CA Foundation 2026 · Paper 3 · Quantitative AptitudeQ67 · 1 markOfficial key verified
The function f(x)=x2−25x−5f(x)=\dfrac{x^2-25}{x-5} is undefined at x=5x=5, what value must be assigned to f(5)f(5) if f(x)f(x) is to be continuous at x=5x=5?
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x2−25x−5=x+5\dfrac{x^2-25}{x-5}=x+5 for x≠5x\ne5; the limit at 5 is 1010, so set f(5)=10f(5)=10.

Step 1 — factor the numerator

x2−25=(x−5)(x+5)x^2-25=(x-5)(x+5), so for x≠5x\ne5

f(x)=(x−5)(x+5)x−5=x+5.f(x)=\frac{(x-5)(x+5)}{x-5}=x+5.

Step 2 — take the limit at x=5x=5

lim⁡x→5f(x)=5+5=10.\lim_{x\to5}f(x)=5+5=10.

Step 3 — condition for continuity

For ff to be continuous at x=5x=5, the defined value must equal the limit:

f(5)=10.f(5)=10. …

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