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If α ≠ β but α^{2} = 5α – 3 and β^{2 }= 5β – 3 then the equation having α/β and β/α as its roots is [2002]
We have α^{2} = 5α – 3 and β^{2} = 5β – 3;
⇒ α & β are roots of equation,
x^{2} = 5x – 3 or x^{2} – 5x + 3 = 0
∴ α + β = 5 and αβ = 3
Thus, the equation havingas its roots is
= 0 or 3x^{2} – 19x +3 = 0
Difference between the corresponding roots of x^{2}+ax+b=0 and x^{2}+bx+a=0 is same and a ≠ b, then [2002]
Let α, β and γ, δ be the roots of the equations
x^{2} + αx + β = 0 and
x^{2} + 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
⇒ α^{2} – 4β = β^{2} – 4α
⇒ (α^{2} – β^{2}) + 4(α – β) = 0
⇒ α + β + 4 = 0 (∵ α ≠ β)
Product of real roots of the equation t^{2}x^{2}+x+9=0 [2002]
Product of real roots =
∴ Product of real roots is always positive.
If p and q are the roots of the equation x^{2}+px+q=0, then
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.
If a, b, c are distinct +ve real numbers and a^{2}+b^{2}+c^{2}=1 then ab + bc + ca is [2002]
∵ (a – b)^{2} + (b – c)^{2} + (c – a)^{2} > 0
⇒ 2(a^{2} + b^{2} + c^{2} – ab – bc – ca) >0
⇒ 2 > 2(ab + bc + ca)
⇒ ab + bc + ca < 1
If the sum of the roots of the quadratic equation 2ax^{2} + bx +c= 0 is equal to the sum of the squares of their reciprocals, then are in [2003]
=
As for given conditon, α + β =
On simplification 2a^{2}c = ab^{2 }+ bc^{2}
are in A.P..
are in H.P
The value of ' a' for which one root of the quadratic equation (a^{2} 5a + 3) x^{2} +(3a  1)x + 2= 0 is twice as large as the other is [2003]
Let the roots of given equation be α and 2α then
=
or 39a = 26 or
The number of real solutions of the equation x^{2}  3 x + 2 = 0 is
No.of solution 4
The real number x when added to its inverse gives the minimum value of the sum at x equal to [2003]
or
For max. or min.,
= 2(+ve minima) ∴x = 1
Let two numbers have arith metic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation[2004]
Let two numbers be a and b then and
∴ Equation with roots a and b is
x^{2}  (a + b)x + ab=0 ⇒ x^{2} 18x + 16= 0
If (1 p) is a root of quadratic equation x^{2} + px + (1 p)=0 th en its r oot are [2004]
Let the second root be α.
Then α + (1  p) =p
⇒ α =1 Also α .(1  p) =1p
⇒ (α  1)(1 p) = 0
⇒ p = 1[∵α = 1]
∴ Roots are α = 1 and p  1=0
If one r oot of th e equation x^{2} + px + 12=0 is 4, wh ile the equation x^{2 }+ px +q= 0 has equal r oots , then th e value of ‘q’ is[2004]
4 is a root of x^{2} + px + 12=0
⇒ 16 + 4p +12 = 0
⇒ p =7
Now, the equation x^{2} + px +q= 0 has equal roots.
∴ p^{2}  4q = ⇒
In a triangle PQR, . If tan and – tan are the roots of ax^{2} + bx + c = 0, a ≠ 0 then [2005]
are the roots of ax^{2} + bx +c= 0
⇒ – b = a – c or c = a + b.
If both the roots of the quadratic equation x^{2}  2kx + k^{2} + k – 5 = 0 are less than 5, then k lies in the interval [2005]
both roots are less than 5 then
(i) Discriminant ≥ 0
(ii) p(5) > 0
(iii)
Hence (i ) 4k^{2}– 4(k^{2} + k – 5) ≥ 0
4k^{2} – 4k^{2} – 4k + 20 ≥ 04k ≤ 20
⇒ k ≤ 5
(ii) f(5) > 0 ; 25 – 10 k + k^{2 }+ k – 5 > 0
or k^{2} – 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 ).
If the roots of the quadratic equation x^{2} + px +q= 0 are tan30° and tan15°, respectively, then the value of 2 + q – p is [2006]
x^{2} + px +q= 0
Sum of roots = tan30° + tan15° = – p
Product of roots = tan30° . tan15° = q
⇒ – p =1q ⇒ qp=1
∴ 2+qp=3
All the values of m for which both roots of the equation x^{2}  2mx +m^{2}  1=0 are greater than – 2 but less then 4, lie in the interval [2006]
Equation x^{2}  2mx +m^{2}  1=0
(x m)^{2} 1=0 or (x m+1)(x m1)= 0
x = m  1,m+1
m – 1 > –2 and m + 1<4
⇒ m > 1 and m < 3 or,, 1 < m<3
If x is real, the maximum value of is [2006]
3x^{2}(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
If the difference between the roots of the equation x^{2} + ax + 1 = 0 is less than , then the set of possible values of a is [2007]
Let α and β are roots of the equation x^{2} + αx + 1 = 0
So, α + β = – α and αβ = 1
given
⇒ a^{2} – 9 < 0 ⇒ a^{2} < 9 ⇒ – 3 < a < 3
⇒ a ∈ (–3, 3)
Statement1 : For every natural number n ≥ 2,
Statement2 : For every natural number n ≥ 2,[2008]
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 statement1.
