Maths Question Combo :-
1) The condition that sinA , cosA may be the roots of ax² + bx + c = 0.
2) If a , b , c ∈ R and a ≠ b , then the roots of the equation 2(a - 6)x² - 11(a + b + c)x - 3(a - b) = 0 are .
3) The number of real roots of √(x - 3) × (x² + 7x + 10) = 0 , x ∈ R.
[ Note :- root is only for (x - 3) ]
Answers
EXPLANATION.
Question = 1.
The conditions that,
sin(A), cos(A) may be the roots of ax² + bx + c.
As we know that,
Let, we assume that.
⇒ One roots = sin(A) = x.
⇒ Other roots = cos(A) = y.
Sum of the zeroes of the quadratic polynomial.
⇒ α + β = - b/a.
⇒ x + y = - b/a.
⇒ sin(A) + cos(A) = - b/a. - - - - - (1).
Products of the zeroes of the quadratic polynomial.
⇒ αβ = c/a.
⇒ xy = c/a.
⇒ sin(A).cos(A) = c/a. - - - - - (2).
From equation (1) & (2), we get.
Squaring equation (1), we get.
⇒ sin(A) + cos(A) = - b/a. - - - - - (1).
⇒ [sin(A) + cos(A)]² = [-b/a]².
As we know that,
Formula of :
⇒ (x + y)² = x² + y² + 2xy.
⇒ sin²x + cos²x = 1.
Using this formula in the equation, we get.
⇒ sin²A + cos²A + 2sin(A).cos(A) = b²/a².
⇒ 1 + 2sin(A).cos(A) = b²/a².
Put the value of sin(A).cos(A) = c/a in the equation, we get.
⇒ 1 + 2[c/a] = b²/a².
⇒ 1 + 2c/a = b²/a².
⇒ a² + [2a²c/a] = b².
⇒ a² + 2ac = b²
Question = 2.
⇒ a, b, c ∈ R and a ≠ b.
Roots of the equation.
⇒ 2(a - 6)x² - 11(a + b + c)x - 3(a - b) = 0.
As we know that,
⇒ D = Discriminant Or b² - 4ac.
⇒ [-11(a + b + c)]² - 4[2(a - 6)][-3(a - b)].
⇒ [121(a + b + c)² + 24(a - 6)(a - b)].
As we can see that,
⇒ (a + b + c)² ≥ 0.
⇒ (a - b) > 0.
So, roots are real and unequal.
Question = 3.
Number of real roots.
⇒ √(x - 3)(x² + 7x + 10) = 0, x ∈ R.
From equation.
⇒ x² + 7x + 10.
Factorizes the equation into middle term splits, we get.
⇒ x² + 5x + 2x + 10.
⇒ x(x + 5) + 2(x + 5).
⇒ (x + 2)(x + 5).
⇒ √(x - 3)(x + 2)(x + 5) = 0.
If we find zeroes, we get.
⇒ (x - 3) ≥ 0.
⇒ x ≥ 3.
⇒ (x + 2) = 0.
⇒ x = - 2.
⇒ (x + 3) = 0.
⇒ x = - 3.
As we can see that only one real roots exists.
-: Question 1 :-
To Find :-
The condition that sinA , cosA may be the roots of ax² + bx + c = 0
Solution :-
Let us assume that ;
x = Sin A
x' = Cos A
Then , x + x' = -b/a = Cos A + Sin A
x . x' = c/a = Cos A . Sin A
=> Here , Many situation arises some of them are :-
i) If x + x' = Cos A + Sin A and x . x' = Cos A . Sin A
ii) We knows that ; Cos²x + Sin²x = 1
=> Cos² A + Sin² A = 1 = x² + x'²
=> [ x² + x'² ] - 1 = 0
iii) [ x² + x'² ] - 1 = 0
=> [ x² + x'² ] = 1
=> [ x + x' ]² - 2xx' = 1
iv) x + x' = -b/a = Cos A + Sin A
x . x' = c/a
=> Squaring both sides of x + x' = -b/a
=> ( x + x' )² = ( -b/a )²
=> x² + x'² + 2xx' = b²/a²
=> ( x² + x'² ) + 2c/a = b²/a²
=> 1 + 2c/a = b²/a²
=> a² × ( 1 + 2c/a ) = b²
=> a² + 2ac = b²
-: Question 3 :-
To Find :-
The number of real roots of √(x - 3) × (x² + 7x + 10) = 0 , x ∈ R.
Solution :-
Here , The equation is :-
=> ( √x - 3 ) ( x² + 7x + 10 ) = 0
=> Multiplying both sides by ( √x - 3 ) ;
=> ( √x - 3 ) ( √x - 3 ) ( x² + 7x + 10 ) = 0 . ( √x - 3 )
=> ( x - 3 ) ( x² + 7x + 10 ) = 0
=> x ( x² + 7x + 10 ) - 3 ( x² + 7x + 10 ) = 0
=> x³ + 7x² + 10x - 3x² - 21x - 30 = 0
=> x³ + 4x² - 21x - 30 = 0
Here the given equation has 3 degree . So , The equation will have 3 values of x such that x ∈ R . Because Imaginary numbers are also subset of Real Numbers because they exist in maths . So , we can say that the equation will have 3 roots .
-: Question 2 :-
To Find :-
If a , b , c ∈ R and a ≠ b , then the roots of the equation 2(a - 6)x² - 11(a + b + c)x - 3(a - b) = 0 are
Solution :-
The Given Equation is :-
2 ( a - 6 ) x² - 11 ( a + b + c ) x - 3 ( a - b ) = 0
Hey Dear User , You don't mentioned that whether we have to find the nature of the roots or we have to find the roots in terms of ' a , b and c ' . So , I'm finding the nature of the roots only !
On Comparing the given equation with " mx² + nx + j = 0 " we get ;
m = 2 ( a - 6 )
n = - 11 ( a + b + c )
j = - 3 ( a - b )
Now , D = n² - 4mj
=> D = [ - 11 ( a + b + c )² ] - 4 × 2 × ( a - 6 ) × ( a - b ) × - 3
=> D = [ - ( 11a + 11b + 11c )² ] + 24 ( a - b ) ( a - 6 )
=> D = [ - ( 11a + 11b + 11c )² ] + 24 ( a² - ab - 6a + 6b )
The Discrimanant can also be written as ;
=> D = [ - ( 11a + 11b + 11c ) ]² + ( √24 )² ( a - b/2 - 3 )² + ( √144 )² ( -24b - 144)²/( √96 )²
Here. ,You can see that Discrimanant is the square of a real number because x ∈ R . So , The roots will be rational , different and real .
Note :- If the answer is wrong or there is some mistake in the answer . So , you can contact me anytime !