TITA Based Questions: Linear and Quadratic Equations

Ans: 3
Case 1: x ≥ 0 ⇒ |x| = x
∴ 2 × x × (x² + 1) = 5x²
⇒ 2x(x² + 1) = 5x²
⇒ 2x(x² + 1) - 5x² = 0
⇒ x[2(x² + 1) - 5x] = 0
⇒ x(2x² - 5x + 2) = 0 
⇒ x(2x² - 4x - x + 2) = 0 
⇒ x[2x(x - 2) - (x - 2)] = 0 
⇒ x(2x - 1)(x - 2) = 0
⇒ x = 0 or ½ or 2.
We need only integral solutions hence acceptable answers are 0 and 2.
Case 2: x < 0 ⇒ |x| = -x
∴ 2 × -x × (x² + 1) = 5x²
⇒ -2x(x² + 1) = 5x²
⇒ 2x(x² + 1) + 5x² = 0
⇒ x[2(x² + 1) + 5x] = 0
⇒ x(2x² + 5x + 2) = 0
⇒ x(2x² + 4x + x + 2) = 0
⇒ x[(2x(x + 2) + (x + 2)] = 0
⇒ x(2x + 1)(x + 2) = 0
⇒ x = 0 or -1/2 or -2
We need only integral solutions hence acceptable answers are 0 and -2.
∴ Acceptable integral solutions are -2, 0 and 2, i.e., 3 integral solutions.
Hence, 3.

Ans: 6
α and β be two distinct root of 2x² - 6x + k = 0,
∴ α + β = -(-6)/2 = 3
And, α × β = k/2 = k/2
Now, 3 and k are the roots of the equation x² + px + p = 0.
∴ Sum of the roots = 3 + k/2 = -(p)/1 = -p    ...(1)
∴ Product of the roots = 3 × k/2 = (p)/1 = p    ...(2)
(1) + (2)
⇒ 3 + k/2 + 3k/2 = p - p = 0
⇒ k = -3/2
⇒ p = -9/4    [from (2)]
Now, we need to find 8(k - p)

= 6
Hence, 6.

Ans: 2
Let p and q be the other two real roots of the given cubic equation.
Product of the three roots of the given cubic equation

⇒ q = 1/p
∴ The three roots are -2, p and 1/p

Sum of the three roots of the given cubic equation 

We know than sum of a number and its reciprocal is either less than or equal to - 2 or greater than or equal to 2.

⇒ 2 ≤ -2r + 1 ≤ -2
⇒ 1 ≤ -2r ≤ -3
⇒ -1/2 ≥ r ≥ 3/2
∴ Least non-negative integral value of r is 2.
Hence, 2.

Ans: 6
The given equation can be written as: x² - 2x + 1 + 2kx + 11 = 0
⇒ x² + (2k - 2)x + 12 = 0
The given quadratic has no real roots, hence discriminant is less than 0.
⇒ (2k - 2)² - 4 × 1 × 12 < 0
⇒ 4(k - 1)² - 48 < 0
⇒ (k - 1)² < 12
Largets integral value of k satisying above inequality is 4.
Now, we have k/4y + 9y
= 4/4y + 9y
= 1/y + 9y
We know AM ≥ GM

⇒ (1/y + 9y)/2 ≥ 3
⇒ 1/y + 9y ≥ 6
∴ Least possible value of 1/y + 9y = 6.
Hence, 6.

Ans: 9
Let the roots of the given quandratic equation be p and q.


Squaring (1) we get


⇒ pq = 9/2 = c
From (2) we get
⇒ 9(p² + q²) = 5p²q²
⇒ 9(p² + q²) = 5(pq)²  
⇒ 9((p + q)² - 2pq) = 5(9/2)²  
⇒ (p + q)² - 9 = 45/4
⇒ (p + q)² = 81/4
⇒ (p + q) = ± 9/2 = - b
largest possible value of b = 9/2
∴ Largest possible value of a + b = 9/2 + 9/2 = 9.
Hence, 9.

Ans: 4 
Given,

For this to be equal to 1
Case 1: x² - 10 = 1 

Rejected as x is not an integer.
Case 2: x² - 3x - 10 = 0
⇒ (x - 5)(x + 2) = 0
⇒ x = 5 or -2 i.e., 2 integral values of x.
Case 3: x² - 10 = -1 and x² - 3x - 10 = even
If x² - 10 = -1 ⇒ x = ±3
For both x = +3 and -3 x² - 3x - 10 is even, hence, 2 integral values of x.
⇒ Total 4 integral values of x are possible.
Hence, 4.