The quadritic equations x^{2 }– 6x + a = 0 and x^{2 }– cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is [2009]
Let the roots of equation x^{2} – 6x + α = 0 be α and 4 β and that of the equation
x^{2 }–cx + 6 = 0 be α and 3β .
Then α + 4β = 6 ; 4aβ = α and α + 3β = c ; 3αβ = 6
⇒ α = 8
∴ The equation becomes x^{2} – 6x + 8 = 0
⇒ (x –2) (x – 4) = 0
⇒ roots are 2 and 4 ⇒ α = 2, β = 1
∴ Common root is 2.
If the roots of the equation bx^{2} + cx + a = 0 be imaginary, then for all real values of x, the expression 3b^{2}x^{2} + 6bcx + 2c^{2} is : [2009]
Given that roots of the equation bx^{2} + cx + a = 0 are imaginary
∴ c^{2} – 4ab < 0 ....(i)
Let y = 3b^{2}x^{2} + 6 bc x + 2c^{2 }
⇒ 3b^{2}x^{2} + 6 bc x + 2c^{2 }– y = 0
As x is real, D ≥ 0
⇒ 36 b^{2}c^{2} – 12 b^{2 }(2c^{2} – y ) ≥ 0
⇒ 12 b^{2} (3 c^{2} – 2 c^{2}+ y ) ≥ 0
⇒ c^{2 }+ y ≥ 0
⇒ y ≥ – c^{2 }
But from eqn.
(i), c^{2} < 4ab or – c^{2} > – 4ab
∴ we get y ≥ – c^{2} > – 4ab
⇒ y > – 4 ab
If then the maximum value of  Z  is equal to : [2009]
Given that
Now Z =
⇒
⇒
If α and β are the roots of the equation x^{2} – x + 1 = 0, then α^{2009} + β^{2009} = [2010]
x^{2} x + 1=0
The equation e^{sinx }– e^{–sinx}– 4 = 0 has : [2012]
Given equation is e^{sinx} – e^{–sinx} – 4 = 0
Put e^{sin x} = t in the given equation, we get t^{2} – 4t – 1 = 0
⇒
⇒ ( ∵ t=e^{sinx})
⇒
⇒and
So rejected So, rejected
Hence given equation has no solution.
∴ The equation has no real roots.
The real number k for which the equation, 2x^{3} + 3x + k = 0 has two distinct real roots in [0, 1] [JEE M 2013]
f (x) = 2x^{3} + 3x + k
f'(x) = 6x^{2} + 3 > 0 ∀ x ∈R(∵ x^{2} > 0)
⇒ f(x) is strictly increasing function
⇒ f(x) = 0 has only one real root, so two roots are not possible.
The number of values of k , for which the system of equations : [JEE M 2013]
(k + 1) x + 8y = 4k
kx + (k + 3) y = 3k – 1
has no solution, is
From the given system, we have
(∵ System has no solution)
⇒ k^{2} + 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.
If the equations x^{2} + 2x + 3 = 0 and ax^{2} + bx + c = 0, a,b,c ∈ R, have a common root, then a : b : c is [JEE M 2013]
Given equations are x^{2} + 2x + 3 = 0 …(i)
ax^{2} + 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
If a ∈ R and the equation 3 ( x  [ x])^{2 }+ 2 ( x [ x])+a ^{2} =0 (where [x] denotes the greatest integer ≤ x ) has no integral solution, then all possible values of a lie in the interval: [JEE M 2014]
Consider –3(x – [x])^{2} + 2 [x – [x]) + a^{2} = 0
⇒ 3{x}^{2} – 2{x} –a^{2} = 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 α and β be the roots of equation px^{2} + qx +r= 0, p ≠ 0. If p, q, r are in A.P. and = 4 then the value of α β is: [JEE M 2014]
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
Let α and β be the roots of equation x^{2} – 6x – 2 = 0. If a_{n} = α^{n} – β^{n}, for n ≥ 1, then the value is equal to :[JEE M 2015]
The sum of all real values of x satisfying the equation (x^{2}  5 x+ 5) ^{x2 +4x 60 }= 1 is : [JEE M 2016]
(x^{2}  5 x + 5) ^{x2 + 4x60}=1
Case I x^{2 }– 5x + 5 = 1 and
x^{2} + 4x – 60 can be any real number
⇒ x = 1, 4
Case II x^{2} – 5x + 5 = –1 and
x^{2} + 4x – 60 has to be an even number
⇒ x = 2, 3 where 3 is rejected because for x = 3,
x^{2} + 4x – 60 is odd.
Case III x^{2} – 5x + 5 can be any real number and
x^{2} + 4x – 60 = 0
⇒ x = –10, 6
⇒ Sum of all values of x = –10 + 6 + 2 + 1 + 4 = 3
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