QUIZ-16 LIMITS AND DERIVATIVES

LIMITS AND DERIVATIVES quiz-16

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. The value of lim_x→0⁡ (a^x-b^x)/x , is

•  log⁡(a/b)
•  log⁡(b/a)
•  log⁡ (ab)
•  -log⁡ (ab)

Solution
We have, lim_x→0⁡ (a^x-b^x)/x =lim_x→0⁡{((a^x-1)/x)-((b^x-1)/x) }=log⁡(a)-log⁡(b)=log⁡(a/b)

Q2.Let f:R→R be a differentiable function such that f(2)=2. Then, the value of lim_x→2⁡ ∫^(f(x))₂ (4 t³)/(x-2) dt, is

•  6f^' (2)
•  12f^' (2)
•  32f^' (2)
•  None of these

Solution
We have, lim_x→2⁡ ∫^(f(x))₂ (4 t³)/(x-2) dt =lim_x→2⁡ [t⁴ ]₂ ^(f(x))/(x-2)=lim_x→2⁡ ({f(x)}⁴-16)/(x-2) =lim_x→2⁡ (4{f(x)}³ f'(x))/1 [Applying L’ Hospital’s Rule] =4(f(2) )³ f^' (2)=32f^' (2)

Q3. The value of lim_x→0⁡ (1+sin⁡x-cos⁡x+log⁡ (1-x) )/x³ , is

•   1/2
•  -1/2
•  0
•  1

Solution
We have, lim_x→0⁡ (1+sin⁡x-cos⁡x+log⁡ (1-x) )/x³ =lim_x→0⁡(1+(x-x³/3!+x5/5!-…)-(1-x²/2!+x⁴/4!-x⁶/6!…)+(-x-x²/2-x³/3-x⁴/4))/x³ =lim_x→0 ⁡(-1/3!-1/3)+x² (1/5!-1/5)+⋯=-1/6-1/3=-1/2

Q4. lim_x→1⁡cos^-1⁡((1-√x)/(1-x)) is equal to

•  π/3
•  π/6
•  π/2
•  π/4

Solution
lim_x→1⁡cos^-1⁡((1-√x)/(1-x)) =lim_x→1⁡cos^-1⁡((1-√x)/(1-√x)(1+√x) ) =lim_x→1⁡cos^-1⁡(1/(1+√x)) =cos^-1⁡(1/2) =π/3

Q5.If lim_x→∞ [(x³+1)/(x²+1)-(ax+b)]=2 , then

•  a=1 and b=1
•  a=1 and b=-1
•  a=1 and b=-2
•  a=1 and b=2

Solution
Given, lim_x→∞⁡ [(x³+1)/(x²+1)-(ax+b)]=2 ⇒ lim_x→∞⁡ [(x(1-a)-b-a/x+((1-b))/x₂ )/(1+1/x² )]=2 This limit will exist, if 1-a=0 and b=-2 ⇒ a=1 and b=-2  

Q6. If lim_x→1⁡ (ax²+bx+c)/(x-1)² =2, then (a, b, c) is

•  (2, -4, 2)
•  (2, 4, 2)
•  (2, 4, -2)
•  (2, -4, -2)

Solution
Given, lim_x→1⁡ (ax²+bx+c)/(x-1)² =2 This limit will exist, if ax²+bx+c=2(x-1)² ⇒ ax²+bx+c=2x²-4x+2 ⇒ a=2,b=-4, c=2

Q7.The value of lim_n→∞⁡ xⁿ/(xⁿ+1), where x<-1 is="" span="">

•  1/2
•  -1/2
•  1
•  None of these

Solution
lim_n→∞ xⁿ/(xⁿ+1)=lim_n→∞⁡ xⁿ/ (1+1/xⁿ)xⁿ =lim_n→∞ 1/(1+1/xⁿ ) =1

Q8. The value of lim_x→0⁡ sin⁡x/√(x² ), is

•  1
•  -1
•  0
•  None of these

Solution
We have, lim_x→0⁡ sin⁡x/√(x² )=lim_x→0⁡ sin⁡x/(|x|) Now, lim_x→0^- ⁡ sin⁡x/(|x|) =lim_x→0⁡ sin⁡x/(x|)=-1 and, lim_x→0⁺ ⁡ sin⁡x/(|x|) =lim_x→0⁡ sin⁡x/x =1 Hence, lim_x→0⁡ sin⁡x/(|x|) does not exist

Q9.The value of lim_h→0⁡ ln⁡ (1+2h)-2 ln⁡ (1+h) /h² is

•  1
•  -1
•  0
•  None of these

Solution
We have, lim_h→0 log⁡(1+2h)-2 log⁡(1+h) /h² =lim_h→0⁡ log⁡{(1+2h)/(1+h)² }/h² =-lim_h→0⁡log⁡{(1+h)²/(1+2h)}/h² =-lim_h→0⁡ log⁡{1+h²/(1+2h)}/({h²/(1+2h)}(1+2h)) =-1=-1

Q10. lim_x→∞⁡((2x-3)(3x-4))/((4x-5)(5x-6)) is equal to

•  1/10
•  0
•  1/5
• 3/10

Solution
lim_x→∞ ((2x-3)(3x-4))/((4x-5)(5x-6)) =lim_x→∞⁡(6x²-17x+12)/(20x²-49x+30) =lim_x→∞⁡ (12x-17)/(40x-49) [using L’ Hospital’s rule] =lim_x→∞⁡ 12/40=3/10 [using L’ Hospital’s rule]