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11th Maths Weekly Test-3 ( Basic Algebra )

St. Britto Hr. Sec. School - Madurai

11th Maths Weekly Test-3 ( Basic Algebra )-Aug 2020

Time : 1.00 Hours

Part - A

Multiple Choice Question - Answer All Question

The condition that the equation ax² + bx + c = 0 may have one root is the double the other is:

2b² = 9ac
b²= ac
b² = 4ac
9b² = 2ac

Find the other root of x²-4x+1=0 given that 2+\(\sqrt{3}\) is a root:

\(\sqrt{3}\)+2
-\(\sqrt{3}\)-\(\sqrt{2}\)
2-\(\sqrt{3}\)
\(\sqrt{3}\)-2

Let \(\alpha\) and \(\beta\) are the roots of a quadratic equation px² + qx + r = 0 then

\(\alpha +\beta=-{p\over r}\)
\(\alpha\beta={p\over r}\)
\(\alpha +\beta={-q\over p}\)
\(\alpha \beta =r\)

Logarithm of 144 to the base 2\(\sqrt{3}\) is

2
3
4
5

The Value of \({ log }_{ 3/4 }^{ (4/3) }\) is

-2
1
2
-1

If |x+2|≤9,then x belongs to

(-∞,-7)
[-11,7]
(-∞,-7)ᴗ[11,\(\infty \))
(-11,7)

If x is a real number and |x| < 5 then

x ≥ 5
-5 < x < 5
x ≤ -5
-5 ≤ x ≤ 5

The solution set of the following inequality |x-1| \(\ge \) |x-3| is

[0,2]
[2,∞]
(0,2)
(-∞,2)

If x is real and k = \(\frac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } \) then

\(k\epsilon \left[ \frac { 1 }{ 3 } ,3 \right] \)
k ≥ 3
\(k\le \frac { 1 }{ 3 } \)
none of these

The equation whose roots are numerically equal but opposite in sign to the roots 3x²-5x-7 =0 is

3x²-5x-7 =0
3x²+5x-7 =0
3x²-5x+7 =0
3x²+x-7

The value of log 1 is

1
0
\(\infty\)
-1

The roots of the equation \(x+{1\over x}=3{1\over 3},x\ne 0\) are

1, 3
\({1\over 3},3\)
\(3,{-1\over 3}\)
\(1,{1\over 3}\)

The number of real solution of |2x-x²-3|=1 is

0
2
3
4

If \(\frac { 1-2x }{ 3+2x-{ x }^{ 2 } } =\frac { A }{ 3-x } +\frac { B }{ x+1 } \) ,then the value of A+B is

\(\frac { -1 }{ 2 } \)
\(\frac { -2 }{ 3 } \)
\(\frac { 1 }{ 2 }\)
\(\frac { 2 }{ 3 } \)

The zero of the polynomial function f(x)=9x²-16 are:

(9,16)
(3,4)
\((\frac{4}{3},-\frac{4}{3})\)
\((\frac{3}{4},-\frac{3}{4})\)

(x²-2x+2)(x²+2x+2) are the factors of the polynomial

(x²-2x)²
x⁴-4
x⁴+4
(x²-2x+2)²

The value of \({ log }_{ 3 }\frac { 1 }{ 81 } \) is

-2
-8
-4
-9

Find a so that the sum and product of the roots of the equation 2x²+(a-3)x+3a-5 = 0 are equal is

1
2
0
4

If a and b are the roots of the equation x²-kx+c = 0 then the distance between the points (a;0) and (b;0)

\(\sqrt { { 4k }^{ 2 }-c } \)
\(\sqrt { { k }^{ 2 }-4c } \)
\(\sqrt { 4c-{ k }^{ 2 } } \)
\(\sqrt { k-8c } \)

If \({ log }_{ \sqrt { x } }\) 0.25 =4 ,then the value of x is

0.5
2.5
1.5
1.25

The value of \({ log }_{ 10 }^{ 8 }+{ log }_{ 10 }^{ 5 }-{ log }_{ 10 }^{ 4 }\)=

\({ log }_{ 10 }^{ 9 }\)
\({ log }_{ 10 }^{ 36 }\)
1
-1

Which whole number is not a natural number?

1
2
3
0

If a and b are the roots of the equation x²-kx+16=0 and a²+b²=32 then the value of k is

10
-8
-8,8
6

The factors of the polynomial \(6\sqrt { { 3x }^{ 2 } } -47x+5\sqrt { 3 } \) are

\((2x-5\sqrt { 3 } )(3\sqrt { 3 } x-1)\)
\((2x-5\sqrt { 3 } )(3\sqrt { 3 } x+1)\)
\((2x+5\sqrt { 3 } )(3\sqrt { 3 } x+1)\)
\((2x+5\sqrt { 3 } )(3\sqrt { 3 } x-1)\)

\(\sqrt [ 4 ]{ { \left( -2 \right) }^{ 4 } } \times { \left( -1000 \right) }^{ \frac { 1 }{ 3 } }\)is

20
-20
2^-10
100

If 8 and 2 are the roots of x²+ax+c=0 and 3,3 are the roots of x²+dx+b=0;then the roots of the equation x²+ax+b = 0 are

1,2
-1,1
9,1
-1,2

The value of log₂3 . log₂₇32:

\(\frac{5}{2}\)
\(\frac{2}{5}\)
\(\frac{5}{3}\)
\(\frac{3}{5}\)

The rationalising factor of \(\frac { 5 }{ \sqrt [ 3 ]{ 3 } } \) is

\(\sqrt [ 3 ]{ 6 } \)
\(\sqrt [ 3 ]{ 3 } \)
\(\sqrt [ 3 ]{ 9 } \)
\(\sqrt [ 3 ]{ 27 } \)

If \(\frac{x}{x^2-5x+6}=\frac{A}{x-2}+\frac{B}{x-3}\) then value of A is:

2
0
3
-2

If \(x={1\over 2+\sqrt{3}}\) then the value of x³ - x² - 11x + 3 is

0
1
2
4

The logarithmic form of 5²=25 is

\({ log }_{ 5 }^{ 2 }=25\)
\({ log }_{ 2 }^{ 5 }=25\)
\({ log }_{ 2 }^{ 25 }=2\)
\({ log }_{ 25 }^{ 5 }=2\)

For x≥2, |x-2|=

2-x
2+x
x-2
x

\(\sqrt [ 4 ]{ 11 } \) is equal to

\(\sqrt [ 8 ]{ 11^{ 2 } } \)
\(\sqrt [ 8 ]{ 11^{ 4 } } \)
\(\sqrt [ 8 ]{ 11^{ 8 } } \)
\(\sqrt [ 8 ]{ 11^{ 6 } } \)

If \(\frac { kx }{ (x+2)(x-1) } =\frac { 2 }{ x+2 } +\frac { 1 }{ x-2 } \) ,then the value of k is

1
2
3
4

If x < 7,then

-x < -7
- x ≤ -7
-x > -7
-x ≥ -7

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