AP ICET 2013

Solution

A/q 

AB = DE = 15m

So CD = CE - AB = 15m

Also given AE = 36m. It implies BD = 36.

Hence in right angle triangle BCD. 

Base = 36m

Height = 15 m 

So according to Pythagoras theorem Hypotenuse (or distance between their tops) will be 39m. Answer

SolutionsThe Polynomial is divisible by ( x² -1 ), so it will have ( x² -1 ) as one of its factors.

Also, ( x² -1 ) = (x+1)(x-1)

So both +1 and - 1 will be the roots of the polynomial.

Hence f(1) and f(-1) will be equal to 0.

f(1) = a(1) + b (1) + 3 (1) - 4 - 4 = a + b + 3 - 4 - 4 = a + b - 5 = 0 ….(i)

f(1) = a(1) + b (-1) + 3 (1) + 4 - 4 = a - b + 3 = 0 ……..(ii)

On solving (i) and (ii)

a = 1 and b = 4 .Hence Answer (1,4)

Solution:

If factors are (x- α) and (x - β) then roots of the expression will be α , β.

Product of roots = α x β = (c/a) ..........Eqn (i)

Sum of roots = α + β = -(b/a)  ..........Eqn (ii)

In the new case,

Let polynomial be Mx² + NX + O

So product of roots = O/M = (1/α) x (1/β) = (1/αβ) = a/c …….from equation (i)

Sum of roots = -N/M = (1/α) + (1/β) = {(α + β) / αβ} = (-b/a) / (c/a) = - b/c…….from equation (ii)

So product of roots = a/c = O/M

Sum of roots = - b/c = - N/M

Solution: 

Principle used to solve :Rem (A x B) = Rem. (A) x Rem. (B)

A/q

When divided by 12

Rem. (63 x 65 x 67 x 69) = Rem. (63) x Rem.(65) x Rem.(67) x Rem.(69) 

The common multiple of 12 samller than all the above nos. is 60. So taking 60 as reference.     

Rem. (63) x Rem.(65) x Rem.(67) x Rem.(69)  = Rem. (3 x 5 x7 x9) = Rem.(945) 

945 leaves a remainder of 9 when divided by 12, as 936 is the closest multiple of 12 before 945.   

Hence Answer 9                             

A polynomial in x leaves remainders 8, 4 when divided by (x + 2) and (x - 2) respectively. If the same polynomialis divided by $$x^{2} - 4$$ then the remainder is

Let fn. be f(x).

As f(x) leaves remainder, 8 when divided by (x+2)

Therefore, f(-2)= 8

When f(x) is divided by (x−2), remainder = 4

Therefore, f(2)=4

Now, polynomial is divided by $$x^{2} - 4$$ or (x+2)(x-2).

the remainder will be linear. in the form of ax+b.

f(x)=q(x)(x+2)(x−2)+r(x)

When, x=2,

⇒f(2)=0.(x−2)q(2)+r(2)=4⇒f(1)=0.(x−2)q(1)+r(1)=4

r(2) = 2a+b = 4......(i)

When, x=-2

⇒f(-2)=0.(x+2)q(-2)+r(-2)=8⇒f(-2)=0.(x+2)q(-2)+r(2)=8

r(-2) = (-2a)+b = 8......(i)

Solving above two equations, we get remainder r(x) = 6-x Answer

given that 

x + y = 12     equation 1 

y + z = 20     equation 2

z + x = 18      equation 3 

on adding equation  1 + 2 + 3 we get 

x + y + z = 25  equation 4   now perform operation eq4 - eq1, eq4 - eq2, and  eq4 - eq3 

we get 

x = 5 ,y = 7 and z = 13 then  put the value in eq following 

x - y + z = 5 - 7 + 13 = 11 Answer

let he have x number of 2 Rs coin and y number of 5 Rs coin 

then we can say that 

2x + 5y = 50        equation 1 

x + y     = 16        equation 1 on solving equation 1 and 2 we get 

x = 10  number of 2 Rs coin 

y = 6    number of 2 Rs coin    

then by question x - y = 10 - 6 = 4   Answer   

let the first term is a and common ratio is r then

$$T_{1}$$= {a}$$\times{r}^{1-1}$$ = a =  $$\frac{2}{3}$$                equation 1

$$T_{6}$$= {a}$$\times{r}^{6-1}$$ = {a}$$\times{r}^{5}$$ = 162      equation 2

equation 1 \equation 2

$$\frac{2}{3}$$\162 = 1\r^5

243 = r^5

3 = r

r = 3 

put the value of a and  r in equation following equation 

$$T_{8}$$= {a}$$\times{r}^{8-1}$$ =  $$\frac{2}{3}$$ $$\times{3}^{7}$$ 
                             $$T_{8}$$         =   $$\frac{2}{3}$$ $$\times243$$ $$\times3$$ $$\times3$$                                 

                              $$T_{8}$$        = 1458     Answer 

Let x-d,x and x+d  be the three numbers.

Then in question given that 

 x - d + x + d + x = 3 or x = 1 and also given that 

 x + d - x + d = 16    or  d = 8 

then we get the sequence = -7,1 and 9  so then multiplication on all term 

we get    -63  Answer 

let  $$(r+1)^{th}$$ term be the independent of x which is given by bio nominal theorem 

$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$  $$a^{(8-r)}$$ $$\times$$ $$b^{r}$$ 

here we have a = 3$$x^{3}$$ and b = -4\x 

put the value we get 

$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$  (3$$x^{3}$$)^(8-r)  $$\times$$ (-4\x)^r

$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$ (3)^(8-r)($$x^{3}$$)^(8-r) $$\times$$ (-4)^r(x)^(-r)

$$T_{r+1}$$ = 8$$c_{r}$$ $$\times$$ (3)^(8-r)($$x^{24-3r}$$) $$\times$$ (-4)^r(x)^(-r)

since  the term is independent of x, we have 

24 - 4r = 0

r  =  6 

so the (r+1) = 7 term is independent so we can say that 

$$T_{7}$$ = 8$$c_{6}$$ $$\times$$ $$3^{3-1}$$ $$\times$$ (-4)^2(6-1)

$$T_{7}$$ = 8 $$\times$$ 7 $$\times$$  $$3^{2}$$ $$\times$$ $$2^{11}$$

$$T_{7}$$ = $$2^{14}$$ $$\times$$ 7 $$\times$$ $$3^{2}$$  Answer