Test: Single Correct MCQs: Quadratic Equation and Inequations (Inequalities) | JEE Advanced

ℓ, m, n are real, ℓ ≠ m

Given equation is

(ℓ - m) x ² - 5(ℓ + m) x - 2(ℓ -m)= 0

D = 25(ℓ + m)² + 8(ℓ - m)² > 0,ℓ, m∈R

∴ Roots are real and unequal.

The given equations are
x  + 2y + 2z = 1 ....(1)
and 2x + 4y + 4z = 9 ....(2)
Subtracting (1) × (2) from (2),
we get 0 =7 (not possible)
∴ No solution.

u = x² + 4y² + 9z² – 6yx – 3zx – 2xy

 ∴ u is always non-negative.

As a, b, c > 0, a, b, c should be real (note that order relation is not defined in the set of complex numbers)
∴ Roots of equation are either real or complex conjugate.
Let α, β be the roots of ax² +bx + c = 0, then

⇒ Either both α, β are – ve (if roots are real) or both α, β have – ve real parts (if roots are complex conjugate)

The given equation is (x -b)(x -c) + (x- a)(x - c) + (x - a)(x -b)= 0
⇒ 3x² - 2(a+ b +c)x + (ab +bc +ca)=0

Discriminant = 4(a + b + c)² – 12(ab + bc + ca)

= 4[a² + b² + c² - ab - bc- ca]

= 2 [(a -b)² +(b -c)² +(c -a)² ] ≥ 0 ∀ a, b, c

∴ Roots of given equation are always real.

Let y = 2 log₁₀ x – log_x 0.01

[Here x  > 1 ⇒ log₁₀ x > 0] Now since sum of a real + ve number and its reciprocal is always greater than or equal to 2.

∴

∴ Least value of y is 4.

As (x² + px + 1) is a factor of ax³ + bx + c, we can assume that zeros of  x² + px + 1 are a, b and that of ax³ + bx + c be α, β, γ so that

α + β = – p .... (i)

αβ = 1 .... (ii)

and  α + β + γ = 0 .... (iii)

Solving (ii) and (v) we get γ = – c / a.
Also from (i) and (iii) we get γ = p

∴ p = g = – c / a Using equations (i) , (ii) and (iv) we get

  (using  γ = p = – c / a)

a² – c²= ab

∴ (c) is the  correct answer.

| x |² – 3 | x | + 2 = 0
Case  I   : x < 0 then | x | = – x
⇒ x² + 3x + 2 = 0  
⇒ (x + 1)  (x + 2) = 0 x = – 1, – 2 (both acceptable as < 0)
Case II: x > 0 then | x | = x
⇒ x² – 3x + 2= 0
⇒ ( x – 1) (x – 2) = 0
x = 1, 2 ( both acceptable as > 0)
∴ There are 4 real solutions.

Let the distance of school from A = x
∴ The distance of the school form B = 60 – x
Total distance covered by 200 students = 2[150 x + 50 (60 – x) ] = 2[100 x + 3000] This is min., when x = 0
∴ school should be built at town  A.

if p = 5, q = 3, r = 2 max (p, q) = 5    ;    max (p, q, r) = 5
⇒ max (p, q) = max (p, q, r)
∴ (a) is not true. Similarly we can show that (c) is not true.

Also min (p,q) = (p + q- |q - p|)
Let p < q then LHS = p

and  R.H.S. = (p + q- q+ p) =p

Given expression x¹² - x⁹ + x⁴ - x + 1= f (x) (say)

For x < 0 put x = – y where y > 0

then we get f ( x) = y¹² + y⁹ +y⁴ +y +1>0 for y > 0

For 0 < x < 1,   x⁹ < x ⁴ ⇒ - x9 +x⁴ > 0

Also  1– x  > 0 and x¹² > 0

⇒ x¹² – x⁹ + x⁴  + 1– x > 0    ⇒ f (x) > 0

For x > 1

f (x) = x (x³  – 1) (x⁸ + 1) + 1 > 0

So f (x) > 0 for –∞< x <∞ 

Given equation is 

Clear ly x ≠ 1 for th e given eq. to be defin ed. If 

x– 1 ≠ 0, we can cancel the common term     on both sides to get x = 1, but it is not possible. So given eq. has no roots.

