1 Crore+ students have signed up on EduRev. Have you? 
If ℓ, m, n are real, ℓ ≠ m, then the roots by the equation: (ℓ – m)x^{2} – 5 (ℓ + m) x – 2 (ℓ – m) = 0 are (1979)
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 equation x + 2y + 2z = 1 and 2x + 4y + 4z = 9 have
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.
If x, y and z are real and different and (1979) u = x^{2} + 4y^{2} + 9z^{2} – 6yz – 3zx – 2xy, then u is always.
u = x^{2} + 4y^{2} + 9z^{2} – 6yx – 3zx – 2xy
∴ u is always nonnegative.
Let a > 0, b >0 and c > 0. Then the roots of the equation ax^{2} + bx + c = 0 (1979)
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 ax^{2} +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)
Both the roots of the equation (x – b) (x – c) + (x – a) (x – c) + (x – a) (x – b) = 0are always
The given equation is (x b)(x c) + (x a)(x  c) + (x  a)(x b)= 0
⇒ 3x^{2}  2(a+ b +c)x + (ab +bc +ca)=0
Discriminant = 4(a + b + c)^{2} – 12(ab + bc + ca)
= 4[a^{2} + b^{2} + c^{2}  ab  bc ca]
= 2 [(a b)^{2} +(b c)^{2 }+(c a)^{2} ] ≥ 0 ∀ a, b, c
∴ Roots of given equation are always real.
The least value of the expression 2 log_{10}x – log_{x}(0.01), for x > 1, is(1980)
Let y = 2 log_{10} x – log_{x }0.01
[Here x > 1 ⇒ log_{10} 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.
If (x^{2} + px + 1) is a factor of (ax^{3 }+ bx + c), then (1980)
As (x^{2} + px + 1) is a factor of ax^{3} + bx + c, we can assume that zeros of x^{2 }+ px + 1 are a, b and that of ax^{3} + 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)
a^{2} – c^{2}= ab
∴ (c) is the correct answer.
The number of real solutions of the equation  x ^{2} – 3  x  + 2 = 0 is (1982  2 Marks)
 x ^{2} – 3  x  + 2 = 0
Case I : x < 0 then  x  = – x
⇒ x^{2} + 3x + 2 = 0
⇒ (x + 1) (x + 2) = 0 x = – 1, – 2 (both acceptable as < 0)
Case II: x > 0 then  x  = x
⇒ x^{2} – 3x + 2= 0
⇒ ( x – 1) (x – 2) = 0
x = 1, 2 ( both acceptable as > 0)
∴ There are 4 real solutions.
Two towns A and B are 60 km apart. A school is to be built to serve 150 students in town A and 50 students in town B. If the total distance to be travelled by all 200 students is to be as small as possible, then the school should be built at (1982  2 Marks)
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, q, r are any real numbers, then (1982  2 Marks)
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
The largest interval for which x^{12} – x^{9} + x^{4} – x + 1 > 0 is (1982  2 Marks)
Given expression x^{12}  x^{9} + x^{4}  x + 1= f (x) (say)
For x < 0 put x = – y where y > 0
then we get f ( x) = y^{12} + y^{9} +y^{4} +y +1>0 for y > 0
For 0 < x < 1, x^{9} < x^{ 4} ⇒  x9 +x^{4 }> 0
Also 1– x > 0 and x^{12} > 0
⇒ x^{12} – x^{9} + x^{4} + 1– x > 0 ⇒ f (x) > 0
For x > 1
f (x) = x (x^{3} – 1) (x^{8} + 1) + 1 > 0
So f (x) > 0 for –∞< x <∞
The equation has (1984  2 Marks)
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.
If a^{2} + b^{2} + c^{2} = 1, then ab + bc + ca lies in the interval (1984  2 Marks)
Given that a^{2} + b^{2} + c^{2} = 1 ....(1) We know (a + b +c)^{2} ≥ 0
⇒ a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca ≥ 0
⇒ 2 ( ab + bc + c a ) ≥1 [Using (1)]
⇒ ab + bc + ca ≥1 / 2....(2) ....(2) Also we know that
Þ a^{2 }+ b^{2} + c^{2} 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.
