## LIMITS AND DERIVATIVES quiz-19

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. limx→∞ [√(x2+2x-1)-x] is equal to
•
•  1/2
•  4
•  1
Solution
limx→∞ [√(x2+2x-1)-x] =limx→∞ [(x2+2x-1-x2)/(√(x2+2x-1)+x)] =limx→∞ [(2-1/x)/(√(1+2/x-1/x2 )+1)]=1

Q2.If Ai=(x-ai)/(|x-ai |),i=1,2,…,n and if a1<a2<a3<⋯<an. Then, limx→am ⁡ (A1 A2…An),1≤m≤n
•  Is equal to -1m
•  Is equal to -1m+1
•  Is equal to -1m-1
•  Does not exist
Solution
We have, Ai=(x-ai)/(|x-ai |),i=1,2,…,n and a1<a2<⋯<an-1<an Let x be in the left neighbourhood of am. Then, x-ai<0 for i=m,m+1,…,n and, x-ai>0 for i=1,2,…,m-1 Ai={ (=((x-ai ))/(-(x-ai ) )=-1 for i=m,m+1,…,n =(x-ai)/(x-ai )=1 for i=1,2,…,m-1) Similarly, if x is in the right neighbourhood of am. Then, x-ai<0 for i=m+1,…,n and x-ai>0 for i=1,2,….,m ∴Ai={ (Ai=(x-ai)/(-(x-ai ) )=-1 for i= m+1,…,n Ai=(x-ai)/(x-ai )=1 for i=1,2,…,m) Thus, we have limx→am- ⁡ (A1 A2…An)=-1n-m+1 and, limx→am+ )⁡ (A1 A2…An)=-1n-m Hence, limx→am ⁡(A1 A2…An ) does not exist

Q3. limx→0⁡ {(1+tan⁡x)/(1+sin⁡x )}cosec x is equal to
•   1/e
•  1
•  e
•  e2
Solution
limx→0⁡ {(1+tan⁡x)/(1+sin⁡x )}cosec x =limx→0⁡ [(1+sin⁡x/cos⁡x )(cos⁡x/sin⁡x ) ](1/cos⁡x )/(1+sin⁡x )(1/sin⁡x ) =e(limx→0 1/cos⁡x /e =e/e=1

Q4. limn→∞⁡(12+22+32+...+n2)/n3 is equal to
•  1/2
•  2/3
•  1/3
•  1/6
Solution
limn→∞⁡ (12+22+32+⋯+n2)/n3 =limn→∞ (∑ n2 )/n3 =limn→∞ (n(n+1)(2n+1))/(6n3 ) =limn→∞ 1/6 (1+1/n)(2+1/n) =1/3

Q5.If l1=limx→-2⁡ (x+|x|) ,l2=limx→-2⁡ (2x+|x|) and l3=limx→π/2⁡ cos⁡x/(x-π/2), then
•  l1<l2<l3
•  l2<l3<l1
•  l3<l2<l1
•  l1<l3<l2
Solution
We have, l1=limx→-2 (x+|x|) =-2+2=0 l2=limx→-2⁡ (2x+|x|) =-4+2=-2 and l3=limx→π/2⁡ cos⁡x/(x-π/2) =limx→π/2 ⁡sin⁡(π/2)/(-(π/2-x))=-1 ∴l2<l3<l1

Q6. limx→2⁡ =(√(1+√(2+x)) -√3)/(x-2) is equal to
•  1/(8√3)
•  1/√3
•  a28√3
•  √3
Solution
limx→2 =(√(1+√(2+x)) -√3)/(x-2) =limx→2⁡ ((1+√(2+x)-3))/((x-2)(√(1+√(2+x)) +√3)) =limx→2 (x-2)/((x-2)(√(1+√(2+x)) +√3)(√(2+x)+2)) =limx→2 1/((√(1+√(2+x)) +√3)(√(2+x)+2)) =1/((√(1+2)+√3)(√(2+2)+2))=1/(8√3)

Q7.If a=min⁡ {x2+4x+5:x∈R} and b=limθ→0⁡ (1-cos⁡2θ)/θ2 , then the value of ∑nr=0 n Cr ar bn-r, is
•  2n
•  3n
•  2n+1
•  2n-1
Solution
We have, x2+4x+5=(x+2)2+1≥1 for all x ∴a=1 b=limθ→0)⁡ (1-cos2⁡θ)/θ2 =1 ∴∑nr=0 n Cr ar bn-r=(a+b)n=2n

Q8.If limx→0⁡{(x3+1)/(x2+1)-(ax+b) }=2, then
•  a=1,b=1
•  a=1,b=2
•  a=1,b=-2
•  None of these
Solution
We have, limx→∞{(x3+1)/(x2+1)-(ax+b) }=2 ⇒limx→∞⁡ (x3 (1-a)-bx2-ax+(1-b))/(x2+1)=2 ⇒1-a=0 and -b=2⇒a=1,b=-2

Q9.The value of limn→∞⁡ (√(n2+1)+√n)/(∜(n3+n-) ∜n), is
•  0
•  1
•  -1
•  None of these
Solution
We have, limn→∞⁡ (√(n2+1)+√n)/(∜(n3+n-) ∜n) =limn→∞⁡ n{√(1+1/n2 )+1/√n}/(n3/4 {∜(1+1/n2 )-1/√n} ) =limn→∞ n1/4 {√(1+1/n2 )+1/√n}/{∜(1+1/n2 )-1/√n} =∞×2=∞

Q10. Let f(a)=g(a)=k and their nth derivatives fn (a),gn (a) exist and are not equal for somen. Further if limx→a⁡ (f(a)g(x)-f(a)-g(a)f(x)+g(a))/(g(x)-f(x))=4, then the value of k is equal to
•  4
•  2
•  1
•  0
Solution
Given, limx→a⁡ (f(a)g(x)-f(a)-g(a)f(x)+g(a))/(g(x)-f(x) )=4 Applying L’ Hospital’s rule, ⇒ limx→a⁡ (f(a)g'(x)-g(a)f'(x))/(g'(x)-f'(x))=4 ⇒ limx→a⁡ (kg'(x) -kf'(x))/(g'(x)-f'(x))=4 ⇒ k=4 ## Want to know more

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