## LIMITS AND DERIVATIVES quiz-14

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 xn+1=√(2+xn ),n≥1, then limn→∞⁡ xn is equal to
•   -1
•  2
•  √5
•  3
Solution
We have, x1=3,xn+1=√(2+xn ) ∴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,…xn,… is a monotonically decreasing sequence bounded below by 2. So, it is convergent. Let lim⁡ xn =x. Then, xn+1=√(2+xn ) ⇒lim⁡ xn+1 =√(2+lim⁡xn ) ⇒x=√(2+x) ⇒x2-x-2=0 ⇒(x-2)(x+1)=0 ⇒x=2 [∵xn>0 for all n∈N ∴x>0]

Q2.limx→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=limh→0 f(1+h) =limh→0 √(1-cos⁡2h )/h =limh→0 (√2 sin⁡h)/h=√2 LHL=limh→0 f(1-h) limh→0 √(1-cos⁡ (-2h) )/h =limh→0 √2 sin⁡h/(-h)=-√2 Here, LHL≠RHL So, limit does not exist

Q3. limx→0⁡ sin⁡ |x| /x is equal to
•   1
•  0
•  positive infinity
•  does not exist
Solution
LHL=limx→0⁡ (-sin⁡x)/x =-1 RHL=limx→0 sin⁡x/x =1 ⇒ LHL≠RHL ⇒ limx→0⁡ sin⁡ |x| /x Does not exist

Q4. limx→Ï€/6⁡[(3 sin⁡ x-√3 cos⁡x )/(6x-Ï€)]
•  √3
•  1/√3
•  -1/√3
•  -1/3
Solution
limx→Ï€/6 (3 sin⁡ x-√3 cos⁡x )/(6x-Ï€)=limx→Ï€/6⁡ (3 cos⁡ x+√3 cos⁡x )/6 =(3 cos⁡ Ï€/6+√3 sin⁡ Ï€/6 )/6 =(3.√3/2+√3/2)/6 =1/√3

Q5.limx→∞⁡(x3/(3x2-4)-x2/(3x+2)) is equal to
•  -1/4
•  -1/2
•  0
•  2/9
Solution
limx→∞⁡(x3/(3x2-4)-x2/(3x+2)) =limx→∞ (x3 (3x+2)-x2 (3x2-4))/((3x2-4)(3x+2)) =limx→∞ (2x3+4x2)/(9x3+6x2-12x-8) =limx→∞ (2+4/x)/(9+6/x-12/x2-8/x3 )=2/9

Q6. The value of the constant Î± and Î² such that limx→∞⁡((x2+1)/(x+1)-Î±x-Î²)=0 are respectively
•  (1, 1)
•  (-1,1)
• (1,-1)
•  (0,1)
Solution
Given, limx→∞⁡((x2+1)/(x+1)-Î±x-Î²)=0 ⇒ limx→∞⁡((x2+1-Î±(x2+x)-Î²(x+1))/(x+1))=0 ⇒ limx→∞⁡((2x-Î±(2x+1)-Î²(1))/1)=0 [by L’ Hospital’s rule] If this limit is zero, then the function 2x-Î±(2x+1)-Î²=0 or x(2-2Î±)-(Î±+Î²)=0 Equating the coefficient of x and constant terms, we get 2-2Î±=0 and Î±+Î²=0 ⇒ Î±=1, Î²=-1

Q7.The value of limx→0⁡((∫(x2)0 sec2⁡ t dt )/(x sin⁡x )) is
•  3
•  2
•  1
•  0
Solution
limx→0⁡((∫(x2)0 sec2⁡ t dt )/(x sin⁡x ))=limx→0⁡ (sec2 x2.2x)/(sin⁡x+x cos⁡x ) =limx→0⁡ ( 2x.sec 2 x2)/x(sin⁡x/x+cos⁡x ) =(2×1)/(1+1)=1[∵ limx→0⁡ sin⁡x/x=1 ]

Q8. The value of limx→2⁡ (x(5x-1))/(1-cos⁡x ), is
•  5 log⁡2
•  2 log⁡5
•  1/2 log⁡5
•  1/5 log⁡2
Solution
We have, limx→2⁡ (x(5x-1))/(1-cos⁡x ) =limx→0⁡ (((5x-1)/x))/((1-cos⁡x)/x2 )=2log⁡5

Q9.The value of limx→∞⁡ {log(n-1) )⁡n. logn⁡(n+1).log(n+1)⁡(n+2)…{…log(nk-1)⁡ (nk)}, is
•
•  n
•  k
•  None of these
Solution
We have, logb⁡a×logc⁡b=logc⁡a ∴limx→∞⁡ {log(n-1)⁡n. logn⁡(n+1).log(n+1)⁡ (n+2)…log(nk-1) ⁡(nk )} =limn→∞⁡{log(n-1)⁡ nk } =limn→∞⁡ loge⁡ nk /loge⁡ (n-1) =k limn→∞⁡ loge⁡ n/loge⁡ (n-1) =k limn→∞⁡ (1/n)/(1/n-1) [Using L’ Hospital’s Rule] =k limn→∞⁡ (n-1)/n=k

Q10. If f(x)=cot-1⁡[(3x-x3)/(1-3x2)] and g(x)=cos-1⁡ [(1-x2)/(1+x2)], then limx→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 ))=Ï€/2-3tan-1⁡x and g(x)=cos-1⁡((1-x2)/(1+x2 ))=2tan-1⁡x ∴ limx→a⁡ (f(x)-f(a))/(g(x)-g(a)) =limx→a ( Ï€/2-3 tan-1⁡ x-Ï€/2+3 tan-1⁡a )/(2 tan-1⁡ x-2 tan-1⁡a ) =-3/2 limx→a tan-1⁡ x-tan-1⁡a /tan-1⁡x-tan-1⁡a =-3/2

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