Assertion (A): f(x)={cases} 3x-8, & x ≤ 5 \\ ...

CBSE Class 12 Math 2025 All Sets Solved Paper

Question : 19 of 20

Marks: +1, -0

Assertion (A):

f(x)=\begin{cases} 3x-8, & x \le 5 \\ 2k, & x>5 \end{cases}

is continuous at x=5 for

k = \frac{52}{.}

Reason (R): For a function f to be continuous at

x = a, \lim\limits_{x \to a^{-}} f(x) = \lim\limits_{x \to a^{+}} f(x) = f(a).

Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.

Solution:

We have,

f(x)=\begin{cases} 3x-8, & x \le 5 \\ 2k, & x>5 \end{cases}

since, f(x) is continuous at x = 5

\therefore \lim\limits_{x \to 5^{-}} f(x) = \lim\limits_{x \to 5^{+}} f(x) = f(5)

Now, LHL

= \lim\limits_{x \to 5^{-}} (3x-8) = \lim\limits_{h \to 0} [3(5-h)-8]

= 15 - 8

= 7

\text{RHL} = \lim\limits_{x \to 5^{+}} 2k

= \lim\limits_{h \to 0} 2k = 2k

Also,

f(5) = 3(5) - 8 = 7

\therefore 2k = 7

k = \frac{7}{2}

Reason is correct as for continuity at

x = a, \lim\limits_{x \to a^{+}} f(x) = \lim\limits_{x \to a^{+}} f(x) = f(a)

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