∴ (a) is the correct answer.

 Given that a² + b² + c² = 1 ....(1) We know (a + b +c)²  ≥ 0

⇒ a² + b² + c² + 2ab + 2bc + 2ca ≥ 0

⇒ 2 ( ab + bc + c a ) ≥-1 [Using (1)]

⇒ ab + bc + ca ≥-1 / 2....(2) ....(2) Also we know that

Þ a² + b² + c²- ab- bc - ca≥ 0

Þ ab + bc + ca ≤ 1     [Using (1)]  .......(2)

Combining (2) and (3), we get

-1/ 2 ≤ ab + bc + ca≤1 ∴ab + bc + ca ∈ [-1/ 2,1]

∴ (c) is the correct answer.

First of all for log (x – 1) to be defined, x – 1 > 0 ⇒ x  >  1 ....(1)
Now, log_0.3 (x – 1) < log_0.09  (x – 1)
⇒ log_0.3 (x – 1) < log _(0.3)²  (x – 1)
⇒ log_0.3 (x – 1) < log_0.3  (x – 1)
⇒ 2 log_0.3 (x – 1) < log_0.3  (x – 1)
⇒ log_0.3 (x – 1)² < log_0.3  (x – 1)
⇒ (x – 1)²  >  (x – 1) NOTE THIS STEP
[The inequality is reversed since base lies between 0 and 1]
⇒ (x – 1)²  –  (x – 1) > 0 ⇒ (x – 1)  (x – 2) > 0  ....(2)
Combining (1) and (2) we get x  >  2
∴ x ∈ (2,∞)

From the first method,

q = αβ, r = α⁴+β⁴ ...(3)

Product of the roots of the equation x² – 4qx + (2q² – r) = 0

= 2q² – r = 2α²β² – α⁴ – β⁴ = – (α² – β²)² [From (3)]

= – (positive quantity) = – ve quantity
⇒  one root is positive and other is negative

KEY CONCEPT : If f (α) and f (β) are of opposite signs then there must lie a value γ between α and β such that f (γ) = 0.
a, b, c are real numbers and a ≠ 0.
As α is a root of a²x² + bx + c = 0
∴ a²α² + bα + c = 0 ....(1)
Also b is a root of a2x2 – bx – c = 0  ∴
a²β² – bβ –  c = 0 .... (2)
Now, let f (x) = a²x² + 2bx + 2c
Then f (α) = a²α²  + 2bα + 2c = a²α² + 2(bα + c)
= a²α² + 2(– a²α²) [Using eq. (1)]
= – a²α².
and f (β) = a²β² + 2bβ + 2c
= a²β² + 2(bβ + c)
= a²β² + 2(a²β²) [Using eq. (2)]
= 3a²β² > 0.
Since f (α) and f (β) are of opposite signs and γ is a root of equation f (x) = 0
∴ γ must lie between α and β
Thus α < γ < β.     ∴    (d) is the correct option.

The given eq. is sin (ex) = 5^x + 5^–x We know 5^x and 5^–x both are +ve real numbers using

AM ≥ GM 

∴ R.H.S. of given eq. ≥ 2 While sin ex ∈ [-1,1] i . e . LHS ∈ [-1,1]
∴ The equation is not possible for any real value of x.
Hence (a) is the correct answer.

α, β are roots of the equation (x – a) (x – b) = c, c ≠ 0
∴ (x – a) (x – b) – c = (x – α)(x – β)
⇒ (x – α) (x – β) + c = (x – a)(x – b)
⇒ roots of (x – α) (x – β) + c = 0 are α and β.
∴ (c) is the correct option.

