Question

Question Number 15.

No direct answers

Complete answers required! ​

Answer 1

Solution by finding zeroes :

The given polynomial is

    p(x) = 4x² - 3x - 1

    = 4x² - (4 - 1)x - 1

    = 4x² - 4x + x - 1

    = 4x (x - 1) + 1 (x - 1)

    = (x - 1) (4x + 1)

So, we can say that x = 1 and x = -1/4 are the zeroes of p(x)

    Let, α = 1 and β = -1/4

    or, 1/α = 1, 1/β = - 4

We need to find the polynomial whose zeroes are reciprocal of the zeroes of p(x)

If q(x) be the required polynomial, then

    q(x) = (x - 1/α) (x - 1/β)

    = (x - 1) {x - (-4)}

    = (x - 1) (x + 4)

    = x² + 4x - x - 4

    = x² + 3x - 4

So, the required polynomial is

           q(x) = x² + 3x - 4

Solution by relation between zeroes and coeffecients method :

The given polynomial is

    f(x) = 4x² - 3x - 1

Let, α and β be the zeroes of f(x)

Then by the relation between zeroes and coefficients, we get

    α + β = 3/4

    αβ = -1/4

We need to find the polynomial whose zeroes are 1/α and 1/β

Now, 1/α + 1/β

    = (α + β)/(αβ)

    = {(3/4)} / {-(1/4)}

    = - 3

and 1/α * 1/β

    = 1/(αβ)

    = 1/{-(1/4)}

    = - 4

So, the required polynomial be

q(x) = x² - (α + β)x + αβ

or, q(x) = x² - (-3)x + (-4)

     or, q(x) = x² + 3x - 4

Answer 2

Question : D, E and F are respectively the midpoints of the sides BC, CA and AB of ∆ ABC. Find the ratio of areas of ∆ABC and ∆DEF.

Answer : Ratio 1 : 4

Step by step explanation :

We have to find the area of ∆ABC and ∆DEF.

Let's prove the triangles similar.

In ∆ ABC,

D and F are midpoints of AB and AC respectively.

Thus, DF || BC

Also, DF || BE...(1)

Now, Similarly,

E and F are midpoints of BC and AC.

Thus, EF || AB

Also, EF || DB...(2)

From (1) and (2) ,

It is proved that opposite sides of quadrilateral are parallel.

So, DBEF is a parallelogram.

Now, there's a property that :

Opposite angles of Parallelogram are congruent.

So,

Angle DFE = Angle ABC...(3)

Now,

Similarly,

It is proved that DEFC is a parallelogram.

Now,

Angle EDF = Angle ACB..(4) opposite angles of Parallelogram are congruent.

Thus, in ∆ EDF and ∆ ABC,

From (3) and (4) ,

∆DEF ~ ∆ABC

Now, we know that :

When two triangles are similar, the ratio of their areas is equal to the square of ratio of their corresponding sides.

Therefore,

A(∆DEF)/A(ABC) = (DE)^2/ (AC)^2

As it's a parallelogram, DE = FC

So,

A(∆DEF)/A(ABC) = (FC)^2/ (AC)^2

As F is the midpoint of AC,

FC = AC/2

So,

A(∆DEF)/A(ABC) = (AC/2)^2/ (AC)^2

A(∆DEF)/A(ABC) = (AC)^2/4)/ (AC)^2

A(∆DEF)/A(ABC) = (1/4)/1

A(∆DEF)/A(ABC) = (1/4)

A(∆DEF) : A(ABC) = 1 : 4

Hence, Ratio of area of ∆DEF and ∆ABC is 1 : 4