Free Objective Test 03 Practice Test - 11th and 12th

Question 1

If every pair of the equations $x^{2} + p x + q r = 0, x^{2} + q x + r p = 0, x^{2} + r x + p q = 0$ have a common root, then the sum of three common roots is

$\frac{- (p + q + r)}{2}$

$\frac{- p + q + r}{2}$

-(p+q+r)

-p+q+r

SOLUTION

Solution : A

Let the roots be α, β; β, γ and γ, α respectively.

∴ α + β = -p, β + γ = -q, γ + α = -r

Adding all, we get ∑α = $\frac{- (p + q + r)}{2}$ etc.

Question 2

If $x^{2} - h x - 21 = 0, x^{2} - 3 h x + 35 = 0 (h > 0)$ has a common root, then the value of h is equal to

SOLUTION

Solution : D

Subtracting, we get 2hx = 56 or hx = 28

Putting in any, $x^{2} = 49$

∴ $[\frac{28}{h}]^{2} = 7^{2}$ ⇒ h = 4(h > 0)

Question 3

If the two equations $x^{2} - c x + d = 0$ and $x^{2} - a x + b = 0$ have one common root and the second has equal roots, then $2 (b + d) =$ )

a+c

-ac

SOLUTION

Solution : C

Let roots of $x^{2} - c x + d = 0$ be α, β then roots of $x^{2} - a x + b = 0$ be α, α

∴ α + β = c, αβ = d, α + α = α, $α^{2}$ = b

Hence $2 (b + d) = 2 (α^{2} + α β) = 2 α (α + β) = a c$

Question 4

If the ratio of the roots of the equation $a x^{2} + b x + c = 0$ be $p : q$ , then

$p q b^{2} + (p + q)^{2} a c = 0$

$p q b^{2} - (p + q)^{2} a c = 0$

$p q a^{2} - (p + q)^{2} b c = 0$

$p q b^{2} - (p - q)^{2} a c = 0$

SOLUTION

Solution : B

Let pα,qα be the roots of the given equation $a x^{2} + b x + c = 0$

Then pα + qα = $- \frac{b}{a} a n d p α . q α = \frac{c}{a}$

From first relation, α = $- \frac{b}{a (p + q)}$

Substituting this value of α in second relation, we get

$\frac{b^{2}}{a^{2} (p + q)^{2}} \times p q = \frac{c}{a}$ ⇒ $b^{2} p q - a c (p + q)^{2} = 0$

Note : Students should remember this question as a fact.

Question 5

If α and β are the roots of the equation $x^{2} - a (x + 1) - b = 0$ , then $(α + 1) (β + 1) =$

-b

1-b

b-1

SOLUTION

Solution : C

Given equation $x^{2} - a (x + 1) - b = 0$

⇒ $x^{2} - a x - a - b$ ⇒ α + β = α, αβ = -(a + b)

Now $(α + 1) (β + 1) = α β + α + β + 1 = - (a + b) + a + 1 = 1 - b$

Question 6

If α and β are the roots of the equation $2 x^{2} - 3 x + 4 = 0$ , then the equation whose roots are $α^{2}$ and $β^{2}$ is

$4 x^{2} + 7 x + 16 = 0$

$4 x^{2} + 7 x + 6 = 0$

$4 x^{2} + 7 x + 1 = 0$

$4 x^{2} - 7 x + 16 = 0$

SOLUTION

Solution : A

α + β = $\frac{3}{2}$ and αβ = 2

$α^{2} + β^{2} = (α + β)^{2} - 2 α β = \frac{9}{4} - 4 = - \frac{7}{4}$

hence required equation $x^{2} - (α^{2} + β^{2}) x + α^{2} β^{2} = 0$

⇒ $x^{2} + \frac{7}{4} x + 4 = 0$ ⇒ $4 x^{2} + 7 x + 16 = 0$

Question 7

If α and β are the roots of the equation $x^{2} + 6 x + λ = 0$ and $3 α + 2 β = - 20$ , then λ =