If log_{0.3} (x – 1) < log_{0.09}(x – 1), then x lies in the interval – (1985  2 Marks)
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,∞)
If α and β are the roots of x^{2} + px + q = 0 and α^{4}, β^{4} are the roots of x^{2} – rx + s = 0, then the equation x^{2} – 4qx + 2q^{2} – r = 0 has always (1989  2 Marks)
From the first method,
q = αβ, r = α^{4}+β^{4} ...(3)
Product of the roots of the equation x^{2} – 4qx + (2q^{2} – r) = 0
= 2q^{2 }– r = 2α^{2}β^{2} – α^{4 }– β^{4} = – (α^{2} – β^{2})^{2} [From (3)]
= – (positive quantity) = – ve quantity
⇒ one root is positive and other is negative
Let a, b, c be real numbers, a ≠ 0. If α is a r oot of a^{2}x^{2 }+ bx + c = 0 . β is the root of a^{2}x^{2} – bx – c = 0 and 0 < α < β, then the equation a^{2}x^{2} + 2bx + 2c = 0 has a root g that always satisfies (1989  2 Marks)
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 a^{2}x^{2} + bx + c = 0
∴ a^{2}α^{2 }+ bα + c = 0 ....(1)
Also b is a root of a2x2 – bx – c = 0 ∴
a^{2}β^{2} – bβ – c = 0 .... (2)
Now, let f (x) = a^{2}x^{2 }+ 2bx + 2c
Then f (α) = a^{2}α^{2} + 2bα + 2c = a^{2}α^{2} + 2(bα + c)
= a^{2}α^{2} + 2(– a^{2}α^{2}) [Using eq. (1)]
= – a^{2}α^{2}.
and f (β) = a^{2}β^{2} + 2bβ + 2c
= a^{2}β^{2} + 2(bβ + c)
= a^{2}β^{2} + 2(a^{2}β^{2}) [Using eq. (2)]
= 3a^{2}β^{2} > 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 number of solutions of the equation sin(e)^{x} = 5^{x} + 5^{–x} is (1990  2 Marks)
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.
Let α, β be the roots of the equation (x – a) (x – b) = c, c ≠ 0.Then the roots of the equation (x – α) (x – β) + c = 0 are (1992  2 Marks)
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.
The number of poin ts of intersection of two curves y = 2 sinx and y = 5x^{2} + 2x + 3 is (1994)
We have
while y =
⇒ The two curves do not meet at all.
If p, q, r are +ve and are in A.P., the roots of quadratic equation px^{2} + qx + r = 0 are all real for (1994)
For real roots q^{2 } 4 pr≥0
(∵ p, q, r are in A.P.)
Let p,q ∈ {1,2, 3,4} . The number of equations of the form px^{2} + qx + 1 = 0 having real roots is (1994)
For the equation px^{2 }+ qx + 1 = 0 to have real roots
D ≥ 0 ⇒ q^{2} ≥ 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 q^{2} ≥ 12 ⇒ q =4
If p = 4 then q^{2} ≥ 16 ⇒ q = 4
∴ No. of req. equations = 7.
If the roots of the equation x^{2} – 2ax + a^{2} + a – 3 = 0 are real and less than 3, then (1999  2 Marks)
KEY CONCEPT : If both roots of a quadratic equation ax^{2 }+ bx + c = 0 are less than k then af(k)
> 0, D ≥ 0, α + β < 2 k.
f(x) = x^{2} – 2ax + a^{2} + a – 3 = 0,
f(3) > 0, α + β < 6, D ≥ 0
⇒ a^{2} – 5a + 6 > 0, a < 3, – 4a + 12 ≥ 0
⇒ a < 2 or a > 3, a < 3, a < 3 ⇒ a < 2.
If α and β (α < β) are the roots of the equation x^{2} + bx + c = 0, where c < 0 < β, then (2000S)
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
If a, b, c, d are positive real numbers such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation
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
If b > a, then the equation (x – a) (x – b) –1 = 0 has (2000S)
The given equation is ( x – a) (x – b) – 1 = 0, b > a.
or x^{2} – (a + b) x + ab –1= 0
Let f(x) = x^{2} – (a + b) x + ab – 1
Since coeff. of x^{2} 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 xaxis 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, ∞).
For the equation 3x^{2} + px + 3 = 0, p > 0, if one of the root is square of the other, then p is equal to (2000S)
Let α, α^{2} be the roots of 3x^{2} + px + 3.
∴ α + α^{2} = – p / 3 and α^{3 }= 1
⇒ (α – 1) (α^{2} + α + 1) = 0 ⇒ α =1 or α^{2} + α = – 1
If α = 1, p = – 6 which is not possible as p > 0
If α^{2} + α = –1 ⇒ – p/3= –1 ⇒ p = 3.
If a_{1},a_{2}.....,a_{n} are positive real numbers whose product is a fixed number c, then the minimum value of a_{1} + a_{2} + .......+a_{n1} + 2a_{n} is (2002S)
We have
[Using A.M. ≥ G.M.]
