Math, asked by PRATHEEV9B34, 1 year ago

please help me to solve these tomorrow exam

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Answered by honey8076
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all Questions.........?

Answered by LuciferMorningstar99
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(2) mx³ + x² - 2x + n = 0...... (equation-1)

we have two values of x = (-1) & 1

By putting these values in So we got two equation-1, we get

-m + n = - 3 & m + n = 1

By adding these equation, we get

m = 2 & n = - 1

(3) y = 2 & 1/2

Putting these in the main equation, we get two equations

(4m + 10y + n) and (m/4 + 5/2 + n)

Equating these, we get

m = n = - 30/7

(4) x + 1/x = 7..........(equation-1)

Squaring both sides in equation-1, we get

=> x² + 1/x² + 2x*1/x = 49

=> x² + 1/x² = 47

Now Cubing both sides in equation-1, we get

=> x³ + 1/x³ + 3x*1/x (x² + 1/x²) = 7³

=> x³ + 1/x³ + 3(47) = 343

=> x³ + 1/x³ = 343 - 141

=> x³ + 1/x³ = 202

(5) We have two values of a = 1,(-3)

Putting two values in main equation, we get two equation

(-m + n = 14) & (-9m + n = - 12)

Equating these two, we get

m = 13/4 & n = 69/4

(6) x³ - 3x² - 10x + 24

=> x²(x - 3) - 2(5x + 12)

=> (x² - 2)(x - 3)(5x + 12)

(7) Let f(p) = p²(q-r) + q²(r-p) + r²(p-q)

To show (p-q) is a factor of f(b), we find the zero of the binomial (p - q). (p - q) = 0 => p = q f(q) = q²(q-r) + q²(r-q) + r²(q-q) = q2(q-r) - q2(q-r) + 0 = 0

Therefore, (p - q) is a factor of p²(q-r) + q²(r-p) + r²(p-q).

The given expression is a cyclic expression in p, q and r, so the factors of the expression are also cyclic.

Therefore (q - r), (r - p) are the other factors of the polynomial.

Hence, (p - q), (q - r) and (r - p) are the factors of p²(q-r) + q²(r-p) +r²(p-q).

(8) Let p(x)=q(y)=r(z)= (x+y+z)³-x³-y³-z³

Now, if (x+y) (y+z) (z+x) are the factors of the polynomial, then (x+y), (y+z) & (z+x) individually are also its factors.

Now, let x+y=0

=> x=-y

Given that, p(x)=(x+y+z)³-x³-y³-z³

Then, p(-y)=(-y+y+z)³-(-y)³-y³-z³ = z³+y³-y³-z³=0

∴ by factor theorem (x+y) is a factor of p(x)

Again, let y+z=0

=> y=-z

Given that, q(y) =(x+y+z)³-x³-y³-z³

Then, q(-z) = (x-z+z)³-x³-(-z)³-z³=x³-x³+z³-z³=0

∴ by factor theorem (y+z) is a factor of q(y) [=p(x)]

Similarly, we can easily prove that (z+x) is a factor of r(z) [=q(y)=p(x)]

(9) x³ + x/4 - 1/8 = 0

=> 8x³ + 2x - 1 = 0

=> 4x(2x² + 1) - 1(2x +1) = 0

=> (4x - 1)(2x² + 1)(2x + 1) = 0

=> x = 1/4, 1/root2 & 1/2

(10) x² + x -6 ) x³ - 3x² -12x  + 19 ( x -4

                x³ + x² -6x

    --------------------------------

              -4x² -6x + 19 

              -4x²-4x + 24 

-------------------------------------

                   -2x - 5 

hence ,

   x³ - 3x² -12x + 19 = (x² +x - 6)(x - 4) -(2x + 5)

   x³ - 3x² -12x + 19 + (2x + 5)  = (x² + x - 6)(x - 4)

hence if we add (2x + 5) we'll get the required equation...

(11) Let p (x) = x3 - 6x2 - 15x + 80 and q (x) = x2 + x - 12

By division algorithm, when p (x) is divided by q (x), the remainder is a linear expression in x.

So, let r (x) = ax + b is subtracted to p (x) so that p (x) + r (x) is divisible by q (x).

Let, f (x) = p (x) – r (x)

⇒ f(x) = x3 - 6x2 - 15x + 80 – (ax + b)

⇒ f(x) = x3 - 6x2 – (a + 15)x + (80 – b)

We have,

q(x) = x2 + x – 12

⇒ q(x) = (x + 4) (x - 3)

Clearly, q (x) is divisible by (x + 4) and (x - 3) i.e. (x + 4) and (x - 3) are factors of q (x)

Therefore, f (x) will be divisible by q (x), if (x + 4) and (x - 3) are factors of f (x).

i.e. f(-4) = 0 and f(3) = 0

f (3) = 0

⇒ (3)3 – 6(3)2 – 3 (a + 15) + 80 – b = 0

⇒ 27 – 54 – 3a – 45 + 80 – b = 0

⇒ 8 – 3a – b = 0 (i)

f (-4) = 0

⇒ (-4)3 – 6 (-4)2 – (-4) (a + 15) + 80 – b = 0

⇒ -64 – 96 + 4a + 60 + 80 – b = 0

⇒ 4a – b – 20 = 0 (ii)

Subtract (i) from (ii), we get

⇒ 4a – b – 20 – (8 – 3a – b) = 0

⇒ 4a – b – 20 – 8 + 3a + b = 0

⇒ 7a = 28

⇒ a = 4

Put value of a in (ii), we get

⇒ b = -4

Putting the value of a and b in r (x) = ax + b, we get

r (x) = 4x – 4

Hence, p (x) is divisible by q (x), if r (x) = 4x – 4 is subtracted from it.

(15) y² - 1 =0

So y = 1 & (-1)

Putting it in the main equation we get two equations - (m + n = 0) & (-m + n =-6)

Equating these two equations, we get

=> m = 3 & n = - 3

(16)First

we determine all the factors of the given polynomial by splitting middle term and then consider any of

the factor as any of the dimension.

Solution:

 

(i) Volume : 3x² – 12x

Since,volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid.

= 3x²–12x

= 3x(x – 4)

 

Hence,

Possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (x – 4)

 

(ii) Volume : 12ky² + 8ky – 20k

Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid.

12ky²+ 8ky – 20k

=4k(3y² + 2y – 5)

[By middle term splitting]

=4k(3y² +5y-3y-5)

= 4k[y(3y+5)-1(3y+5)]

= 4k (3y +5) (y – 1)

 

Hence,

Possible expression for length = 4k

Possible expression for breadth = (3y +5)

Possible expression for height = (y – 1)

(17) Just put a = 1 & (-1) in the main equation and you will see that p + r = q+s+t = 0

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