QUIZ-13 LIMITS AND DERIVATIVES

LIMITS AND DERIVATIVES quiz-13

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→0 (x cos⁡x-log⁡(1+x))/x² equals

•  1/2
•  0
•  1
•  -1/2

Solution
Using expressions of cos⁡x and log⁡ (1+x), the given limit is equal to lim_x→0 (x{1-x²/2!+x⁴/4!-x⁶/6!…}-{x-x²/2+x³/3-x⁴/4…})/x² =lim_x→0⁡(1/2-x/2!-x/3+⋯)=1/2

Q2.If f(x) is differentiable and strictly increasing function, then the value of lim_x→0⁡ (f(x² )-f(x))/(f(x)-f(0)) is

•  1
•  0
•  -1
•  2

Solution
Here, lim_x→0⁡ (f(x² )-f(x))/(f(x)-f(0)) =lim_x→0)⁡(f^' (x² ).2x-f^' (x))/(f^' (x) ) =(-f^' (0))/(f^' (0))=-1

Q3. The value of lim_n→∞⁡cos⁡ (x/2) cos⁡ (x/4) cos⁡ (x/8)…cos⁡(x/2ⁿ ) is

•   x/sin⁡x
•  x/cos⁡x
•  (sin⁡x)/x
•  (cos⁡x)/x

Solution
We know that cos⁡ A cos⁡ 2A cos⁡4A …cos⁡ 2^n-1 A=sin⁡ 2ⁿ A/(2ⁿ sin⁡A ) lim_n→∞ ⁡cos⁡ (x/2) cos⁡ (x/4)…cos⁡ (x/2^n-1 ) cos⁡(x/2ⁿ ) =lim_n→∞ sin⁡x/(2ⁿ sin⁡ (x/2ⁿ) ) [put A=x/2ⁿ ] =lim_n→∞ sin⁡x/x.((x/2ⁿ))/sin⁡ (x/2ⁿ) =sin⁡x/x

Q4. The value of lim_x→1⁡ (log₂⁡2x )^log_x⁡5 is

•  5/2
•  e^log₂⁡5
•  log⁡5/log⁡2
•  e^log₅⁡2

Solution
We have, lim_x→1⁡ (log₂⁡2x )^log_x⁡5 =lim_x→1⁡ (log₂⁡2+log₂⁡x )^log_x⁡5 =lim_x→1⁡ (1+log₂⁡x )^1/log_5⁡x =e^{(lim_x→1⁡log₂⁡x .1/log₅⁡x )} =e^{lim_x→1⁡log₂⁡5} =e^log₂⁡5

Q5.The value of lim_x→7⁡ (2-√(x-3))/(x²-49) is

•  2/9
•  -2/49
•  1/64
•  -1/56

Solution
lim_x→7⁡ (2-√(x-3))/(x²-49)=lim_x→7⁡ (-1/(2√(x-3)))/2x =-1/4.7.2=-1/56  

Q6. lim_x→∞⁡ (∫^2x₀ xe^(x2 dx)/e^4x2 equals

•  0
•  ∞
•  2
•  1/2

Solution
lim_x→∞⁡ (∫^2x₀ xe^x2 dx)/e^4x2 =lim_x→∞⁡ (∫^2x₀ e^x2 dx² )/(2e^4x2 ) ) =lim_x→∞⁡ ([e^(x2 ) ]^2x₀/(2e^4x2 ) =lim_x→∞⁡ (e^4x2 -1)/(2e^4x2 ) =lim_x→∞⁡ (1/2-1/e^4x2 )=1/2

Q7.If f(x)=sin⁡(e^x-2-1)/log⁡(x-1) , then lim_x→2⁡ f(x) is given by

•  -2
•  -1
•  0
•  1

Solution
We have, ⇒lim_x→2>⁡ f(x)=lim_x→2⁡ sin⁡ e^x-2-1)/log⁡ (x-1) ⇒lim_x→2⁡ f(x)=lim_x→2⁡{sin⁡(e^x-2-1)/(e^x-2-1).(e^x-2-1)/(x-2).(x-2)/log⁡(1+(x-2) ) } ⇒lim_x→2⁡ f(x)=1×1×1=1

Q8. The value of lim_x→∞⁡√(a² x²+ax+1)-√(a² x²+1), is

•  1/2
•  1
•  2
•  None of these

Solution
We have, lim_x→∞⁡√(a² x²+ax+1)-√(a² x²+1) =lim_x→∞⁡ ax/(√(a² x²+ax+1)+√(a² x²+1)) =lim_x→∞⁡ a/(√(a²+a/x+1/x² )+√(a²+1/x² ))=a/(2 a)=1/2

Q9.The value of lim_x→1⁡(log₅⁡5x )^log_x⁡5 is

•  1
•  e
•  -1
•  None of these

Solution
We have, lim_x→1⁡(log₄⁡5x )^log_x⁡5 =lim_x→1⁡(log₅⁡5+log₅⁡x )^log_x⁡5 =lim_x→1⁡ (1+log_5⁡x )^{(1/log₅⁡x )} =e^{lim_x→1⁡ log₅⁡x.1/log_x⁡5} =e¹=e

Q10. The value of lim_x→0⁡ (e^x+log⁡ (1+x)-(1-x)^-2 )/x² is equal to

•  0
•  -3
•  -1
• Infinity

Solution
lim_x→0⁡ (e^x+log⁡ (1+x)-(1-x)^-2 )/x² [0/0 form] =lim_x→0⁡ (e^x+(1+x)^-1⁡ -2(1-x)^-3 )/2x [by L’ Hospital’s rule] =lim_x→0 ⁡(e^x-(1+x)^-2 -6(1-x)^-4 )/2 [by L’ Hospital’s rule] =(e⁰-1-6)/2=-3