Test: 35 Year JEE Previous Year Questions: Quadratic Equation and Inequations (Inequalities)

  We have α² = 5α – 3 and β² = 5β – 3;
⇒ α & β are roots of equation,
x² = 5x – 3 or x² – 5x + 3 = 0
∴ α + β  = 5 and  αβ = 3

Thus, the equation havingas its roots is

= 0  or  3x² – 19x +3 = 0

Let α, β and γ, δ be the roots of the equations
x² + αx + β = 0 and
x² + βx + α = 0 respectively.
∴ α + β = –α, αβ = β and γ + δ = –β, γ δ = α.
Given |α – β| = |γ – δ|
⇒ (α – β)² = (γ – δ)²
⇒ (α + β)² – 4αβ = (γ + δ)² – 4γδ
⇒ α²  – 4β = β² – 4α
⇒ (α² – β²) + 4(α – β) = 0
⇒ α + β + 4 = 0 (∵ α ≠ β)

Product of real roots = 

∴ Product of real roots is always positive.

p + q = – p and pq = q
⇒ q (p – 1) = 0 ⇒ q = 0 or p = 1.
If q = 0, then p = 0. i.e.p = q
∴ p = 1 and q = –2.

∵ (a – b)² +  (b – c)² + (c – a)² > 0
⇒ 2(a² + b² + c² – ab – bc – ca) >0
⇒ 2 > 2(ab + bc + ca)
⇒ ab + bc + ca < 1

 = 

As for given conditon, α + β =

On simplification 2a²c = ab² + bc²

 are in A.P..

 are in H.P

Let the roots of given equation be α and 2α then

 = 

or  39a = 26 or 

 No.of solution 4

 or 

For max. or min., 

 = 2(+ve minima)            ∴x = 1

Let two numbers be a and b then  and

∴ Equation with roots a  and b is

x² - (a + b)x + ab=0 ⇒ x² -18x + 16= 0

Let the second root be α.
Then α + (1 - p) =-p
⇒ α =-1 Also α .(1 - p) =1-p
⇒ (α - 1)(1- p) = 0
⇒ p = 1[∵α = -1]
∴ Roots are α = -1 and p - 1=0

4 is a root of x² + px + 12=0
⇒ 16 + 4p +12 = 0
⇒ p =-7
Now, the equation x² + px +q= 0 has equal roots.

∴ p² - 4q = ⇒

are the roots of ax² + bx +c= 0

⇒ – b = a – c   or   c = a + b.

both roots are less than 5 then
(i) Discriminant ≥ 0
(ii) p(5) > 0
(iii) 

Hence (i ) 4k²– 4(k² + k – 5) ≥ 0  
4k² – 4k² – 4k + 20 ≥ 04k ≤ 20
⇒ k ≤ 5

(ii) f(5) > 0 ; 25 – 10 k + k² + k – 5 > 0
or  k² – 9k + 20 > 0
or  k (k – 4) –5(k – 4) > 0
or  (k – 5) (k – 4) > 0
⇒ k∈( –∞, 4 ) U ( –∞, 5)

(ii) 

The interection of (i), (ii) & (iii) gives k ∈ ( – ∞, 4 ).

x² + px +q= 0
Sum of roots = tan30° + tan15° = – p
Product of roots = tan30° . tan15° = q

⇒ – p =1-q ⇒ q-p=1
∴ 2+q-p=3

Equation x² - 2mx +m² - 1=0  
(x -m)² -1=0 or (x -m+1)(x -m-1)= 0
x = m - 1,m+1
m – 1 > –2  and m + 1<4
⇒ m >- 1 and m < 3 or,,    -1 < m<3

3x²(y -1) + 9x(y -1) + 7y -17=0

D ≥ 0 ∵ x is real

81( y - 1)² - 4 x 3( y - 1)(7y - 17)≥ 0
⇒ ( y - 1)(y - 41) ≤ 0
⇒ 1 ≤ y ≤ 41
∴ Max value of y is 41

Let α and β are roots of the equation x² + αx + 1 = 0
So, α + β = – α and αβ = 1
given

⇒ a² – 9 < 0 ⇒ a² < 9 ⇒ – 3 < a < 3
⇒ a ∈ (–3, 3)

Statement 2 is 

 which is true

Now 

Also      ∴ Adding all, we get

Hence both the statements are correct and statement 2 is a correct explanation of statement-1.