Ans: 34
Let the price of small, medium and large cups be Rs. 2x, 5x and p respectively.
⇒ 2x × 5x × p = 800   ...(1)
Also, (2x + 6) × (5x + 6) × p = 3200   ...(2)
(2) = (1) × 4
⇒ (2x + 6) × (5x + 6) × p = 2x × 5x × p × 4
⇒ 10x² + 42x + 36 = 40x²
⇒ 30x² - 42x - 36 = 0
⇒ 5x² - 7x - 6 = 0
⇒ 5x² - 10x + 3x - 6 = 0
⇒ (5x + 3)(x - 2) = 0
⇒ x = 2
∴ Price of small cup = 2x = 4
Price of medium cup = 5x = 10
Price of large cup = 800/(4 × 10) = 20
⇒ Sum of the prices of three cups = 4 + 10 + 20 = 34.
Hence, 34.

Ans: 1
Let x + 1/x = y
⇒ y² - 3y + 2 = 0
⇒ (y - 1)(y - 2) = 0
⇒ y = 1 or 2
We know, sum of a number and its reciprocal are either ≤ 2 or ≥ 2.
∴ y = 2⇒ x + 1/x = 2
This is only possible when x = 1 hence, only one real root.
Hence, 1.

Ans: 10
Payment for 12 months = 90 + t {Assuming t as the value of a turban}
Payment for 9 months should be ¾(90 + t)
Payment for 9 months is given to us as 65 + t
Equating the two values we get
¾(90 + t) = 65 + t
270 + 3t = 260 + 4t
t = 10 Rs.

Ans: 15
3H + 5M + 1F = 23.50
5H + 9M + 1F = 39.50
2H + 2M + 2F = ?
Calculate 2(Equation 1) - (Equation 2)
H + M + F = 2×23.5 - 39.5
H + M + F = 7.5
2H + 2M + 2F = 15.

Ans: 28 
x can take any integer value from [-7,7].
So, there are 15 valid values of x.
For each of these values, there are 2 corresponding values of y. eg: For x = 3; y can be 4 or -4.
Except when x = 7 or -7; where the only possible value of y is 0.
Total valid values of x = 13×2 + 1 + 1 = 28.

Ans: 79
We have to find out a + b + c
f(1) is 1 + a + b + c
So, we need to find out f(1) - 1
Let the 4^th root be r
Coefficient of x³ is - (Sum of the roots)
0 = - (r + 2 -3 + 5)
r = - 4
So, f(x) = (x - 2) (x + 3) (x + 4)(x - 5)
f(1) = (-1)×4×5×(-4) = 80
a + b + c = f(1) - 1 = 79. 

Ans: 2
x =2+2^2/3+2^1/3
x - 2 = 2^2/3+2^1/3
(x - 2)³ = (2^2/3+2^1/3)³
x³ - 6x² + 12x - 8 = 2² + 3. 2^4/3.2^1/3 + 3. 2^2/3.2^2/3 + 2
x³ - 6x² + 12x - 8 = 4 + 3.2^5/3 + 3.2^4/3 + 2
x³ - 6x² + 12x - 8 = 6 + 6.2^2/3 + 6.2^1/3 = (12 + 6.2^2/3 + 6.2^1/3) - 6
x³ - 6x² + 12x - 8 = 6x - 6
x³ - 6x² + 6x = 2. 

Ans: 2

as the value of the expression will be positive, we can reject x=-1.
Hence, x=2.

Ans: 98
Writing the equation as
(x - α)(x - β)(x - γ) = 0, expanding and equating coefficients we get :
αβγ = -3
αβ + αγ + βγ = -7
α + β + γ = 0
From
α²+ β² +γ² = (α + β + γ)²-2(αβ + αγ + βγ) = 14
α²β² + α²γ² + α²β² = (αβ + αγ + βγ)²- 2 (α²βγ + αβ²γ + αβγ²)
= (αβ + αγ + βγ)² - 2αβγ(α + β + γ)
=49
Then α⁴ + β⁴ + γ⁴ = (α² + β² + γ²)² - 2 (α²β² + α²γ² + α²β²)
=14²-2.49 = 98