QUIZ-14 LIMITS AND DERIVATIVES

LIMITS AND DERIVATIVES quiz-14

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. If x₁=3 and x_n+1=√(2+x_n ),n≥1, then lim_n→∞⁡ x_n is equal to

•   -1
•  2
•  √5
•  3

Solution
We have, x₁=3,x_n+1=√(2+x_n ) ∴x₂=√(2+x₁ )=√(2+3)=√5,x₃=√(2+x₂ )=√(2+√5) ∴x₁>x₂>x₃ It can be easily shown by mathematical induction that the sequence x₁,x₂,…x_n,… is a monotonically decreasing sequence bounded below by 2. So, it is convergent. Let lim⁡ x_n =x. Then, x_n+1=√(2+x_n ) ⇒lim⁡ x_n+1 =√(2+lim⁡x_n ) ⇒x=√(2+x) ⇒x²-x-2=0 ⇒(x-2)(x+1)=0 ⇒x=2 [∵x_n>0 for all n∈N ∴x>0]

Q2.lim_x→1⁡ √(1-cos⁡ 2(x-1) )/(x-1)

•  Exists and is equals √2
•  Exists and is equals -√2
•  Does not exist because x-1→0
•  Does not exist because left hand limit is not equal to right hand limit

Solution
RHL=lim_h→0 f(1+h) =lim_h→0 √(1-cos⁡2h )/h =lim_h→0 (√2 sin⁡h)/h=√2 LHL=lim_h→0 f(1-h) lim_h→0 √(1-cos⁡ (-2h) )/h =lim_h→0 √2 sin⁡h/(-h)=-√2 Here, LHL≠RHL So, limit does not exist

Q3. lim_x→0⁡ sin⁡ |x| /x is equal to

•   1
•  0
•  positive infinity
•  does not exist

Solution
LHL=lim_x→0⁡ (-sin⁡x)/x =-1 RHL=lim_x→0 sin⁡x/x =1 ⇒ LHL≠RHL ⇒ lim_x→0⁡ sin⁡ |x| /x Does not exist

Q4. lim_x→π/6⁡[(3 sin⁡ x-√3 cos⁡x )/(6x-π)]

•  √3
•  1/√3
•  -1/√3
•  -1/3

Solution
lim_x→π/6 (3 sin⁡ x-√3 cos⁡x )/(6x-π)=lim_x→π/6⁡ (3 cos⁡ x+√3 cos⁡x )/6 =(3 cos⁡ π/6+√3 sin⁡ π/6 )/6 =(3.√3/2+√3/2)/6 =1/√3

Q5.lim_x→∞⁡(x³/(3x²-4)-x²/(3x+2)) is equal to

•  -1/4
•  -1/2
•  0
•  2/9

Solution
lim_x→∞⁡(x³/(3x²-4)-x²/(3x+2)) =lim_x→∞ (x³ (3x+2)-x² (3x²-4))/((3x²-4)(3x+2)) =lim_x→∞ (2x³+4x²)/(9x³+6x²-12x-8) =lim_x→∞ (2+4/x)/(9+6/x-12/x²-8/x³ )=2/9  

Q6. The value of the constant α and β such that lim_x→∞⁡((x²+1)/(x+1)-αx-β)=0 are respectively

•  (1, 1)
•  (-1,1)
• (1,-1)
•  (0,1)

Solution
Given, lim_x→∞⁡((x²+1)/(x+1)-αx-β)=0 ⇒ lim_x→∞⁡((x²+1-α(x²+x)-β(x+1))/(x+1))=0 ⇒ lim_x→∞⁡((2x-α(2x+1)-β(1))/1)=0 [by L’ Hospital’s rule] If this limit is zero, then the function 2x-α(2x+1)-β=0 or x(2-2α)-(α+β)=0 Equating the coefficient of x and constant terms, we get 2-2α=0 and α+β=0 ⇒ α=1, β=-1

Q7.The value of lim_x→0⁡((∫^(x2)₀ sec²⁡ t dt )/(x sin⁡x )) is

•  3
•  2
•  1
•  0

Solution
lim_x→0⁡((∫^(x2)₀ sec²⁡ t dt )/(x sin⁡x ))=lim_x→0⁡ (sec² x².2x)/(sin⁡x+x cos⁡x ) =lim_x→0⁡ ( 2x.sec ² x²)/x(sin⁡x/x+cos⁡x ) =(2×1)/(1+1)=1[∵ lim_x→0⁡ sin⁡x/x=1 ]

Q8. The value of lim_x→2⁡ (x(5^x-1))/(1-cos⁡x ), is

•  5 log⁡2
•  2 log⁡5
•  1/2 log⁡5
•  1/5 log⁡2

Solution
We have, lim_x→2⁡ (x(5^x-1))/(1-cos⁡x ) =lim_x→0⁡ (((5^x-1)/x))/((1-cos⁡x)/x² )=2log⁡5

Q9.The value of lim_x→∞⁡ {log_(n-1) )⁡n. log_n⁡(n+1).log_(n+1)⁡(n+2)…{…log_(n^k-1)⁡ (n^k)}, is

•  ∞
•  n
•  k
•  None of these

Solution
We have, log_b⁡a×log_c⁡b=log_c⁡a ∴lim_x→∞⁡ {log_(n-1)⁡n. log_n⁡(n+1).log_(n+1)⁡ (n+2)…log_(n^k-1) ⁡(n^k )} =lim_n→∞⁡{log_(n-1)⁡ n^k } =lim_n→∞⁡ log_e⁡ n^k /log_e⁡ (n-1) =k lim_n→∞⁡ log_e⁡ n/log_e⁡ (n-1) =k lim_n→∞⁡ (1/n)/(1/n-1) [Using L’ Hospital’s Rule] =k lim_n→∞⁡ (n-1)/n=k

Q10. If f(x)=cot^-1⁡[(3x-x³)/(1-3x²)] and g(x)=cos^-1⁡ [(1-x²)/(1+x²)], then lim_x→a⁡ (f(x)-f(a))/(g(x)-g(a)) (0<a<1/2) is

•  -3/2
•  1/2
•  3/2
• None of these

Solution
f(x)=cot^-1⁡((3x-x³)/(1-3x² ))=π/2-3tan^-1⁡x and g(x)=cos^-1⁡((1-x²)/(1+x² ))=2tan^-1⁡x ∴ lim_x→a⁡ (f(x)-f(a))/(g(x)-g(a)) =lim_x→a ( π/2-3 tan^-1⁡ x-π/2+3 tan^-1⁡a )/(2 tan^-1⁡ x-2 tan^-1⁡a ) =-3/2 lim_x→a tan^-1⁡ x-tan^-1⁡a /tan^-1⁡x-tan^-1⁡a =-3/2