Hi catcracker! I have a question.
For question 1, after you have proved that ‘c’ can only take only value – 1,
in the alternate method, you have solved using options.
In general, if CBD is one of the options, can we still solve using options?
If I find 1 option that satisfies the conditions, wouldn’t I still have to ensure that no other values can satisfy the equation?
Thanks for the wonderfully detailed posts!
Normally, no. But in this case if 1 satisfies and 2 doesn’t, then 3, 4, etc will also not satisfy, wouldn’t you agree?
Ah yes, because the value of ‘a’ would keep decreasing.
Can you help me with this question discussed in IMS CAT workshop?
Find the number of positive roots of f(x) = a*x^2 + b*x + c ( a not equal to 0) .
Conditions given : 1. f(x) has real roots
2. c > 0
3. the minimum value that f(x) can take is 9a + 3b + c.
Options : 1) 0 2) 1 3) 2 4) 1 or 2
My solution : From condition 3 , we see that minimum value is at x = 3.
So the roots must be symmetrical about 3.
For eg: 2 and 4 ( where both roots are positive)
-1 and 7 ( where only 1 is positive)
Therefore, it is option 4.
The official answer was option 2 ( I only thought of this now, while at home).
Is my approach/answer right? Would you suggest substituting different values of roots for questions such as these?
The official answer was not option 2 but option 3 (value 2). The thing is it is asking for POSITIVE roots. In the case you gave of -1 and 7 as roots, c would be negative (think about it!). Both roots will have to be positive.
Ok, I got it..to have a minimum value, it must be an upward drawn parabola ( a> 0) , but product of roots is negative.
Thank you sir! 🙂
Could you post topics about questions to do with different types of minimum and maximum questions? I have been trying to search, but, couldn’t find those many to solve. It would be really helpful if you could or suggest a website that has those types of questions.
I don’t know of any such source. I have posted a few very basic ones, but the very fancy ones rarely come in the exam so I haven’t bothered too much. (And this year, I am not getting any spare time to research and post….have already posted on all the main topics where I had something useful to say)
I tried solving question no. 2 without putting in options. Just like the way you answered question no. 1.
Could you please correct where I went wrong as I ain’t getting 13 as an answer.
Given – Section C has 8 questions and there are 100 questions
a+b = 92
Section A contains at least 60% of total marks and Section B contains at least 20% of total marks
means, a>=3/5(a+2b+3c) and b>=1/5(a+2b+3c)
means, 5a>=3(92 + b + 24) and 5b >= 92 + b + 24
means, 5(92 – b) >= 348 + 3b and 5b >= 116 + b
How is this possible if both the conditions about the least % of marks from Section A and B have to be true?
Also, how would I know what is the answer if it was asked as a TITA question, ex – write the sum of both possible options for number of questions in section B.
Looking forward towards your reply J and T.
PS – your blog is amazing..!!
Thank you. 🙂
I think the error is here “b>=1/5(a+2b+3c)” because the marks for b questions are 2b. so it should have been “2b>=1/5(a+2b+3c)”. See whether that gives you a correct answer?
YES…I got b>=12.9 in one case and b<=14 in another.
Thanks J for the quick revert. 🙂
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