Mathematics Test

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Mathematics Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

If p,q,r are real and p≠q, then roots of the equation (p−q)x²+5(p+q)x−2(p−q)=0 are:

A
Real and equal

B
Complex

C
Real and unequal

D
None of these

Solution

Discriminant =b²−4ac

=25(p+q)²+8(p−q)²

Hence, D≥0

Hence, roots are real.

Also p and q are real.

∴D≠0

Hence, roots are not equal.

Hence option C is the correct answer.
Question 2
1 / -0

The roots of a²x²+abx=b²,a≠0,b≠0 are:

A
Equal

B
Non- real

C
Unequal

D
None of these

Solution

Here, \(a^{2} x^{2}+a b x=b^{2}, a \neq 0, b \neq 0\)
\(\Rightarrow a^{2} x^{2}+a b x-b^{2}=0\)
Now, roots of the quadratic equation \(=\frac{-a b \pm \sqrt{(a b)^{2}-4 a^{2}\left(-b^{2}\right)}}{2 a^{2}}\) \(=\frac{-a b \pm \sqrt{5 a^{2} b^{2}}}{2 a^{2}}\)
\(=\frac{-a b \pm \sqrt{5} a b}{2 a^{2}}\)
\(=\frac{-b \pm \sqrt{5} b}{2 a}\)
Thus, the roots are real and unequal.

Hence option C is the correct answer.
Question 3
1 / -0

Each of the following quadratic equations represents the graph shown. Which equation reveals the exact values of the x−intercepts of the graph?

A
y=(1/2)(2x−5)(x+1)

B

\(y=2 x^{2}-\frac{3}{2} x-5\)

C

\(y=\left(x-\frac{3}{4}\right)^{2}\)

D

\(y=\left(x-\frac{3}{2}\right)^{2}-\frac{9}{16}\)

Solution

From the graph the roots are \(x=-1, \frac{5}{2},\) Now at \(x=0\) we have \(y=\frac{-5}{2}\)
\(\Rightarrow y=a\left(x-\frac{5}{2}\right)(x+1)\)
\(x=0 \Rightarrow y=\frac{-5}{2}=a \frac{-5}{2}(1) \Rightarrow a=1\)
\(\therefore y=\left(x-\frac{5}{2}\right)(x+1)=\frac{(2 x-5)(x+1)}{2}\)

Hence option A is the correct answer.
Question 4
1 / -0

All the values of m for which both the roots of the equation x²−2mx+m²−1=0 are greater than −2 but less than 4 lie in the interval :

A
−2

B
m>3

C
−1

D
1

Solution

The given equation is,

x²−2mx+m²−1=0

or (x−m)²−1=0

or (x−m+1)(x−m−1)=0

or x=m−1,m+1

From given condition,

m−1>−2 and m+1<4

⇒m>−1 and m<3

Hence, −1

Hence option C is the correct answer.
Question 5
1 / -0

The graph of y=4x²+3x+7 :

A
lies entirely below the x-axis.

B
lies entirely above the x-axis.

C
cuts the x-axis.

D
touches the x-axis and lies below it.

Solution

\(y=4 x^{2}+3 x+7\)
consider \(4 x^{2}+3 x+7=0\)
\(\Delta=b^{2}-4 a c=9-4(4) 7=-1920\)
\(\therefore\) the equation has no roots.
\(\therefore y=0 \Rightarrow\) it doesn't cut the x-axis anywhere and lies above it.

Hence option B is the correct answer.
Question 6
1 / -0

If \(\alpha, \beta\) are the roots of \(x^{2}+p x+q=0\) and \(\gamma, \delta\) are the roots of \(x^{2}+p x+r=0\), then \(\frac{(\alpha-\gamma)(\alpha-\delta)}{(\beta-\gamma)(\beta-\delta)}=\)?

A
1

B
q

C
r

D
q+r

Solution

since, \(\alpha, \beta\) are the roots of \(x^{2}+p x+q=0\)
\(\alpha+\beta=-p\) and \(\alpha \beta=q\)
Also, \(\gamma, \delta\) are roots of \(x^{2}+p x+r=0\)
\(\gamma+\delta=-p\) and \(\gamma \delta=r\)
Consider \(\frac{(\alpha-\gamma)(\alpha-\delta)}{(\beta-\gamma)(\beta-\delta)}\)
\(=\frac{\alpha^{2}-(\gamma+\delta) \alpha+\gamma \delta}{\beta^{2}-(\gamma+\delta) \beta+\gamma \delta}\)
\(=\frac{\alpha^{2}+p \alpha+r}{\beta^{2}+p \beta+r}\)
\(=\frac{-q+r}{-q+r}\left(\because \alpha, \beta\right.\) are roots of \(\left.x^{2}+p x+q=0\right)\)
\(=1\)

Hence option A is the correct answer.
Question 7
1 / -0

If \(\alpha, \beta, \gamma\) are the roots of \(x^{3}+p x^{2}+q x+r=0,\) then \(\sum \frac{\beta^{2}+\gamma^{2}}{\beta \gamma}=?\)

A

\(\frac{p q}{r}-1\)

B

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

C

\(\frac{p q}{r}-3\)

D

\(\frac{p q}{r}-4\)

Solution

As \(\alpha, \beta, \gamma\) are the roots of \(x^{3}+p x^{2}+q x+r=0\)
Gives
\(\alpha+\beta+\gamma=-p\)
\(\alpha \beta+\beta \gamma+\gamma \alpha=q\)
\(\alpha \beta \gamma=-r\)
Now \(\sum \frac{\beta^{2}+\gamma^{2}}{\beta \gamma}=\frac{\beta^{2}+\gamma^{2}}{\beta \gamma}+\frac{\alpha^{2}+\gamma^{2}}{\gamma \alpha}+\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\)
\(=\frac{\alpha \beta^{2}+\gamma^{2} \alpha+\alpha^{2} \beta+\beta \gamma^{2}+\gamma \alpha^{2}+\beta^{2} \gamma}{\alpha \beta \gamma}\)
\(=\frac{(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma}{\alpha \beta \gamma}\)
\(=\frac{-p q-3(-r)}{-r}=\frac{p q}{r}-3\)

