Maths

p and q are the roots of the equation x² – 4ax + 3 = 0

pq = 3

Similarly the constant term of x² – 2bx + c = 0

Product of $\left(\frac{2}{p}-q\right)and\left(\frac{2}{q}-p\right)$

Hence, c = $\frac{4}{pq}-2-2+pq=\frac{1}{3}$

f (x) = $\frac{x}{4x^2+7x+9}$ ………. (1)

Divide by x in numerator and denominator equation (1) can be written as f (x) = $\frac{1}{4x+7x+\frac{9}{x}}$

f (x) is maximum, when the denominator is minimum we have to find the minimum value of $4x+\frac{9}{x}$ , so, denomination will also minimum

we know that, AM ≥ GM

$\frac{4x+\frac{9}{x}}{2}\ge \sqrt[]{4x*\frac{9}{x}}$

$\frac{4x+\frac{9}{x}}{2}=6$

$4x+\frac{9}{x}=12$

The minimum value of $4x+\frac{9}{x}$ is 12

Therefore, the maximum value of f (x) = $\frac{1}{12+7}=\frac{1}{19}$

For Aman,

SP of 1 book = 14Rs.

CP of 1 book = $\frac{100}{(100+x)}*14$

For Baman,

SP of 1 book = 12Rs.

CP of 1 book = $\frac{100}{(100+y)}*12$

Price of all notebooks is same, therefore.

$\frac{100}{(100+x)}*14$ = $\frac{100*12}{100+y}$

100 + 7y = 6a

100 = 12y – 7y; [x=2y]

5y = 100

y = 20

a = 40

CP each notebook = $\frac{100*12}{100+20}$ = 10Rs.

In his sales with Daman, he sold 10 books to her for Rs. 130

CP of 10books = Rs. 100

Profit percent = $\frac{130-100}{30}*100$

=30percent

a + b + c + d = 7

a ³ 0, b³ 0, c ³ 1, d ³ 1

(a, b, c, d are the numbers of balls)

Let c = c' + 1 = 0, 0 ≤ c' ≤ 5

d = d' + 1 = 0, 0 ≤ d' ≤ 5

a + b + c' + d' = 5

Numbers of solution = ^{5 + 4 – 1}C_4–1

= ⁸C₃ = 56

Let f (x) = ax² + bx + c

f (0)= 2

2 = a (0)² + b (0) + c

c = 2

f (– 2) = 3

a (–2)² +b (–2) + 2 = 3

4a – 2b = 1                    ………. (1)

The given equation has a maximum value at x = –2

$\frac{-b}{2a}=-2$

b = 4a

put in equation 1

we get, $a=\frac{-1}{4},b=-1$

f (6)= $\frac{-1}{4}{(6)}^2+(-1)6+2$

= –13

P₁ + P₂ + P₃ = 59

P₁ and P₂ → minimize

Then P₃ → minimum i.e. 47

So, only possible value of P₃ can take → 29, 31, 37, 41, 43, and 47

Total 6 values

$\frac{1}{x}=\frac{1}{15}-\frac{6}{y}$

$\frac{1}{x}=\frac{y-90}{15y}$

$x=\frac{15y}{y-90}$ ______equation 1

i.e. y = 90+y,

1≤ y₁ ≤9

y₁ is odd

Equation 1 becomes

$x=\frac{15(90+y_1)}{y_1}$

$=15+\frac{90*15}{y_1}$

y₁ can take 1,3,5,9

y=91, 93, 95, 99

1 + x = 3.5

x= 2.5

Required percent y = $\frac{2.5}{3}*100$  = 83.33 percent

x, a, b, y are in GP

a = xr

b = xr²

y = xr³

r = ${\left(\frac{y}{x}\right)}^{1/3}$

rewrite,

a = x^2/3.y^1/3

b = x^1/3.y^2/3

a * b = x * y

a³ + b³ = x²y + y² x

= xy (x+y)

= xy (2A) …………… $\left[\frac{x+y}{2}=A\right]$

Hence, n = 2