We have

while y = 

⇒ The two curves do not meet at all.

For real roots q² - 4 pr≥0

(∵ p,  q, r are in A.P.)

For the equation px² + qx + 1 = 0 to have real roots

D ≥ 0 ⇒ q² ≥ 4p

If  p = 1 then  q ² ≥ 4 ⇒ q = 2, 3, 4

If  p = 2 then  q ² ≥ 8 ⇒ q = 3, 4

If  p = 3 then  q² ≥ 12 ⇒ q =4

If  p = 4 then  q² ≥ 16 ⇒ q = 4

∴ No. of req. equations = 7.

KEY CONCEPT : If both roots of a quadratic equation ax² + bx + c = 0 are less than k then af(k)

> 0, D ≥ 0, α + β < 2 k.

f(x) = x² – 2ax + a² + a – 3 = 0,

f(3) > 0, α + β < 6, D ≥ 0

⇒ a² – 5a + 6 > 0, a < 3, – 4a + 12 ≥ 0

⇒ a < 2 or a > 3, a < 3, a < 3  ⇒ a < 2.

Given c < 0 < b   and α + β = – b ....(1) αβ = c
....(2)

From (2), c < 0 ⇒ αβ < 0 ⇒ either a is  -ve or β is - ve and second ;quantity is positive.

from (1), b > 0 ⇒ – b < 0  ⇒ α + β < 0 ⇒ the sum is negative

⇒ modules of nengative quantity is > modulus of positive quantity but α< β is given.
Therefore, it is clear that α is negative and β is positive and modulus of α is greater than modulus of β ⇒ α< 0 <β< |α|

As A.M. ≥ G.M.  for positive real numbers,  we get

(Putting values)

Also (a + b) (c + d) > 0             [∴    a,b,c,d > 0]

∴ 0 ≤ M≤1

The given equation is ( x – a) (x – b) – 1 = 0, b > a.

or x² – (a + b) x + ab –1= 0

Let  f(x) = x² – (a + b) x + ab – 1

Since coeff. of x² i.e. 1 > 0,
∴ it represents upward parabola, intersecting x - axis at two points. (corresponding to two real roots, D being +ve).

Also f (a) = f (b) = – 1 ⇒ curve is below x-axis at a and b

⇒ a and b both lie between the roots.
Thus the graph of given eqn is as shown.

from graph it is clear that one root of the equation lies in (– ∞, a) and other in (b, ∞).

Let α, α² be the roots of  3x² + px + 3.

∴ α + α² = – p / 3 and α³ = 1

⇒ (α – 1) (α² + α + 1) = 0 ⇒ α =1 or  α² + α  = – 1

If α = 1, p = – 6 which is not possible as p > 0

If α² + α = –1  ⇒ – p/3= –1 ⇒ p = 3.

We have

[Using A.M. ≥ G.M.]

 For x < – 2 ,   | x + 2 | = – (x + 2) and the  inequality becomes

x² + x + 2 + x  > 0     ⇒ (x + 1)²  + 1 > 0

which is valid ∀ x∈R but x < – 2

∴ x ∈ ( - ∞,- 2) ....(1)

For x ≥ 2, | x + 2 |=x+2 and the inequality becomes

x²  – x – 2 + x > 0 ⇒ x² > 2 

but  ....(2)

From (1) and (2)

Let   and

then using AM GM, we get 

⇒

⇒

KEY CONCEPT : f(x) = ax² + bx + c has same sign as that of a if  D < 0.

x² + 2ax + 10 - 3a >0∀x

⇒ D < 0 ⇒ 4a² - 4(10 - 3a)< 0 ⇒ a² + 3a - 10<0

⇒ (a + 5)(a - 2) < 0 ⇒a ∈ (-5, 2)

x² + px +q= 0

Let roots be a and a²

∴ (q)^1/3 + (q^1/3)² =-p

Taking cube of both sides, we get

∵ a, b, c are sides of a triangle and a ≠ b≠c

∴

Similarly, we have

b² + c² - 2bc < a² ; c² + a² -2ca<b²

On adding, we get

a² + b² + c² < 2(ab + bc+ ca)

