LIMITS AND DERIVATIVES quiz-13

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. limx→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 limx→0 (x{1-x2/2!+x4/4!-x6/6!…}-{x-x2/2+x3/3-x4/4…})/x2 =limx→0⁡(1/2-x/2!-x/3+⋯)=1/2

Q2.If f(x) is differentiable and strictly increasing function, then the value of limx→0⁡ (f(x2 )-f(x))/(f(x)-f(0)) is
•  1
•  0
•  -1
•  2
Solution
Here, limx→0⁡ (f(x2 )-f(x))/(f(x)-f(0)) =limx→0)⁡(f' (x2 ).2x-f' (x))/(f' (x) ) =(-f' (0))/(f' (0))=-1

Q3. The value of limn→∞⁡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⁡ 2n-1 A=sin⁡ 2n A/(2n sin⁡A ) limn→∞ ⁡cos⁡ (x/2) cos⁡ (x/4)…cos⁡ (x/2n-1 ) cos⁡(x/2n ) =limn→∞ sin⁡x/(2n sin⁡ (x/2n) ) [put A=x/2n ] =limn→∞ sin⁡x/x.((x/2n))/sin⁡ (x/2n) =sin⁡x/x

Q4. The value of limx→1⁡ (log2⁡2x )logx⁡5 is
•  5/2
•  elog2⁡5
•  log⁡5/log⁡2
•  elog5⁡2
Solution
We have, limx→1⁡ (log2⁡2x )logx⁡5 =limx→1⁡ (log2⁡2+log2⁡x )logx⁡5 =limx→1⁡ (1+log2⁡x )1/log5⁡x =e(limx→1⁡log2⁡x .1/log5⁡x ) =elimx→1⁡log2⁡5 =elog2⁡5

Q5.The value of limx→7⁡ (2-√(x-3))/(x2-49) is
•  2/9
•  -2/49
•  1/64
•  -1/56
Solution
limx→7⁡ (2-√(x-3))/(x2-49)=limx→7⁡ (-1/(2√(x-3)))/2x =-1/4.7.2=-1/56

Q6. limx→∞⁡ (∫2x0 xe(x2 dx)/e4x2 equals
•  0
•
•  2
•  1/2
Solution
limx→∞⁡ (∫2x0 xex2 dx)/e4x2 =limx→∞⁡ (∫2x0 ex2 dx2 )/(2e4x2 ) ) =limx→∞⁡ ([e(x2 ) ]2x0/(2e4x2 ) =limx→∞⁡ (e4x2 -1)/(2e4x2 ) =limx→∞⁡ (1/2-1/e4x2 )=1/2

Q7.If f(x)=sin⁡(ex-2-1)/log⁡(x-1) , then limx→2⁡ f(x) is given by
•  -2
•  -1
•  0
•  1
Solution
We have, ⇒limx→2>⁡ f(x)=limx→2⁡ sin⁡ ex-2-1)/log⁡ (x-1) ⇒limx→2⁡ f(x)=limx→2⁡{sin⁡(ex-2-1)/(ex-2-1).(ex-2-1)/(x-2).(x-2)/log⁡(1+(x-2) ) } ⇒limx→2⁡ f(x)=1×1×1=1

Q8. The value of limx→∞⁡√(a2 x2+ax+1)-√(a2 x2+1), is
•  1/2
•  1
•  2
•  None of these
Solution
We have, limx→∞⁡√(a2 x2+ax+1)-√(a2 x2+1) =limx→∞⁡ ax/(√(a2 x2+ax+1)+√(a2 x2+1)) =limx→∞⁡ a/(√(a2+a/x+1/x2 )+√(a2+1/x2 ))=a/(2 a)=1/2

Q9.The value of limx→1⁡(log5⁡5x )logx⁡5 is
•  1
•  e
•  -1
•  None of these
Solution
We have, limx→1⁡(log4⁡5x )logx⁡5 =limx→1⁡(log5⁡5+log5⁡x )logx⁡5 =limx→1⁡ (1+log5⁡x )(1/log5⁡x ) =elimx→1⁡ log5⁡x.1/logx⁡5 =e1=e

Q10. The value of limx→0⁡ (ex+log⁡ (1+x)-(1-x)-2 )/x2 is equal to
•  0
•  -3
•  -1
• Infinity
Solution
limx→0⁡ (ex+log⁡ (1+x)-(1-x)-2 )/x2 [0/0 form] =limx→0⁡ (ex+(1+x)-1⁡ -2(1-x)-3 )/2x [by L’ Hospital’s rule] =limx→0 ⁡(ex-(1+x)-2 -6(1-x)-4 )/2 [by L’ Hospital’s rule] =(e0-1-6)/2=-3

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