Quantitative Aptitude
QUADRATIC EQUATIONS MCQs
Quadratic Equations, 10th And 12th Grade Quadratic Equations
Total Questions : 102
| Page 7 of 11 pages
Answer: Option C. -> 2
:
C
Equation|3x2+12x+6|=5x+16 ......(i)
when3x2+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.
:
C
Equation|3x2+12x+6|=5x+16 ......(i)
when3x2+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.
Answer: Option A. -> 4, - 5
:
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≤0or(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≥0ory≥4y+5≤0ory≤−5
This is not possible.
Case 2:y−4≤0ory≤4y+5≥0ory≥−5 Both of these are satisfied if -5≤y≤4
hence maximum value of y is 4 and minimum value is -5
:
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≤0or(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≥0ory≥4y+5≤0ory≤−5
This is not possible.
Case 2:y−4≤0ory≤4y+5≥0ory≥−5 Both of these are satisfied if -5≤y≤4
hence maximum value of y is 4 and minimum value is -5
Answer: Option B. -> (−113,13)
:
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≥0or(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.
:
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≥0or(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.
Answer: Option D. -> 0
:
D
Given equationesinx−e−sinx−4=0
Letesinx=y, then given equation can be written as
y2−4y−1=0 ⇒y=2±√5
But the value of y=esinx is always positive, so
y=2+√5 (∵ 2 < √5
⇒logey=loge(2+√5)⇒ sin x = loge(2+√5)>1
no solution
:
D
Given equationesinx−e−sinx−4=0
Letesinx=y, then given equation can be written as
y2−4y−1=0 ⇒y=2±√5
But the value of y=esinx is always positive, so
y=2+√5 (∵ 2 < √5
⇒logey=loge(2+√5)⇒ sin x = loge(2+√5)>1
no solution
Answer: Option D. -> roots doesnot exist
:
D
given equation is√3x+1+1 = √x
⇒√3x+1 = √x - 1
Squaring both the sides, we get 3x+ 1 = x + 1 -2√x
⇒2√x + 2x = 0 (irrational function)
Thusx≠ 0 andx≠ 1, Since equation is non-quadratic equation.
:
D
given equation is√3x+1+1 = √x
⇒√3x+1 = √x - 1
Squaring both the sides, we get 3x+ 1 = x + 1 -2√x
⇒2√x + 2x = 0 (irrational function)
Thusx≠ 0 andx≠ 1, Since equation is non-quadratic equation.
Answer: Option D. -> (5, 4)
:
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
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)
Answer: Option D. -> 40
:
D
Letα 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 ofα 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
:
D
Letα 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 ofα 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
Answer: Option A. -> Non-negative
:
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
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.
Answer: Option B. -> {4}
:
B
xlogx(1−x)2=9
⇒logx(9)=logx(1−x)2(;∵ ax=N ⇒logaN=x)
⇒9=(1−x2) ⇒1+x2−2x−9=0
⇒x2−2x−8=0 ⇒(x + 2)(x - 4) = 0 ⇒x = -2, 4
But x>0;∴x=4
:
B
xlogx(1−x)2=9
⇒logx(9)=logx(1−x)2(;∵ ax=N ⇒logaN=x)
⇒9=(1−x2) ⇒1+x2−2x−9=0
⇒x2−2x−8=0 ⇒(x + 2)(x - 4) = 0 ⇒x = -2, 4
But x>0;∴x=4
Answer: Option B. -> 0,24
:
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=8a3andx=a3
⇒ (a3)2=8a3⇒ a29=8a3⇒ a = 0, 24.
Alternate Method : We can check by putting the values of a from the options.
:
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=8a3andx=a3
⇒ (a3)2=8a3⇒ a29=8a3⇒ a = 0, 24.
Alternate Method : We can check by putting the values of a from the options.
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