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Q30:
Find the values of $a$ and $b$ such that the function defined by $f(x) = \begin{cases} 5, & \text{if } x \le 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \ge 10 \end{cases}$ is a continuous function.

Solution :

Step 1: Identifying Continuity Criteria

For a function $f(x)$ to be continuous on its entire domain $(-\infty, \infty)$, it must be continuous at every point in the domain. Specifically, it must be continuous at the transition points $x = 2$ and $x = 10$.

A function $f(x)$ is continuous at $x = c$ if and only if:

$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$ [Definition of continuity at a point]

Step 2: Analysis at $x = 2$

For $f(x)$ to be continuous at $x = 2$:

  • Left-hand limit (LHL): $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (5) = 5$
  • Right-hand limit (RHL): $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax + b) = 2a + b$
  • Value of the function: $f(2) = 5$

Equating LHL and RHL [By definition of continuity]:

$2a + b = 5$ --- (Equation 1)

Step 3: Analysis at $x = 10$

For $f(x)$ to be continuous at $x = 10$:

  • Left-hand limit (LHL): $\lim_{x \to 10^-} f(x) = \lim_{x \to 10^-} (ax + b) = 10a + b$
  • Right-hand limit (RHL): $\lim_{x \to 10^+} f(x) = \lim_{x \to 10^+} (21) = 21$
  • Value of the function: $f(10) = 21$

Equating LHL and RHL [By definition of continuity]:

$10a + b = 21$ --- (Equation 2)

Step 4: Solving the Simultaneous Linear Equations

We have the system of equations:

(1) $2a + b = 5$
(2) $10a + b = 21$

Subtract (1) from (2) [By algebraic elimination method]:

$(10a + b) - (2a + b) = 21 - 5$

$8a = 16$

$a = 2$

Substitute $a = 2$ into Equation (1):

$2(2) + b = 5$

$4 + b = 5$

$b = 1$

Step 5: Visualization of the Piecewise Function

x-axis y-axis

Final Solution: The values for the constants are $a = 2$ and $b = 1$.


More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.1


CBSE Solutions for Class 12 Mathematics Continuity and Differentiability


Chapters in CBSE - Class 12 Mathematics


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