The set of all real numbers x for which x^{2} –  x + 2  + x > 0, is (2002S)
For x < – 2 ,  x + 2  = – (x + 2) and the inequality becomes
x^{2} + 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
x^{2} – x – 2 + x > 0 ⇒ x^{2} > 2
but ....(2)
From (1) and (2)
If then is always greater than or equal to (2003S)
Let and
then using AM GM, we get
⇒
⇒
For all ‘x’, x^{2} + 2ax + 10– 3a > 0, then the interval in which ‘a’ lies is (2004S)
KEY CONCEPT : f(x) = ax^{2} + bx + c has same sign as that of a if D < 0.
x^{2} + 2ax + 10  3a >0∀x
⇒ D < 0 ⇒ 4a^{2}  4(10  3a)< 0 ⇒ a^{2} + 3a  10<0
⇒ (a + 5)(a  2) < 0 ⇒a ∈ (5, 2)
If one root is square of the oth er root of the equation x^{2} + px + q = 0, then the relation between p and q is (2004S)
x^{2} + px +q= 0
Let roots be a and a^{2}
∴ (q)^{1/3} + (q^{1/3})^{2} =p
Taking cube of both sides, we get
Let a, b, c be the sides of a triangle where a ≠ b ≠ c and λ ∈ R. If the roots of the equation x^{2} + 2(a + b + c)x + 3λ (ab + bc + ca) = 0 are real, then (2006  3M, –1)
∵ a, b, c are sides of a triangle and a ≠ b≠c
∴
Similarly, we have
b^{2} + c^{2}  2bc < a^{2} ; c^{2} + a^{2} 2ca<b^{2}
On adding, we get
a^{2 }+ b^{2} + c^{2 }< 2(ab + bc+ ca)
....(1)
∵ Roots of the given equation are real
∴
....(2)
From (1) and (2), we get
Let α, β be the roots of the equation x^{2} – px + r = 0 and be the roots of the equation x^{2} – qx + r = 0. Then the value of r is (2007 3 marks)
As α, β are the roots of x^{2} – px + r = 0
∴ α + β = p ....(1)
and αβ = r ....(2)
Also are the roots of x^{2} – 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
Let p and q be real numbers such that p ≠ 0, p^{3 }≠q and p^{3} ≠q . If a and b are nonzero complex numbers satisfying α + β = – p and α^{3} + β^{3} = q, then a quadratic equation having as its roots is (2010)
Given that α + β= – p and α^{3} + β^{3}= q
⇒ (α + β)^{3 }– 3αβ (α + β)= q
⇒ – p^{3 }– 3αβ (– p)=q ⇒
Now for required quadratic equation,
sum of roots =
and product of roots
∴ Required equation is
or ( p^{3 }+ q)x^{2 }– ( p^{3} – 2q)x + ( p^{3} + q)= 0
Let (x_{0}, y_{0}) be the solution of the following equations
(2x)^{ℓn2} = (3y)^{ℓn3}
3^{ℓnx }=2^{ℓny}
Then x_{0 }is (2011)
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 =
Let α and β be the roots of x^{2} – 6x – 2 = 0, with α > β. If a_{n }= α^{n}  β^{n} for n ≥ 1, then the value of is (2011)
∵ α, β are the roots of x^{2} – 6x – 2 = 0
∴ α^{2 }– 6α – 2 = 0
⇒ α^{10 }– 6α^{9}– 2α^{8 }= 0
⇒ α^{10} – 2α^{8 }= 6α^{9} ...(1)
Similarly β^{10} – 2β^{8} = 6β^{9} ...(2)
From equation (1) and (2
α^{10 }– β^{10} – 2 (α^{8}– β^{8}) = 6 (α^{9}– β^{9})
⇒ a_{10} – 2a_{8} = 6a_{9 }
A value of b for which the equations
x^{2} + bx – 1 = 0
x^{2} + x + b = 0
have one root in common is (2011)
Let α be the common root of given equations, then
α^{2} + bα – 1 = 0 ...(1)
and α^{2} + 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 b^{3} + 3b = 0
The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has (JEE Adv. 2014)
Quadratic equation with real coefficients and purely imaginary roots can be considered as
p(x) = x^{2} + a = 0 where a > 0 and a ∈R
The p[ p(x)] = 0 ⇒ (x^{2} + a)^{2} + a = 0
⇒ x^{4} + 2ax^{2} + (a^{2 }+ a) = 0
where a, b≠0
∴ p[p(x)] = 0 has complex roots which are neither purely real nor purely imaginary.
Let Suppose α_{1} and B_{1}are the roots of the equation x^{2} – 2x sec a + 1 = 0 and α_{2 }and β_{2} are the roots of the equation x^{2 }+ 2x tanθ – 1 = 0. If α_{1} > b_{1 }and α_{2} > β_{2}, then α_{1} + β_{2} equals (JEE Adv. 2016)
x^{2} – 2x secθ + 1 = 0 ⇒ x = secθ ± tanθ and x^{2} + 2x tanθ – 1 = 0 ⇒ x = –tanθ ± secθ
α_{1},β_{1} are roots of x^{2} – 2x secθ + 1 = 0 and α_{1}> b_{1}
∴ α_{1 }= secθ – tanθ and b_{1 }= secθ + tanθ α_{2},
β_{2} are roots of x^{2} + 2x tanθ – 1 = 0 and α_{2} > β_{2}
∴ α_{2} = tanθ + secθ, β_{2 }= – tanθ – secθ
∴ α_{1} + β_{2} = secθ – tanθ – tanθ – secθ = – 2tanθ
130 videos359 docs306 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
130 videos359 docs306 tests





