Please enter the following details
Free Daily Targets & Mocks
Study Material & Formulas PDFs
1 on 1 Mentorship
Solution
We can see that at n = 2, $$n^3-11n^2+32n-28=0$$ i.e. (n-2) is a factor of $$n^3-11n^2+32n-28$$
$$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$$
We can further factorize n^2-9n+14 as (n-2)(n-7).
$$n^3-11n^2+32n-28=(n-2)^2(n-7)$$
$$\Rightarrow$$ $$n^3-11n^2+32n-28>0$$
$$\Rightarrow$$ $$(n-2)^2(n-7)>0$$
Therefore, we can say that n-7>0
Hence, n$$_{min}$$ = 8
Video Solution
Create a FREE account and get:
- All Quant CAT complete Formulas and shortcuts PDF
- 35+ CAT previous year papers with video solutions PDF
- 5000+ Topic-wise Previous year CAT Solved Questions for Free
CAT Quant Questions | CAT Quantitative Ability
CAT Linear Equations QuestionsCAT Percentages QuestionsCAT Profit And Loss QuestionsCAT Probability, Combinatorics QuestionsCAT Time Speed Distance Questions
CAT DILR Questions | LRDI Questions For CAT
CAT DI with connected data sets QuestionsCAT Venn Diagrams QuestionsCAT Table with Missing values QuestionsCAT Puzzles QuestionsCAT Routes And Networks Questions