Let the roots of equation x² – 6x + α = 0 be α and 4 β and that of the equation
x² –cx + 6 = 0  be α and 3β .
Then α + 4β = 6 ; 4aβ = α and α + 3β = c ; 3αβ = 6
⇒ α = 8
∴ The equation becomes  x² – 6x + 8 = 0
⇒ (x –2) (x – 4) = 0
⇒ roots are 2 and 4 ⇒ α = 2, β = 1
∴ Common root is 2.

Given that roots of the equation bx² + cx + a = 0 are imaginary
∴ c² – 4ab < 0 ....(i)

Let y = 3b²x² + 6 bc x + 2c²
⇒ 3b²x² + 6 bc x + 2c² – y = 0
As x is real, D ≥ 0
⇒ 36 b²c² – 12 b² (2c² – y ) ≥ 0
⇒ 12 b² (3 c² – 2 c²+ y ) ≥ 0
⇒ c² + y ≥ 0
⇒ y ≥ – c²
But from eqn.
(i), c² < 4ab   or  – c² > – 4ab
∴ we get y ≥ – c² > – 4ab
⇒ y > – 4 ab

Given that  

Now  |Z| = 

⇒ 

⇒ 

x² -x + 1=0 

Given equation is e^sinx – e^–sinx – 4 = 0
Put e^{sin x} = t in the given equation, we get t² – 4t – 1 = 0

⇒

⇒ ( ∵ t=e^sinx)

⇒

⇒and 

So rejected                       So, rejected

Hence given equation has no solution.
∴ The equation has no real roots.

f (x) = 2x³ + 3x + k
f'(x) = 6x² + 3 > 0 ∀ x ∈R(∵ x² > 0)
⇒ f(x) is strictly increasing function
⇒ f(x) = 0 has only one real root, so two roots are not possible.

From the given system, we have

(∵ System has no solution)

⇒ k² + 4k + 3 = 8k
⇒ k = 1, 3
If  k = 1  then which is false

And if k = 3

then which is true, therefore k = 3

Hence for only one value of k. System has no solution.

Given equations are x² + 2x + 3 = 0 …(i)
ax² + bx + c = 0 …(ii)
Roots of equation (i) are imaginary roots.
According to the question (ii) will also have both roots same as (i). Thus

  (say) 
Hence, required ratio is 1 : 2 : 3

Consider –3(x – [x])² + 2 [x – [x]) + a² = 0
⇒ 3{x}² – 2{x} –a² = 0 (∵ x – [x] = {x})

⇒

Now, {x} ∈ (0,1) and (by graph)

Since , x is not an integer
∴ a ∈ (-1,1)- {0} ⇒ a ∈ (-1, 0)∪(0,1)

Let p, q, r are in AP ⇒ 2q = p + r ...(i)

Given  = 4 ⇒​

We have a + b = – q/p and ab = 

  4 ⇒​ q = -4r        ....(ii)

From (i), we have 2( – 4r) = p + r  
⇒  p = –9r
q = – 4r

Now 

(x² - 5 x + 5) ^{x2 + 4x-60}=1
Case I x² – 5x + 5 = 1 and
x² + 4x – 60 can be any real number
⇒ x = 1, 4
Case II x² – 5x + 5 = –1 and
x² + 4x – 60 has to be an even number
⇒ x = 2, 3 where 3 is rejected because for x = 3,
x² + 4x – 60 is odd.
Case III x² – 5x + 5 can be any real number and
x² + 4x – 60 = 0
⇒ x = –10, 6
⇒ Sum of all values of x = –10 + 6 + 2 + 1 + 4 = 3