Question

Given \(f(x) = x^2 - 5x + 4\). Out of first 20 natural numbers, if a number \(x\) is chosen at random, then the probability that the chosen \(x\) satisfies the inequality \(f(x)>10\) is

• A. \(\dfrac{1}{2}\)
• B. \(\dfrac{3}{4}\)
• C. \(\dfrac{7}{20}\)
• D. \(\dfrac{13}{20}\)

Hint:

Solve inequality and count favorable outcomes within given range. Watch domain carefully.

Answer 1

The Correct Option is C

We solve \(f(x) = x^2 - 5x + 4>10\)
\[ x^2 - 5x - 6>0 \Rightarrow (x - 6)(x + 1)>0 \Rightarrow x<-1 \text{ or } x>6 \] Only natural numbers from 1 to 20 are considered. So valid \(x>6\): 7 to 20.
Count = \(20 - 6 = 14\). Total choices = 20.
So required probability = \(\dfrac{14}{20} = \dfrac{7}{10}\) But wait — correction: original condition was \(f(x)>10\), so: \[ x^2 - 5x + 4 - 10>0 \Rightarrow x^2 - 5x - 6>0 \Rightarrow (x - 6)(x + 1)>0 \Rightarrow x>6 \text{ or } x<-1 \] Only natural values satisfying \(x>6\): 7 through 20 → 14 values
Probability = \(\dfrac{14}{20} = \dfrac{7}{10}\) Seems the correct answer in image is (3), but it shows 7/20 — must be a typo. Final correct is: