Quiz-18 CONTINUITY AND DIFFERENTIABILITY

CONTINUITY AND DIFFERENTIABILITY-18

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 f(x)={ (3, x <0 ) (2x+1,x≥0), then

•  Both f(x) and f(|x|) are differentiable at x=0
•  f(x) is differentiable but f(|x|) is not differentiable at x=0
•  f(|x|) is differentiable but f(x) is not differentiable at x=0
•  Both f(x) and f(|x|) are not differentiable at x=0

Solution
We have, 

f(x)={ (3,x < 0) (2x+1,x≥0) 

Clearly, f is continuous but not differentiable at x=0 

Now, f(|x|)=2|x|+1 for all x 

Clearly, f(|x|) is everywhere continuous but not differentiable at x=0

Q2. Let f(x)=[2x³-5],[] denotes the greatest integer function. Then number of points (1, 2) where the function is discontinuous, is

•  0
•  13
•  10
•  3

Solution
Since ϕ(x)=2x³-5 is an increasing function on (1, 2) such that ϕ(1)=-3 and ϕ(2)=11 

Clearly, between -3 and 11 there are thirteen points where f(x)=[2x³-5] is discontinuous

Q3. A function f:R→R satisfies the equation f(x+y)=f(x)f(y) for all x,y∈R and f(x)≠0 for all x∈R. If f(x) is differentiable at x=0 and f' (0)=2, then f'(x) equals

•   f(x)
•  -f(x)
•  2f(x)
•  None of these

Solution

Q4. The set of points, where f(x)=x/(1+|x|) is differentiable, is

•  (-∞,-1)∪(-1,∞)
•  (-∞,∞)
•  (0,∞)
•  (-∞,0)∪(0,∞)

Solution
Let f(x)=(g(x))/(h(x))=x/(1+|x|) 

It is clear that g(x)=x and h(x)=1+|x| are differentiable on (-∞,∞) and (-∞,0)∪(0,∞) respectively 

Thus, f(x) is differentiable on (-∞,0)∪(0,∞).Now, we have to check the differentiable at x=0 

Q5. Consider f(x)={ (x²/(|x|),x≠0) (0, x=0)

•  f(x) is discontinuous everywhere
•  f(x) is continuous everywhere
•  f'(x) exists in (-1,1)
•  f'(x) exists in (-2,2)

Solution
 

⇒lim_(x→0)-f(x)=lim_(x→0)⁡ -x =0,lim_(x→0) + ⁡f(x)=lim_(x→0)⁡x=0 and f(0)=0 So, f(x) is continuous at x=0. Also, f(x) is continuous for all other values of x Hence, f(x) is everywhere continuous Clearly, Lf' (0)=-1 and Rf' (0)=1 Therefore, f(x) is not differentiable at x=0

Q6. A function f on R into itself is continuous at a point a in R, iff for each ∈>0, there exists, δ>0 such that

•  |f(x)-f(a)|<∈⇒|x-a|<δ
•  |f(x)-f(a) |>∈⇒|x-a|>δ
•  |x-a|>δ|f(x)-f(a) |>∈
•  |x-a|<δ|f(x)-f(a) |<∈

Solution
A function f on R into itself is continuous at a point a in R, iff for each ∈>0 there exist δ>0, such that |f(x)-f(a)| < ∈ ⇒ |x-a| < δ

Q7. If a function f(x) is given by f(x)=x/(1+x)+x/((x+1)(2x+1))+x/((2x+1)(3x+1))+⋯∞ then at x=0,f(x)

•  Has no limit
•  Is not continuous
•  Is continuous but not differentiable
•  Is differentiable

Solution

Q8. A function f(x) is defined as fallows for real x,

•  f(x), is not continuous at x=1
•  f(x) is continuous but not differentiable at x=1
•  f(x) is both continuous and differentiable at x=1
•  None of the above

Solution

Q9. f(x)={ (sin⁡3x/sin⁡x , x≠0) (k, x=0) is continuous, if k is

•  3
•  0
•  -3
•  -1

Solution

Q10. If a function f(x) is defined as f(x)={ (x/√(x²),x≠0) (0,x=0) then

•  f(x) is continuous at x=0 but not differentiable at x=0
•  f(x) is continuous as well as differentiable at x=0
•  f(x) is discontinuous at x=0
•  None of these

Solution