Continuity And Differentiability

The correct option is (B) : 10


By adding this …...

The given functional equation is:

Step 1: Substitute Specific Values of
For :

For :

For :

Step 2: Solve the System of Equations
Add equations (1), (2), and (3):

Simplify:

Divide through by 2:

Substitute (from equation (2)) into equation (4):

Final Result: .

The correct answer is 2039
Let fog(x)=h(x)

⇒h−1(x)=(2x−7​)31​

⇒h(x)= fog (x)=2x3+7
fog (x)=a(xb+c)−3

⇒a=2,b=3,c=5

⇒fog(ac)=fog(10)=2007
g(f(x)=(2x−3)3+5
⇒gof(b)=gof(3)=32
⇒sum=2039

The correct answer is : 50

Number of points of non-differentiability for 4+[13 sinx] is 4 12+2=50 (by graph)

Step 1: Rewrite the given differential equation:2(y+2)loge​(y+2)dx+(x+4−2loge​(y+2))dy=0

Step 2: Separate the variables:x+4−2loge​(y+2)2(y+2)loge​(y+2)​dx=−dy

Step 3: Integrate both sides:∫x+4−2loge​(y+2)2(y+2)loge​(y+2)​dx=−∫dy

Step 4: Use the initial condition x(e⁴−2)=1 to find the constant of integration.

Step 5: Solve for the particular solution using the initial condition.

Step 6: Evaluate the particular solution at y=e⁹−2 to find x(e⁹−2).

Final Answer: C.

Given:

Step 1: Differentiate both sides with respect to x:

Using Leibniz rule:

Simplify:

Let :

Step 2: Rewrite the equation:

This is a first-order linear differential equation. The integrating factor (I.F.) is:

Step 3: Solve the differential equation:

Multiply through by :

Step 4: Apply Initial Condition: Given , substitute and :

Thus:

Step 5: Find :

Step 6: Find :

Final Answer: .

Step 1: Using the given equation. We are given that is a root of the equation: Substitute into this equation: Step 2: Substituting into the function . The function is given as: Substitute from equation (1): Simplify the expression inside the cosine: Expanding this gives: Now the function becomes: Step 3: Apply L'Hôpital's Rule.
Since the expression is in the form as , we can apply L'Hôpital's Rule. First, differentiate the numerator and denominator with respect to .
Numerator: Denominator: Thus, the limit becomes: Step 4: Simplifying the limit.
At , the numerator becomes and the denominator also becomes , so we apply L'Hôpital's Rule again. The limit simplifies to: Step 5: Conclusion.
Thus, the correct value of the limit is , and the correct answer is option (3).

To determine , we need to ensure is continuous at . For continuity, we have:

1. Left-hand limit: As , we analyze . Near , and approach zero. However, for (where is an integer), a suitable expansion leads to: Matching terms gives as . Thus, the left-hand limit is .

2. At :

3. Right-hand limit: As , consider with approaching zero, so

Continuity at implies . Solving , we find .

For , continuity requires ; since , any integer satisfies this.

Thus, allow smallest solution .

Then, .

Verification: lies within the given range Therefore,

To ensure continuity of at , the following condition must hold:

Step 1: Evaluating the right-hand limit
We consider:

Step 2: Rationalizing the numerator
Multiply the numerator and denominator by the conjugate of the numerator:

Step 3: Simplifying the expression
This becomes:

Step 4: Canceling and simplifying further

Step 5: Substituting

Step 6: Applying the continuity condition
Since the limit must equal , we have:

To ensure continuity at , we require .

Left-hand limit:

This gives .

Right-hand limit:

Calculating :

To determine the number of points where the function is not continuous, we analyze the potential discontinuities caused by the greatest integer function, denoted by . Discontinuities for such expressions typically occur at integer points where the floor function or transitions.

Let's first identify the key transition points within the domain :

1. The points where changes are integer values: .
2. For , consider where is an integer: within the domain , these points are , corresponding to . However, since is the domain, feasible values are .

By evaluating the potential discontinuities at these points:

1. : Check left and right hand values for continuity: , . Moving across within the domain's scope shows a discontinuity due to the piecewise nature of and .
2. : The function becomes at both points. Hence, potential discontinuity due to a jumping point in .
3. : Both and change, further implying a discontinuity.
4. : The transition of at an edge of the domain indicates potential discontinuity.

Summing these, the points of discontinuity are , leading to a total of 4 points where the function is not continuous.

Step 1: Evaluate the limit w.r.t.
Consider the factor as . Case I:
Then , so Case II:
Then . Dividing numerator and denominator by , Case III:
Left-hand limit at : Right-hand limit at : Since LHL RHL, Hence, Statement 1 is false
. Case IV:
Here . Left and right limits are same (finite), hence is continuous at
. But since the functional form changes across , the limit definition fails to match smoothly. Thus, Statement 2 is also false
.
Final Conclusion:
Both statements are false.