AP ICET 2015

Soution:

The diagonals of rhombus form right angle at the centre.

AO = 10/2 = 5 cm

AD = 13 cm

So DO = $$\sqrt{\ \left(AD^2\right)\ -\ \left(AO^2\right)}$$

= $$\sqrt{\ \left(13^2\right)\ -\ \left(5^2\right)}\ =12\ cm$$

Length of diagonal = 24 

Area of diagonals = $$\frac{1}{2\ }\Pr oduct\ of\ diagonals\ =\ \frac{1}{2\ }\times\ \ 10\ \times24\ =120\ cm^2\ $$

Solution:

Any point on X-axis is represented as (x,0)

Distance of this point from (7,6) = $$\sqrt{\ \left(7-x\right)^2\ +\left(6-0\right)^2}$$= 

Distance of this point from (-3,4) = $$\sqrt{\ \left(-3-x\right)^2\ +\left(4-0\right)^2}$$

Both the distances should be same.

$$\sqrt{\ \left(7-x\right)^2\ +\left(6-0\right)^2}$$ = $$\sqrt{\left(\left(-3-x\right)^2\ \ +\left(4\right)^2\right)\ }$$

$$\sqrt{\ \left(49+x^2\ -14x\right)+36}\ =\ \sqrt{\ 9+x^2+6x+16}$$

$$\left(49+x^2\ -14x\right)+36\ =\ 9+x^2+6x+16$$

$$60=20x$$

$$x\ =\ 3$$

So, point = $$\left(3,0\right)$$ Answer

Solution

Let the required point be (x,0). Any point on x axis will have y ordinate 0

The two given points are

 (3,-4) and (-2,5)

Using the formula

As y =0, so equating y part to 0.

0 = $$\frac{\left(m1\left(5\right)\ +\ m2\ \left(-4\right)\right)}{m1+m2}\ =\ \frac{\left(5m1+-4m2\right)}{m1+m2}=0$$
$$\ \left(5m1+(-4m)2\right)=0\ ;Hence\ \frac{m1}{m2}\ =\frac{4}{5}$$  External division. Answer

Solution:

inverse of  $$p \rightarrow (q \wedge r)$$

~(p -> (q^r))

~ p -> ~(q^r)

~ p ->(~q) v (~r) Answer

Solution

A/q 

So correct option C

Solution

No. of reflexive relation = $$2^{n\left(n-1\right)}$$

here n = 5

A/q

No. of reflexive relation = $$2^{5\left(5-1\right)}$$ = $$2^{20}$$

A/q

A = (-3,3)

B = (-4,-2) U (2,4)

So $$A \cap B = $$ (-3,-2) U (2,3) Answer

Solution

n(A)=n(A-B)+n(A∩B)
=40+16
=56
Therefore, n(A) is 56.

Now n(B)=n(A∩B)+n(B-A)
Therefore, n(B) is 16+57 = 73.

n(B)=73.

Hence n(B)=73 Answer

Solution:

The graph will be similar to that of $$x^3$$, just displaced 8 units upwards.

As, a function is a one-to-one function when it will pass the vertical line test (to make it a function) and also a horizontal line test (to make it one-to-one).

So it is one-one.

Also,

Also, as you progress along the line, every possible y-value is used.So it is also Onto.

Hence it is bijection.

Solution

Given $$y\ =\ \frac{\left(3x-1\right)}{10}$$

$$10\ y\ =\ 3x-1$$

Hence$$f^{-1} (x) =$$  = $$\frac{\left(10x+1\right)}{3}$$

Putting x = 5

$$\frac{\left(10x+1\right)}{3}\ =\ \frac{\left(10\cdot5+1\right)}{3}=\frac{51}{3}=17$$