 ....(1)

∵ Roots of the given equation are real

∴

....(2)

From (1) and (2), we get 

As α, β are the roots of  x² – px + r = 0

∴ α + β = p   ....(1)

and αβ = r   ....(2)

Also are the roots of  x² – qx + r = 0

∴....(3)

Solving (1) and (3) for α and β ,we get

 and

Substituting values of α and β, in equation (2),

we get

Given that α + β= – p and α³ + β³= q

⇒ (α + β)³ – 3αβ (α + β)= q

⇒ – p³ – 3αβ (– p)=q ⇒

Now for required quadratic equation,

sum of roots =

and product of roots  

∴ Required equation is 

or  ( p³ + q)x² – ( p³ – 2q)x + ( p³ + q)= 0

We have (2x)^ℓn2 = (3y)^ℓn3
⇒ ℓn2. ℓn2x = ℓn3. ℓn3y
⇒ ℓn2. ℓn2x = ℓn3. (ℓn3 + ℓny) ...(1)
Also given 3ℓnx = 2lny

⇒ ℓnx. ℓn3 = ℓny. ℓn2 ⇒ ℓny =

Substituting this value of ℓny in equation (1), we get

ℓn2. ℓn2x = ℓn3 

⇒ (ℓn2)² ℓn2x = (ℓn3)² ℓn2 + (ℓn3)² ℓnx

⇒ (ℓn2)² ℓn2x = (ℓn3)² (ℓn2 + ℓnx)

⇒ (ℓn2)² ℓn2x – (ℓn3)² ℓn2x = 0

⇒ [(ℓn2)² – (ℓn3)²] ℓn2x = 0  ⇒ ℓn2x = 0

⇒ 2x = 1   or x = 

∵ α, β are the roots of x² – 6x – 2 = 0

∴ α² – 6α – 2 = 0

⇒ α¹⁰ – 6α⁹– 2α⁸ = 0

⇒ α¹⁰ – 2α⁸ = 6α⁹ ...(1)

Similarly β¹⁰ – 2β⁸  = 6β⁹ ...(2)

From equation (1) and (2

α¹⁰ – β¹⁰ – 2 (α⁸– β⁸) = 6 (α⁹– β⁹)

⇒ a₁₀ – 2a₈ = 6a₉ 

Let α be the common root of given equations, then
α² + bα – 1 = 0 ...(1)
and α² + α + b = 0 ...(2)
Subtracting (2) from (1),
we get (b – 1) α – (b + 1) = 0

Substituting this value of α in equation (1), we get

= 0  or  b³ + 3b = 0

Quadratic equation with real coefficients and purely imaginary roots can be considered as

p(x) = x² + a = 0 where a > 0 and a ∈R
The p[ p(x)] = 0 ⇒ (x² + a)² + a = 0
⇒ x⁴ + 2ax² + (a² + a) = 0

where a, b≠0

∴ p[p(x)] = 0 has complex roots which are neither purely real nor purely imaginary.

x² – 2x secθ + 1 = 0 ⇒ x = secθ ± tanθ and x² + 2x tanθ – 1 = 0 ⇒ x = –tanθ ± secθ

α₁,β₁ are roots of x² – 2x secθ + 1 = 0 and α₁> b₁

∴ α₁ = secθ – tanθ and b₁ = secθ + tanθ α₂,
β₂ are roots of x² + 2x tanθ – 1 = 0 and α₂ > β₂
∴ α₂ = -tanθ + secθ, β₂ = – tanθ – secθ
∴ α₁ + β₂ = secθ – tanθ – tanθ – secθ = – 2tanθ