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Let f(x) = ax^{2 }+ bx + c, where a, b and c are certain constants and a ≠ 0. It is known that f (5) = −3f(2) and that 3 is a root of f(x) = 0.
What is the other root of f(x) = 0?
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
Let f(x) = ax^{2} + bx + c, where a, b and c are certain constants and a ≠ 0. It is known that f (5) = −3f(2) and that 3 is a root of f(x) = 0.
What is the value of a+b+c?
We know the roots are 3 and 4.
Hence, the equation is (x  3) (x + 4) = 0, or x^{2} + x – 12 = 0.
However, even 2(x^{2} +x – 12) = 0 or 100 (x^{2} +x – 12) = 0 and so on will have the same roots.
Hence, we cannot find unique values of a,b,c.
If f (x) = x^{3 }– 4x + p, and f (0) and f(1) are of opposite signs, then which of the following is necessarily true?
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.
Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is 240 + bx + cx^{2}, where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.66%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.
How many units should Mr. David produce daily ?
Cost price function f(x) = cx^{2} + 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) = 0.1x^{2} + 10x + 240.
Selling price = 30 * x. ( Each unit SP is 30)
So, profit = SP – CP = 30x – (0.1x^{2} + 10x + 240) = x^{2}/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 d^{2}p / dt^{2}, we get a negative number, so profit is maximum.
So, profit is maximum when 100 units are produced daily.
Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is 240 + bx + cx2, where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.66%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.
What is the maximum daily profit, in rupees, that Mr. David can realize from his business?
Cost price function f(x) = cx^{2} + 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) = 0.1x^{2} + 10x + 240.
Selling price = 30*x. ( Each unit SP is 30)
So, profit = SP – CP = 30x – (0.1x^{2} + 10x + 240) = x^{2}/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 d^{2}p / dt^{2}, 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 x^{2}/10 + 20x – 240, with x = 100, and it is 760.
f1 (x) = x 0 ≤ x ≤ 1
= 1 x ≥ 1
= 0 otherwise
f2 (x) = f1(–x) for all x
f3 (x) = –f2(x) for all x
f4 (x) = f3(–x) for all x
How many of the following products are necessarily zero for every x:
f1(x) f2 (x), f2 (x) f3 (x), f2(x) f4 (x)
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 function f(x) =  x  2 +  2.5  x + 3.6  x, where x is a real number, attains a minimum at
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.
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