-8

-16

SOLUTION

Solution : B

α + β = -6 .....(i)

αβ = λ ......(ii)

and given 3α + 2β = -20 ......(iii)

Solving (i) and (iii), we get β = 2, α = -8

Substituting these values in (ii), we get λ = -16

Question 8

If the sum of the roots of the equation $λ x^{2} + 2 x + 3 λ = 0$ be equal to their product, then λ =

-4

$- \frac{2}{3}$

SOLUTION

Solution : D

Under condition $- \frac{2}{λ} = 3$ ⇒ $λ = - \frac{2}{3}$

Question 9

If $2 + i \sqrt{3}$ is a root of the equation $x^{2} + p x + q = 0$ , where p and q are real, then (p, q) =

(-4,7)

(4,7)

(-4,-7)

SOLUTION

Solution : A

Since $2 + i \sqrt{3}$ is a root, therefore $2 - i \sqrt{3}$ will be other root. Now sum of the roots = 4 = -p and product of roots = 7 = q. Hence (p, q) = (-4, 7).

Question 10

If α and β be the roots of the equation $2 x^{2} + 2 (a + b) x + a^{2} + b^{2} = 0$ , then the equation whose roots are $(α + β)^{2}$ and $(α - β)^{2})$ is

$x^{2} - 2 a b x - (a^{2} - b^{2})^{2} = 0$

$x^{2} - 4 a b x - (a^{2} - b^{2})^{2} = 0$

$x^{2} - 4 a b x + (a^{2} - b^{2})^{2} = 0$

None of these

SOLUTION

Solution : B

Sum of roots α + β = -(a + b) and αβ = $\frac{a^{2} + b^{2}}{2}$

⇒ $(α + β)^{2} = (a + b)^{2} a n d (α - β)^{2} = α^{2} + β^{2} - 2 α β$

$= 2 a b - (a^{2} + b^{2}) = - (a - b)^{2}$

Now the required equation whose roots are

$(α + β)^{2} a n d (α - β)^{2}$

$x^{2} - {(α + β)^{2} + (α - β)^{2}} x + (α + β)^{2} (α - β)^{2} = 0$

⇒ $x^{2} - {(a + b)^{2} + (a - b)^{2}} x + (a + b)^{2} (a - b)^{2} = 0$

⇒ $x^{2} - 4 a b x - (a^{2} - b^{2})^{2} = 0$

Question 11

If a root of the equation $a x^{2} + b x + c = 0$ be reciprocal of the equation then $a^{'} x^{2} + b^{'} x + c^{'} = 0$ , then

$(c c^{'} - a a^{'})^{2} = (b a^{'} - c b^{'}) (a b^{'} - b c^{'}) .$

$(b b^{'} - a a^{'})^{2} = (c a^{'} - b c^{'}) (a b^{'} - b c^{'}) .$

$(c c^{'} - a a^{'})^{2} = (b a^{'} - c b^{'}) (a b^{'} - b c^{'}) .$

$(c c^{'} + a a^{'})^{2} = (b a^{'} - c b^{'}) (a b^{'} - b c^{'}) .$

SOLUTION

Solution : A

Let α be a root of first equation, then $\frac{1}{α}$ be a root of second equation.

therefore $a α^{2} + b α + c = 0$ and $a^{'} \frac{1}{α^{2}} + b^{'} \frac{1}{α} + c^{'} = 0$ or $c^{'} α^{2} + b^{'} α + d = 0$

Hence $\frac{α^{2}}{b a^{'} - b^{'} c} = \frac{α}{c c^{'} - a a^{'}} = \frac{1}{a b^{'} - b c^{'}}$

$(c c^{'} - a a^{'})^{2} = (b a^{'} - c b^{'}) (a b^{'} - b c^{'}) .$

Question 12

If $α, β$ are the roots of the equation $a x^{2} + b x + c = 0$ then the equation whose roots are $α + \frac{1}{β}$ and $β + \frac{1}{α}$ , is

