11th And 12th > Mathematics
QUADRATIC EQUATIONS MCQs
:
B
Given, x + 2 > √x+4 ⇒ (x+2)2>(x+4) but inequality sign will be remained same if both sides are positive or negative.
R.H.S is positive and so L.H.S will also be positive.
Thus x>-2
⇒x2+4x+4>x+4 ⇒x2+3x>0
⇒ x(x+3)>0⇒ x<−3 or x>0 ⇒ x>0
:
C
Equation |3x2+12x+6|=5x+16 ......(i)
when 3x2+12x+6≥0 ⇔ x2+4x≥−2
⇔ |x+2|2≥4−2⇔ |x+2|≥(√2)2
⇔ x + 2 ≤ - √2 or x + 2 ≥ √2.......(ii)
Then (i) becomes 3x2+12x+6=5x+16
⇔ 3x2+7x−10=0 ⇒ x = 1, -103
But x = -103 does not satisfy (ii)
When 3x2+12x+6<0 ⇒ x2+4x>−2
⇒|x+2|≤√2⇒−√2−2≤x≤−2+√2....(iii)
⇒ 3x2+17x+22=0 ⇒ x=−2,−113
But x = - 113 does not satisfy (iii). So, 1 and -2 are the only solutions.
:
A
x, y, z, ∈ R and distinct.
Now, u = x2+4y2+9z2−6yz−3zx−2xy
=12(2x2+8y2+18z2−12yz−6zx−4xy)
=12{(x2−4xy+4y2)+(x2−6zx+9z2)+(4y2−12yz+9z2)}
=12{(x−2y)2+(x−3z)2+(2y−3z)2}
Since it is sum of squares. So U is always non-negative.
:
A
Let y=x2+14x+9x2+2x+3
⇒ y(x2+2x+3)−x2−14x−9=0
⇒ (y−1)x2+(2y−14)x+3y−9=0
For real x, its discriminant ≥ 0
i.e. 4(y−7)2−4(y−1)3(y−3)≥0
⇒ y2+y−20≤0 or (y−4)(y+5)≤0
Now, the product of two factors is negative if these are of opposite signs. So following two cases arise:
Case 1: y−4≥0 or y≥4 y+5≤0 or y≤−5
This is not possible.
Case 2: y−4≤0 or y≤4 y+5≥0 or y≥−5 Both of these are satisfied if -5≤y≤4
hence maximum value of y is 4 and minimum value is -5
:
B
If the given expression be y, then
y=2x2y+(3y−1)x+(6y−2)=0
If y ≠ 0 then Δ ≥ 0 for real x i.e. B2−4AC≥0
or −39y2+10y+1≥0 or (13y+1)(3y−1)≤0
⇒ −113≤y≤13
if y = 0 then x = -2 which is real and this value of y is included in the above range.
:
D
x2−3x+2 be factor of x4−px2+q=0
Hence (x2−3x+2)=0 ⇒ (x-2)(x-1) = 0
⇒ x = 2, 1, putting these values in given equation
so, 4p - q -16 = 0 ......(i)
and, p - q - 1 = 0 .......(ii)
Solving (i) and (ii), we get (p, q) = (5, 4)
:
D
Let α be a common root, then
α2+αα+10=0
and α2+bα−10=0
from (i) - (ii),
(a−b)α+20=0 ⇒ α = −20a−b
Substituting the value of α in (i), we get
(−20a−b)2+a(−20a−b)+10=0
⇒400 - 20 a(a - b) + 10(a−b)2 = 0
⇒40 -2a2 + 2ab + a2 + b2 -2ab = 0
⇒a2 - b2 = 40
:
B
Exprssions are x2−11x+a and x2−14x+2a will have a common factor, then
⇒ x2−22a+14a=xa−2a=1−14+11
⇒ x2−8a=x−a=1−3 ⇒ x2=8a3 and x=a3
⇒ (a3)2=8a3 ⇒ a29=8a3 ⇒ a = 0, 24.
Alternate Method : We can check by putting the values of a from the options.
:
A
Let α is the common root, so α2+pα+q=0
and α2+qα+p=0
⇒ (p - q) α + (q - p) = 0 ⇒ α=1
Put the value of α , p + q + 1 = 0
:
C
The condition for common roots gives (bc−a2)2=(ca−b2)(ab−c2) On simplification gives a3+b3+c3=3abc