Mathematics Test - 5

Let a₁, a₂, …… an be n A.M`s between 3 and 17.

3, a₁, a₂, a₃ ………… an, 17.

^an/a₁ = 3: 1 an = 3a1

3 + nd = 3a₁ … (1)

For the last term of AP

tn = a + (n - 1)d (here n is the number of term)

17 = 3 + (n + 1)d

d = 14/(n + 1) …(2)

From (1),

3 + nd = 3(3 + d)

On solving, n = 6.

Akshay can call one, two, three, four or five of his friends at a time and arrange them accordingly.

So, the required number of ways is (⁵C₁ × 0!) + (⁵C₂ × 1!) + (⁵C₃ × 2!) + (⁵C₄ × 3!) + (⁵C₅ × 4!)
= 5 + 10 + 20 + 30 + 24 = 89

Case 1: 4 batsmen, 6 bowlers, 1 wicket-keeper

Number of ways = ⁸C₄ × ⁶C₆ × ²C₁ = 140

Case 2: 5 batsmen, 5 bowlers, 1 wicket-keeper

Number of ways = ⁸C₅ × ⁶C₅ × ²C₁ = 672

Case 3: 6 batsmen, 4 bowlers, 1 wicket-keeper

Number of ways = ⁸C₆ × ⁶C₄ × ²C₁ = 840

Total number of ways = 140 + 672 + 840 = 1652

Given that 7^log₇(x² -4x+5) = _x-1

⇒x² - 4x + 5 = (x-1) Þ x² - 5x + 6 = 0 (x - 2) (x - 3) = 0

x = 2, 3

To get an equation from the given equation which has root equal in magnitude but opposite in sign, we have to put X = -x

Putting this in given equation, we get 2(-x)² - 7(-x) + 3 = 0

2x² + 7x + 3 = 0

Focal distance of any point P (x, y) on the ellipse is equal to SP = a + ex.

Here, x = a cos

Hence, SP = a + ae cos q = a(1 + e cosq).