QUIZ-19 LIMITS AND DERIVATIVES

LIMITS AND DERIVATIVES quiz-19

Dear Readers,

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. lim_x→∞ [√(x²+2x-1)-x] is equal to

•  ∞
•  1/2
•  4
•  1

Solution
lim_x→∞ [√(x²+2x-1)-x] =lim_x→∞ [(x²+2x-1-x²)/(√(x²+2x-1)+x)] =lim_x→∞ [(2-1/x)/(√(1+2/x-1/x² )+1)]=1

Q2.If A_i=(x-a_i)/(|x-a_i |),i=1,2,…,n and if a₁<a₂<a₃<⋯<a_n. Then, lim_x→am ⁡ (A₁ A₂…A_n),1≤m≤n

•  Is equal to -1^m
•  Is equal to -1^m+1
•  Is equal to -1^m-1
•  Does not exist

Solution
We have, A_i=(x-a_i)/(|x-a_i |),i=1,2,…,n and a₁<a₂<⋯<a_n-1<a_n Let x be in the left neighbourhood of a_m. Then, x-a_i<0 for i=m,m+1,…,n and, x-a_i>0 for i=1,2,…,m-1 A_i={ (=((x-a_i ))/(-(x-a_i ) )=-1 for i=m,m+1,…,n =(x-a_i)/(x-a_i )=1 for i=1,2,…,m-1) Similarly, if x is in the right neighbourhood of a_m. Then, x-a_i<0 for i=m+1,…,n and x-a_i>0 for i=1,2,….,m ∴A_i={ (A_i=(x-a_i)/(-(x-a_i ) )=-1 for i= m+1,…,n A_i=(x-a_i)/(x-a_i )=1 for i=1,2,…,m) Thus, we have lim_x→am^- ⁡ (A₁ A₂…A_n)=-1^n-m+1 and, lim_x→am⁺ )⁡ (A₁ A₂…A_n)=-1^n-m Hence, lim_x→am ⁡(A₁ A₂…A_n ) does not exist

Q3. lim_x→0⁡ {(1+tan⁡x)/(1+sin⁡x )}^{cosec x} is equal to

•   1/e
•  1
•  e
•  e²

Solution
lim_x→0⁡ {(1+tan⁡x)/(1+sin⁡x )}^{cosec x} =lim_x→0⁡ [(1+sin⁡x/cos⁡x )^{(cos⁡x/sin⁡x )} ]^{(1/cos⁡x )}/(1+sin⁡x )^{(1/sin⁡x )} =e^{(lim_x→0 1/cos⁡x} /e =e/e=1

Q4. lim_n→∞⁡(1²+2²+3²+...+n²)/n³ is equal to

•  1/2
•  2/3
•  1/3
•  1/6

Solution
lim_n→∞⁡ (1²+2²+3²+⋯+n²)/n³ =lim_n→∞ (∑ n² )/n³ =lim_n→∞ (n(n+1)(2n+1))/(6n³ ) =lim_n→∞ 1/6 (1+1/n)(2+1/n) =1/3

Q5.If l₁=lim_x→-2⁡ (x+|x|) ,l₂=lim_x→-2⁡ (2x+|x|) and l₃=lim_x→π/2⁡ cos⁡x/(x-π/2), then

•  l₁<l₂<l₃
•  l₂<l₃<l₁
•  l₃<l₂<l₁
•  l₁<l₃<l₂

Solution
We have, l₁=lim_x→-2 (x+|x|) =-2+2=0 l₂=lim_x→-2⁡ (2x+|x|) =-4+2=-2 and l₃=lim_x→π/2⁡ cos⁡x/(x-π/2) =lim_x→π/2 ⁡sin⁡(π/2)/(-(π/2-x))=-1 ∴l₂<l₃<l₁  

Q6. lim_x→2⁡ =(√(1+√(2+x)) -√3)/(x-2) is equal to

•  1/(8√3)
•  1/√3
•  a²8√3
•  √3

Solution
lim_x→2 =(√(1+√(2+x)) -√3)/(x-2) =lim_x→2⁡ ((1+√(2+x)-3))/((x-2)(√(1+√(2+x)) +√3)) =lim_x→2 (x-2)/((x-2)(√(1+√(2+x)) +√3)(√(2+x)+2)) =lim_x→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=lim_θ→0⁡ (1-cos⁡2θ)/θ² , then the value of ∑ⁿ_r=0 ⁿ C_r a^r b^n-r, is

•  2n
•  3ⁿ
•  2ⁿ⁺¹
•  2^n-1

Solution
We have, x²+4x+5=(x+2)²+1≥1 for all x ∴a=1 b=lim_θ→0)⁡ (1-cos²⁡θ)/θ² =1 ∴∑ⁿ_r=0 ⁿ C_r a^r b^n-r=(a+b)ⁿ=2ⁿ

Q8.If lim_x→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, lim_x→∞{(x³+1)/(x²+1)-(ax+b) }=2 ⇒lim_x→∞⁡ (x³ (1-a)-bx²-ax+(1-b))/(x²+1)=2 ⇒1-a=0 and -b=2⇒a=1,b=-2

Q9.The value of lim_n→∞⁡ (√(n²+1)+√n)/(∜(n³+n-) ∜n), is

•  0
•  1
•  -1
•  None of these

Solution
We have, lim_n→∞⁡ (√(n²+1)+√n)/(∜(n³+n-) ∜n) =lim_n→∞⁡ n{√(1+1/n² )+1/√n}/(n^3/4 {∜(1+1/n² )-1/√n} ) =lim_n→∞ n^1/4 {√(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 lim_x→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, lim_x→a⁡ (f(a)g(x)-f(a)-g(a)f(x)+g(a))/(g(x)-f(x) )=4 Applying L’ Hospital’s rule, ⇒ lim_x→a⁡ (f(a)g'(x)-g(a)f'(x))/(g'(x)-f'(x))=4 ⇒ lim_x→a⁡ (kg^'(x) -kf'(x))/(g'(x)-f'(x))=4 ⇒ k=4