$a c x^{2} + (a + c) b x + (a + c)^{2} = 0$

$a b x^{2} + (a + c) b x + (a + c)^{2} = 0$

$a c x^{2} + (a + b) c x + (a + c)^{2} = 0$

$N o n e o f t h e s e$

SOLUTION

Solution : A

$H e r e α + β =$ $- \frac{b}{a}$ $a n d α β =$ $\frac{c}{a}$

If roots are $α + \frac{1}{β}, β + \frac{1}{α}$ , then sum of roots are

$= (α + \frac{1}{β}) + (β + \frac{1}{a l p h a}) = (α + β) + \frac{α + β}{α β} = - \frac{b}{a c} (a + c)$

and product $= (α + \frac{1}{β}) (β + \frac{1}{α})$

$= α β + 1 + 1 + \frac{1}{α β} = 2 + \frac{c}{a} + \frac{a}{c}$

$= \frac{2 a c + c^{2} + a^{2}}{a c} = \frac{(a + c)^{2}}{a c}$

Hence required equation is given by

$x^{2} + \frac{b}{a c} (a + c) x + \frac{(a + c)^{2}}{a c} = 0$

$\Rightarrow$ $a c x^{2} + (a + c) b x + (a + c)^{2} = 0$

Trick : Let a = 1, b = -3, c = 2,then $α$ = 1, $β$ = 2

$∴ α + \frac{1}{β} = \frac{3}{2} a n d β + \frac{1}{α} = 3$

$∴$ required equation must be

$(x - 3) (2 x - 3) = 0$ i.e. $2 x^{2} - 9 x + 9 = 0$

Here (a) gives this equation on putting

a = 1, b = -3, c = 2

Question 13

If the roots of the equation $a x^{2} + b x + c = 0$ be $α a n d β$ ,, then the roots of the equation $c x^{2} + b x + a = 0$ are

$- α, - β$

$α, \frac{1}{β}$

$\frac{1}{α}, \frac{1}{β}$

$N o n e o f t h e s e$

SOLUTION

Solution : C

$α, β$ are roots of $a x^{2} + b x + c = 0$

$\Rightarrow α + β$ = $- \frac{b}{a} a n d α β = \frac{c}{a}$

Let the roots of $c x^{2} + b x + a = 0$ be $α$ ', $β$ ', then

$α$ ' + $β$ ' $= - \frac{b}{c} a n d α β = \frac{a}{c}$

but $\frac{α + β}{α β} = \frac{- \frac{b}{a}}{\frac{c}{a}} = \frac{- b}{c} \frac{1}{α} + \frac{1}{β} = α^{'} + β^{'}$

Hence $α^{'} = \frac{1}{α} a n d β^{'} = \frac{1}{β}$

Question 14

If $α a n d β$ are the roots of the equation $4 x^{2} + 3 x + 7 = 0$ , then $\frac{1}{α} + \frac{1}{β} =$

$- \frac{3}{7}$

$\frac{3}{7}$

$- \frac{3}{5}$

$\frac{3}{5}$

SOLUTION

Solution : A

Given equation $4 x^{2} + 3 x + 7 = 0$ , $∴$

$α + β = - \frac{3}{4} a n d α β = \frac{7}{4}$

Now $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{\frac{- 3}{4}}{\frac{7}{4}} = \frac{- 3}{4} \times \frac{4}{7} = - \frac{3}{7}$

Question 15

If one root of $5 x^{2} + 13 x + k = 0$ is reciprocal of the other, then k=

$0$

$5$

$\frac{1}{6}$

$6$

SOLUTION

Solution : B

Let first root $= α$ and second root $= \frac{1}{α}$

Product of roots =1
$\Rightarrow \frac{k}{5} = 1 \Rightarrow k = 5$

Free Objective Test 03 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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