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If α ≠ β but α2 = 5α – 3 and β2 = 5β – 3 then the equation having α/β and β/α as its roots is [2002]
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
Difference between the corresponding roots of x2+ax+b=0 and x2+bx+a=0 is same and a ≠ b, then [2002]
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 of the equation t2x2+|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 x2+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 a2+b2+c2=1 then ab + bc + ca is [2002]
∵ (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
If the sum of the roots of the quadratic equation 2ax2 + 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 2a2c = ab2 + bc2
are in A.P..
are in H.P
The value of ' a' for which one root of the quadratic equation (a2 -5a + 3) x2 +(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 x2 - 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
x2 - (a + b)x + ab=0 ⇒ x2 -18x + 16= 0
If (1- p) is a root of quadratic equation x2 + 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) =1-p
⇒ (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 x2 + px + 12=0 is 4, wh ile the equation x2 + px +q= 0 has equal r oots , then th e value of ‘q’ is[2004]
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 = ⇒
In a triangle PQR, . If tan
and – tan
are the roots of ax2 + bx + c = 0, a ≠ 0 then [2005]
are the roots of ax2 + bx +c= 0
⇒ – b = a – c or c = a + b.
If both the roots of the quadratic equation x2 - 2kx + k2 + 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 ) 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 ).
If the roots of the quadratic equation x2 + px +q= 0 are tan30° and tan15°, respectively, then the value of 2 + q – p is [2006]
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
All the values of m for which both roots of the equation x2 - 2mx +m2 - 1=0 are greater than – 2 but less then 4, lie in the interval [2006]
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
If x is real, the maximum value of is [2006]
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
If the difference between the roots of the equation x2 + 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 x2 + αx + 1 = 0
So, α + β = – α and αβ = 1
given
⇒ a2 – 9 < 0 ⇒ a2 < 9 ⇒ – 3 < a < 3
⇒ a ∈ (–3, 3)
Statement-1 : For every natural number n ≥ 2,
Statement-2 : 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 statement-1.
The quadritic equations x2 – 6x + a = 0 and x2 – 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 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.
If the roots of the equation bx2 + cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2 + 6bcx + 2c2 is : [2009]
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
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 x2 – x + 1 = 0, then α2009 + β2009 = [2010]
x2 -x + 1=0
The equation esinx – e–sinx– 4 = 0 has : [2012]
Given equation is esinx – e–sinx – 4 = 0
Put esin x = t in the given equation, we get t2 – 4t – 1 = 0
⇒
⇒ ( ∵ t=esinx)
⇒
⇒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, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1] [JEE M 2013]
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.
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)
⇒ 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.
If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0, a,b,c ∈ R, have a common root, then a : b : c is [JEE M 2013]
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
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]) + 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 α and β be the roots of equation px2 + 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 x2 – 6x – 2 = 0. If an = αn – βn, for n ≥ 1, then the value is equal to :[JEE M 2015]
The sum of all real values of x satisfying the equation (x2 - 5 x+ 5) x2 +4x- 60 = 1 is : [JEE M 2016]
(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
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