Mathematics (Standard) Test 2

Probability that the card drawn is neither a king nor a queen

\(=\frac{52-8}{52}\)

\(=\frac{44}{52}=\frac{11}{13}\)

Outcomes when 5 will come up at least once are-

(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4) and (5,6)

Probability that 5 will come up at least once = \(\frac{11}{36}\)

1+ \(sin^2\alpha\) = 3 sin \(\alpha\) cos \(\alpha\)

\(sin^2\alpha + cos^2\alpha + sin^2\alpha\) = 3 sin \(\alpha\) cos \(\alpha\)

2 \(sin^2\alpha\)- 3sin \(\alpha\) cos \(\alpha\) + \(cos^2\alpha\) = 0

(2sin \(\alpha\) -cos \(\alpha\)) ( sin \(\alpha\) - cos \(\alpha\)) =0

\(\therefore\) cot \(\alpha\) = 2 or cot \(\alpha\) = 1

Since ABCD is a parallelogram, diagonals AC and BD bisect each other,

\(\therefore\) mid point of AC= mid point of BD

\((\frac{\text x+1}2,\frac{6+2}2)\) \(=(\frac{3+4}2,\frac{5+y}2)\)

Comparing the co-ordinates, we get,

\(\frac{\text x+1}2=\frac{3+4}2\), So, x = 6

Similarly, \(\frac{6+2}2=\frac{5+y}2.\) So, y = 3

\(\therefore\) (x, y) = (6, 3)

Any point (x, y) of perpendicular bisector will be equidistant from A & B.

\(\therefore\sqrt{(\text x-4)^2+(y-5)^2}\) \(=\sqrt{(\text x+2)^2+(y-3)^2}\)

Solving we get

-12x – 4y + 28 = 0 or 3x + y – 7 = 0

The smallest number by which \(\frac1{13}\) should be multiplied so that its decimal expansion terminates after two decimal points is

\(\frac{13}{100}\) as \(\frac1 {13} \times \frac{13} {100} = \frac{1}{ 100}\) = 0.01

Since P divides the line segment joining R(-1, 3) and S(9,8) in ratio k : 1

\(\therefore\) coordinates of P are \(\left(\frac{9k-1}{k+1},\frac{8k+3}{k+1}\right)\)

Since P lies on the line x – y + 2=0, then \(\frac{9k-1}{k+1}-\frac{8k+3}{k+1}+2=0\)

9k -1 - 8k -3 +2k + 2 = 0

which gives k = \(\frac23\)

Sum of zeroes = 2 + \(\cfrac12\) = \(\frac{-5}p\)

i.e. \(\frac{5}2 = \frac{-5}p\) . So p= -2

Product of zeroes = 2\(\times\frac12\) = \(\frac{r}p\)

i.e.\(\frac{r}p\) = 1 or r = p = -2

2\(\pi\)r =100. So Diameter = 2r = \(\cfrac{100}\pi\) = diagonal of the square.

side \(\sqrt2\) = diagonal of square = \(\cfrac{100}\pi\)

\(\therefore\) side = \(\frac{100}{\sqrt2\pi}=\frac{50\sqrt2}{\pi}\) 

\(3^{\text x+y}=243=3^5\)

So x + y = 5.....(1)

\(243^{\text x-y}=3\) 

\((3^5)^{\text x-y}=3^1\) 

So, 5x - 5y = 1.......(2)

Since :  \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2},\) so unique solution

Initially, at t = 0, Annie’s height is 48ft

So, at t =0, h should be equal to 48

h(0) = -16\((0)^2\) + 8(0) + k = 48

So k = 48

When Annie touches the pool, her height =0 feet

i.e. -16\(t^2\) + 8t + 48 =0 above water level

2\(t^2\) - t - 6 =0

2\(t^2\) - 4t + 3t -6 = 0

2t(t - 2) + 3(t - 2) = 0

(2t + 3) (t - 2) = 0

i.e. t = 2 or t = -3/2

Since time cannot be negative , so t = 2 seconds

t = -1 & t = 2 are the two zeroes of the polynomial p(t)

Then p(t)=k (t- -1)(t - 2)

= k(t +1)(t - 2)

When t = 0 (initially) \(h_1\) = 48 ft

p(0)=k(\(0^2\) - 0 -2) = 48

i.e. -2k = 48

So the polynomial is -24(\(t^2\) - t -2) = -24 \(t^2\) + 24t + 48.

A polynomial q(t) with sum of zeroes as 1 and the product as -6 is given by q(t) = k(\(t^2\) - (sum of zeroes)t + product of zeroes)

= k(\(t^2\) -1t + -6) ………..(1)

When t = 0 (initially) q(0)= 48ft 

q(0)=k(\(0^2\) - 1(0) -6)= 48

i.e. -6k = 48 or k= -8

Putting k = -8 in equation (1), reqd. polynomial is -8(\(t^2\) -1t + -6)

= -8\(t^2\) + 8t + 48

When the zeroes are negative of each other,

sum of the zeroes = 0

So, -b/a = 0

\(-\frac{(k-3)}{-12}=0\)

\(+\frac{k-3}{12}=0\)

k -3  = 0,

i.e. k = 3.

Centroid of \(\triangle EHJ\) with E(2,1), H(-2,4) and J(-2,-2) is

\(\left(\frac{2+-2+-2}{3},\frac{1+4+-2}{3}\right)\)

= (-2/3, 1)

If P needs to be at equal distance from A(3,6) and G(1,-3), such that A,P and G are collinear, then P will be the mid-point of AG.

So coordinates of P will be \(\left(\frac{3+1}{2},\frac{6+-3}{2}\right)\)

= (2, 3/2)

Let the point on x axis equidistant from I(-1,1) and E(2,1) be (x,0)

then \(\sqrt{(x+1)^2+(0-1)^2}\) \(=\sqrt{(x-2)^2+(0-1)^2}\)

\(x^2+1+2x+1\) \(=x^2+4-4x+1\)

6x = 3

So x \(=\frac{1}{2}\)

\(\therefore\) the required point is \(\left(\frac{1}{2},0\right)\)

Let the coordinates of the position of a player Q such that his distance from K(-4, 1) is twice his distance from E(2, 1) be Q(x, y)

Then KQ : QE = 2 : 1

Q(x, y) \(=\left(\frac{2\times2+1\times-4}{3},\frac{2\times1+1\times1}{3}\right)\)

= (0, 1)

Let the point on y axis equidistant from B(4,3) and C(4,-1) be (0, y)

then \(\sqrt{(4-0)^2+(3-y)^2}\)  \(=\sqrt{(4-0)^2+(y+1)^2}\)

\(16+y^2+9-6y\) \(=16+y^2+1+2y\)

-8y = -8

So y = 1

\(\therefore\) the required point is (0, 1).