  α2/2 (α-β)2
•  2/2 (α-β)2
Solution
We have, limx→a⁡(1-cos⁡ (ax2+bx+c))/(x-α)2 =2limx→α⁡sin2⁡{((ax2+bx+c))/2}/(x-α)2 =2limx→α⁡sin2⁡{(a(x-α)(x-β))/2}/(x-α)2 [(∵α,β are roots of ax2+bx+c=0 ∴ax2+bx+c =a(x-α)(x-β))] =2limx→α⁡[sin⁡{(a(x-α)(x-β))/2}/((a(x-α)(x-β))/2)]2 ×a2/4 (x-β)2 =2(1)2×a2/4 (α-β)2=a2/2 (α-β)2

Q9.limx→0⁡(3x+|x|)/(7x-|x|)
•  2
•  1/6
•  0
•  None of these
Solution
We have, limx→0-⁡ (3x+|x|)/(7x-|x|)=limx→0⁡ (3x-x)/(7x+5x)=1/6 and, limx→0+ (3x+|x|)/(7x-5|x|)=limx→0⁡ (3x+x)/(7x-5x)=2 So, limx→0⁡ (3x+|x|)/(7x-5|x|) does not exist

Q10. The derivative of function f(x) istan4⁡x. If f(x)=0, then limx→0⁡ (f(x))/x is equal to
•  1
•  0
•  -1
• None of these
Solution
limx→0 (f(x))/x=limx→0 (f'(x))/1 =limx→0 tan4⁡x/1=0 ## Want to know more

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