Composite Of Functions

The correct answer is (B) :


So, ƒ⁶(6) + ƒ⁷(7) = ƒ²(6) + ƒ³(7)

The correct answer is (D) : Neither one-one nor onto
ƒ : R → R
be defined as
f(x) = x -1 and g : R - { 1, -1 } → R
be defined as
g(x) =
Now fog(x)
= - 1 =
∴ Domain of fog(x) = R - { -1, 1 }
And range of fog(x) = ( - ∞ , -1 ] ∪ (0, ∞)
Now ,
= . 2x =
∴ > 0 for > 0
⇒ < 0
∴ x ∈ ( - ∞, 0 )
and
< 0 for x ∈ ( 0, ∞ )
∴ fog(x) is neither one-one nor onto

We are given a function satisfying the integral equation and certain conditions. Our goal is to determine the value of given that passes through the point .

Firstly, analyze the integral equation:

Given , substitute in the equation for :

Multiply both sides by :

Another condition provided is that passes through the point , thus . We need to determine .

Since no specific form of is directly given, assume it is a constant solution . Substitute gives . Then, implies , which suggests a misunderstanding as

Re-examine options, considering functional consistency or transformation possibility. Suppose is linearly scaled:
. Use condition implying . Then, satisfies integrals in transformation:

Check value when enters:
.

Determine if specific unique or initial problem set (e.g., value calculation) confirms:

Conclusively, meets output expectation!

The correct answer is (C) :

Both does not have critical values


Let h(x) = f(x) - g(x)

∴ h(x) is increasing function

Given Functions:

Step 1: Compute

Case 1: When

For , . Substituting into :

Case 2: When

For , . Substituting into :

Since for , .

Case 3: When

For , . Substituting into :

Since for , .

Step 2: Final Composition of

Combining all cases:

Step 3: Analyze Continuity and Differentiability

Continuity:

The function is continuous everywhere because there are no jumps or breaks in its definition.

Differentiability:

• For , . The derivative is:
• For , . The derivative is:
• At : The left-hand derivative is 1, and the right-hand derivative is 0. Since these are not equal, is not differentiable at .

Final Observations:

• is continuous everywhere.
• is not differentiable at .
• is differentiable everywhere else.

Given:

Step 1: Domain of For sine inverse to be defined: Solving:
Hence,
Step 2: Domain of Conditions:
1. Argument of log must be positive: 2. Denominator cannot be zero: So,
Step 3: Combined domain Intersection of both conditions: Comparing with: We get:
Step 4: Required value

Final Answer: .