Test: Functions- 2 - CAT MCQ

f(3) = 0. So 9a + 3b + c = 0
f(5) = -3f(2).

So, 25a + 5b + c = -3 (4a + 2b + c).

Solving, we get 37a + 11b + 4c = 0.
So we have 2 equations:
9a + 3b + c = 0                           --------------------> (i)
37a + 11b + 4c = 0                     --------------------> (ii)
Multiply (i) with 4 , we get,

=> 36a + 12b + 4c = 0                --------------------> (iii)
Subtract (iii)  from (ii), we get, a – b = 0 or a = b. So, we got a = b.
Sum of the roots of a quadratic equation is –b/a. Here it is –a/a = -1.
one of the root is 3. Sum of roots is -1.

Hence, the other root is -4

We know the roots are 3 and -4.
Hence, the equation is (x - 3) (x + 4) = 0, or x² + x – 12 = 0.
However, even 2(x² +x – 12) = 0 or 100 (x² +x – 12) = 0 and so on will have the same roots.
Hence, we cannot find unique values of a,b,c.

f(0) = p
f (1) = 1 – 4 + p = p + 3.
p and p+3 are of different signs.
Substitute for the options: 
1st option, if p is a negative number, then p – 3 is also negative, and both are same sign.
For 3rd and 4th option, also it is the same.
For the 2nd option, value of p is between 0 and 3, and p – 3 is negative. Hence it agrees.

Cost price function f(x) = cx² + bx + 240.
When x = 20, f(20) = 400c + 20b + 240.
=> f(40) = 1600c + 40b + 240
66.67% is 2/3.
=> f(40) – f(20) = 2/3 * f(20).
Substituting f(40) and f(20), we get, 1200c+ 20b =(2/3) * (400c + 20b + 240)
Solving, we get 20b + 2800c = 480.

Now, f(60) = 3600c + 60b + 240.
f(60) – f(40) = ½ * f(40)
Substituting, and solving, we get, 360 + 60b + 2400c = 240 + 60b + 3600c
c = 1/10.

Substituting we get, b = 10.

So, the cost function f(x) = 0.1x² + 10x + 240.
Selling price = 30 * x. ( Each unit SP is 30)
So, profit = SP – CP = 30x – (0.1x² + 10x + 240) =  -x²/10 + 20x – 240.

For maximum profit, differentiating this should give 0.

dp/dt = 0, dp/dt = -x/5 + 20
20 – x/5 = 0. x= 100.

Also double differentiating d²p / dt², we get a negative number, so profit is maximum.

So, profit is maximum when 100 units are produced daily.

Cost price function f(x) = cx² + bx + 240.
When x = 20, f(20) = 400c + 20b + 240
f(40) = 1600c + 40b + 240
66.67% is 2/3.
f(40) – f(20) = 2/3 * f(20)
Substituting f(40) and f(20), we get, 1200c+ 20b =(2/3) * (400c + 20b + 240)
Solving, we get 20b + 2800c = 480

Now, f(60) = 3600c + 60b + 240.
f(60) – f(40) = ½ * f(40).
Substituting, and solving, we get, 360 + 60b + 2400c = 240 + 60b + 3600c.
c = 1/10.

Substituting we get, b = 10.

So, the cost function f(x) = 0.1x² + 10x + 240.
Selling price = 30*x. ( Each unit SP is 30)
So, profit = SP – CP = 30x – (0.1x² + 10x + 240) =  -x²/10 + 20x – 240.

For maximum profit, differentiating this should give 0.

dp/dt = 0, dp/dt = -x/5 + 20
=> 20 – x/5 = 0.

=>x= 100.

Also double differentiating d²p / dt², we get a negative number, so profit is maximum. So, profit is maximum when 100 units are produced daily.

The maximum profit can be calculated with the equation -x²/10 + 20x – 240, with x = 100, and it is 760.

We have to consider positive and negative numbers for all the cases.
f1(x) is positive for positive numbers, and 0 for negative numbers. (0 for x = 0)
f2(x) is 0 for positive numbers, and positive for negative numbers. (0 for x = 0)
f3(x) is 0 for positive numbers, and negative for negative numbers. (0 for x = 0)
f4(x) is negative for positive numbers, and 0 for negative numbers. (0 for x = 0)
So, we see that, out of the 3 products in the question,  f1(x)*f2(x) and f2(x)*f4(x) are always zero, for any x.

The best way of doing this question is to substitute the value of x in the option to the given expression in f(x).

We get the minimum value of 1.6 for f(x) when x=2.5.

Now, there is a question whether the minimum value of the expression can be lesser than this.

In order to resolve the ambiguity,plot f(x) vs x in the graph for x=1,2,3,4 and so on.

We find that the minimum value is acheived at x=2.5. 

Hence option(2) is the answer.