77. If

f(x)=

ax²+b,    0≤x≤1

cx+sin(πx/2),    1≤x≤2

is continuous and differentiable at all points in the interval [0,2], and

f(2)=π/4,

then determine the correct option.

(A) a = π/16 and b = π/16 + 1

(B) b = π/16 + 1 and c = π/8

(C) a = π/8 and c = π/8

(D) a = π/8 and b = 3π/16

Continuity and Differentiability of a Piecewise Function

The present question is an excellent illustration of this concept. The function is defined by two different expressions over two intervals, and we are told that it is both continuous and differentiable throughout the interval [0,2]. This immediately tells us that both the function values and their derivatives must agree at the junction point x = 1. Along with the additional condition f(2)=π/4, these properties allow us to determine the unknown constants.

Correct Answer

Option (B)

Understanding the Concept

Whenever a function is defined using different formulas over different intervals, the point where the definition changes becomes extremely important. In this problem, the function changes at x = 1.

Since the function is continuous and differentiable everywhere, two conditions must hold at x = 1.

• The left-hand limit, right-hand limit, and function value must all be equal (continuity).
• The left-hand derivative and right-hand derivative must also be equal (differentiability).

These two equations, together with the given value of f(2), provide enough information to determine the unknown constants.

Step 1: Use the Given Condition f(2)=π/4

For x = 2, we use the second part of the function.

f(x)=cx+sin(πx/2)

Substituting x = 2,

f(2)=2c+sinπ

Since

sinπ=0,

we obtain

2c=π/4

Therefore,

c=π/8.

This immediately eliminates Options (A) and (D).

Step 2: Apply the Continuity Condition at x=1

The value from the left side is

a+b.

The value from the right side is

c+sin(π/2).

Since

sin(π/2)=1,

continuity gives

a+b=c+1.

Substituting

c=π/8,

we obtain

a+b=π/8+1.

Step 3: Apply the Differentiability Condition

Differentiate the first expression.

For

f(x)=ax²+b,

we obtain

f'(x)=2ax.

Hence,

Left derivative at x=1 = 2a.

Now differentiate the second expression.

For

f(x)=cx+sin(πx/2),

the derivative is

f'(x)=c+(π/2)cos(πx/2).

At x=1,

cos(π/2)=0.

Therefore,

Right derivative = c.

Differentiability requires

2a=c.

Since

c=π/8,

we obtain

a=π/16.

Step 4: Calculate b

Using the continuity equation,

a+b=π/8+1.

Substituting

a=π/16,

we obtain

b=π/8+1−π/16

=2π/16−π/16+1

=π/16+1.

Thus,

b=π/16+1.

Verification of the Solution

We have obtained

a=π/16

b=π/16+1

c=π/8.

Check continuity:

a+b=π/16+π/16+1

=π/8+1 ✔

Right side:

c+1=π/8+1 ✔

Check differentiability:

Left derivative:

2a=2×π/16=π/8 ✔

Right derivative:

c=π/8 ✔

Both conditions are satisfied perfectly.

Explanation of Every Option

Option (A)

The values of a and b are individually correct, but this option does not specify the required value of c. Since the question asks for the correct statement among the given options, it is incomplete compared with Option (B).

Option (B)

This option is correct because it gives

b=π/16+1

and

c=π/8,

both of which satisfy the continuity, differentiability, and given boundary condition.

Option (C)

This option is incorrect because

a=π/8

violates the differentiability condition

2a=c.

Here,

2a=π/4≠π/8.

Option (D)

This option is incorrect because neither the value of a nor the value of b satisfies the continuity and differentiability equations simultaneously.

Alternative Method

Instead of solving all three equations together, begin with the easiest condition, namely f(2)=π/4. This immediately gives the value of c. Next use differentiability to determine a, and finally substitute these values into the continuity equation to obtain b. This systematic order reduces algebraic errors and is particularly useful during competitive examinations.

Related Practice Example

Suppose

f(x)=

x²+k,  x≤1

2x,  x≥1

If the function is continuous at x=1, determine k.

From continuity,

1+k=2.

Hence,

k=1.

Such simple examples reinforce the same principles used in the present problem.

Key Takeaways

Whenever a piecewise function is continuous and differentiable, always remember the two essential conditions at the junction point. First, equate the function values to satisfy continuity. Second, equate the derivatives to satisfy differentiability. If additional conditions are given, such as the value of the function at another point, use them first whenever they simplify the calculation.

Final Answer

The required constants are

a=π/16,

b=π/16+1,

c=π/8.

Therefore, the correct statement is

Option (B).