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Assertion (A): f(x) = {(3โข๐‘ฅโˆ’ โข8, ๐‘ฅ โ‰ค 5),(2โข๐‘˜, ๐‘ฅ>5) is continuous at x = 5 for k = 5/2. Reason (R): For a function f to be continuous at x = a, lim_(๐‘ฅโ†’๐‘Žโˆ’)โก ๐‘“โก(๐‘ฅ) = lim_(๐‘ฅโ†’๐‘Ž+) โก๐‘“โก(๐‘ฅ) - Mathematics

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Question

Assertion (A): f(x) = `{(3x -  8",",  x ≤ 5), (2k",",  x > 5):}` is continuous at x = 5 for k = `5/2`.

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

`lim_(x -> a^-)f(x) = lim_(x -> a^+) f(x) = f(a)`

Options

  • 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.

MCQ
Assertion and Reasoning
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Solution

Assertion (A) is false, but Reason (R) is true.

Explanation:

We have, `{(3x -  8",",  x ≤ 5), (2k",",  x > 5):}` 

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

L.H.L. = R.H.L. = f(5)

Now, L.H.L. = `lim_(x -> 5^-)f(3x - 8)`

= `lim_(h -> 0)[3(5 - h) - 8]`

= 3 × 5 × 8

= 15 − 8

= 7

R.H.L. = `lim_(x -> 5^+) 2k`

`lim_(h -> 0) (2k) = 2k`

Also, f(5) = 3(5) − 8

∴ 2k = 7

k = `7/2`

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2024-2025 (March) Delhi Set 1
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