Hence option C is the correct answer.
Question 8
1 / -0

The set of possible values of \(\lambda\) for which \(x^{2}-\left(\lambda^{2}-5 \lambda+5\right) x+\left(2 \lambda^{2}-3 \lambda-4\right)=0\) has roots whose sum and product are both less then 1 is:

A

\(\left(-1, \frac{5}{2}\right)\)

B
(1,4)

C

\(\left(1, \frac{5}{3}\right)\)

D
\(\left(1, \frac{5}{2}\right)\)

Solution

Let \(\alpha, \beta\) be the roots of the given equation, then \(\alpha+\beta=\lambda^{2}-5 \lambda+5<1\), and \(\alpha \beta=2 \lambda^{2}-3 \lambda-4<1\)
\(\Rightarrow \lambda^{2}-5 \lambda+4<0\) and \(, 2 \lambda^{2}-3 \lambda-5<0\)
\(\therefore(\lambda-4)(\lambda-1)<0 \quad\) and \((2 \lambda-5)(\lambda+1)<0\)
\(1<\lambda<4 \quad\) and \(\quad-1<\lambda<\frac{5}{2}\)
\(\therefore\) Required set \(=\left(-1, \frac{5}{2}\right) \cap(1,4)=\left(1, \frac{5}{2}\right)\)

Hence option D is the correct answer.
Question 9
1 / -0

If α,β,γ are the roots of the equation x³+px²+qx+r=0, then α²+β²+γ²=?

A
p²+2q

B
−p²+2q

C
−p²−2q

D
p²−2q

Solution

As \(\alpha, \beta, \gamma\) are roots of \(x^{3}+p x^{2}+q x+r=0\) Gives
\(S_{1}=\alpha+\beta+\gamma=-p\)
\(S_{2}=\alpha \beta+\beta \gamma+\gamma \alpha=q\)
Now, \(\alpha^{2}+\beta^{2}+\gamma^{2}=(\alpha+\beta+\gamma)^{2}-2(\alpha \beta+\beta \gamma+\gamma \alpha)\)
\(=(-p)^{2}-2 q=p^{2}-2 q\)

Hence option D is the correct answer.
Question 10
1 / -0

Let α,β be roots of the equation ax²+bx+c=0 and Δ=b²−4ac. If α+β,α²+β²,α³+β³ are in GP, then :

A
Δ≠0

B
b⋅Δ=0

C
c⋅Δ=0

D
Δ=0

Solution

since, \(\alpha, \beta\) be roots of the equation \(a x^{2}+b x+c=0\) \(\Rightarrow \alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}\)
Now, \(\alpha ^{2}+\beta^{2}=( \alpha +\beta)^{2}-2 \alpha \beta\)
\(\Rightarrow \alpha^{2}+\beta^{2}=\frac{b^{2}-2 a c}{a^{2}}\)
Also, \(\alpha^{3}+\beta^{3}=(\alpha+\beta)\left(\alpha^{2}+\beta^{2}-\alpha \beta\right)\)
\(\Rightarrow \alpha^{3}+\beta^{3}=\left(-\frac{b}{a}\right)\left(\frac{b^{2}-3 a c}{a^{2}}\right)\)
Given, \(\alpha+\beta, \alpha^{2}+\beta^{2}, \alpha^{3}+\beta^{3}\) are in GP
\(\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}=(\alpha+\beta)\left(\alpha^{3}+\beta^{3}\right)\)
Put the values from ( 1 ) and ( 2 ), we get \(\Rightarrow\left(b^{2}-2 a c\right)^{2}=b^{2}\left(b^{2}-3 a c\right)\)
\(\Rightarrow a b^{2} c-4 a^{2} c^{2}=0\)
\(\Rightarrow a c\left(b^{2}-4 a c\right)=0\)
since, \(a \neq 0\)
\(\Rightarrow c . \Delta=0 \quad\) (Given, \(\left.\Delta=b^{2}-4 a c\right)\)

Hence option C is the correct answer.

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Mathematics Test - 10

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Question 1

Real and equal

Complex

Real and unequal

None of these

Solution

Question 2

Equal

Non- real

Unequal

None of these

Solution

Question 3

y=(1/2)(2x−5)(x+1)

\(y=2 x^{2}-\frac{3}{2} x-5\)

\(y=\left(x-\frac{3}{4}\right)^{2}\)

\(y=\left(x-\frac{3}{2}\right)^{2}-\frac{9}{16}\)

Solution

Question 4

−2

m>3

−1

1

Solution

Question 5

lies entirely below the x-axis.

lies entirely above the x-axis.

cuts the x-axis.

touches the x-axis and lies below it.

Solution

Question 6

1

q

r

q+r

Solution

Question 7

\(\frac{p q}{r}-1\)

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

\(\frac{p q}{r}-3\)

\(\frac{p q}{r}-4\)

Solution

Question 8

\(\left(-1, \frac{5}{2}\right)\)

(1,4)

\(\left(1, \frac{5}{3}\right)\)

\(\left(1, \frac{5}{2}\right)\)

Solution

Question 9

p2+2q

−p2+2q

−p2−2q

p2−2q

Solution

Question 10

Δ≠0

b⋅Δ=0

c⋅Δ=0

Δ=0

Solution

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p²+2q

−p²+2q

−p²−2q

p²−2q