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01. If x, y, z are the three factors of a³ - 7a - 6, then value of x + y + z will be?

13a
23
36
4a

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} a^{3} - 7 a - 6 \\ (a + 1) (a^{3} - a - 6) \\ (a + 1) (a + 2) (a - 3) \\ Now, Sum of factors \\ (a + 1) + a + 2 + a - 3 = 3 a \end{aligned}

$\eqalign{ & {a^3} - 7a - 6 \cr & \left( {a + 1} \right)\left( {{a^3} - a - 6} \right) \cr & \left( {a + 1} \right)\left( {a + 2} \right)\left( {a - 3} \right) \cr & {\text{Now}},{\text{Sum of factors}} \cr & \left( {a + 1} \right) + a + 2 + a - 3 = 3a \cr}$

02. If x - 11, then the value of x⁵ - 12x⁴ + 12x³ - 12x² + 12x - 1 is?

111
210
312
4-10

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{5} - 12 x^{4} + 12 x^{3} - 12 x^{2} + 12 x - 1 \\ = x^{5} - 11 x^{4} - x^{4} + 11 x^{3} + x^{3} - 11 x^{2} - x^{2} + 11 x + x - 1 \\ Put x = 11 \\ = 11^{5} - {11.11}^{4} - 11^{4} + {11.11}^{3} + 11^{3} - {11.11}^{2} - 11^{2} + 11.11 + 11 - 1 \\ = 11 - 1 \\ = 10 \end{aligned}

$\eqalign{ & {x^5} - 12{x^4} + 12{x^3} - 12{x^2} + 12x - 1 \cr & = {x^5} - 11{x^4} - {x^4} + 11{x^3} + {x^3} - 11{x^2} - {x^2} + 11x + x - 1 \cr & {\text{Put }}x = 11 \cr & = {11^5} - {11.11^4} - {11^4} + {11.11^3} + {11^3} - {11.11^2} - {11^2} + 11.11 + 11 - 1 \cr & = 11 - 1 \cr & = 10 \cr}$

If a + b + c + d = 4, then the value of $\frac{1}{(1 - a) (1 - b) (1 - c)}$ $\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$ + $\frac{1}{(1 - b) (1 - c) (1 - d)}$ $\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$ + $\frac{1}{(1 - c) (1 - d) (1 - a)}$ $\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$ + $\frac{1}{(1 - d) (1 - a) (1 - b)}$ $\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$ is?

10
21
34
41 + abcd

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{1}{(1 - a) (1 - b) (1 - c)} + \frac{1}{(1 - b) (1 - c) (1 - d)} + \frac{1}{(1 - c) (1 - d) (1 - a)} + \frac{1}{(1 - d) (1 - a) (1 - b)} \\ = \frac{1 - d + 1 - a + 1 - b + 1 - c}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = \frac{4 - (a + b + c + d)}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = \frac{4 - 4}{(1 - a) (1 - b) (1 - c) (1 - d)} \\ = 0 \end{aligned}

$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & = \frac{{1 - d + 1 - a + 1 - b + 1 - c}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - \left( {a + b + c + d} \right)}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - 4}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = 0 \cr}$

04. If x = y + z then x³ - y³ - z³ is?

10
23xyz
3-3xyz
41

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given , x = y + z \\ ∴ x - y - z = 0 \\ Then x^{3} - y^{3} - z^{3} = 3 x y z \end{aligned}

$\eqalign{ & {\text{Given , }}x = y + z \cr & \therefore x - y - z = 0 \cr & {\text{Then }}{x^3} - {y^3} - {z^3} = 3xyz \cr}$

If $x^{2} + \frac{1}{x^{2}} = 98,$ ${x^2} + \frac{1}{{{x^2}}} = 98{\text{,}}$ $(x > 0),$ $\left( {x > 0} \right){\text{,}}$ then the value of $^{3} + \frac{1}{x^{3}}$ $^3 + \frac{1}{{{x^3}}}$ is?

1970
21030
3-970
4-1030

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{2} + \frac{1}{x^{2}} = 98 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 100 \\ \Rightarrow {(x + \frac{1}{x})}^{2} = 100 \\ \Rightarrow x + \frac{1}{x} = 10 \\ \Rightarrow x^{3} + \frac{1}{x^{3}} + 3. x . \frac{1}{x} (x + \frac{1}{x}) = 10^{3} \\ \Rightarrow x^{3} + \frac{1}{x^{3}} = 1000 - 3 \times 10 \\ \Rightarrow x^{3} + \frac{1}{x^{3}} = 970 \end{aligned}

$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 98 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 100 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 100 \cr & \Rightarrow x + \frac{1}{x} = 10 \cr & \Rightarrow {\text{ }}{x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = {10^3} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 1000 - 3 \times 10 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 970 \cr}$

06. If a² + b² + c² = 16, x² + y² + z² = 25 and ax + by + cz = 20 then the value of $\frac{a + b + c}{x + y + z}$ $\frac{{a + b + c}}{{x + y + z}}$ = ?

1 $\frac{3}{5}$ $\frac{3}{5}$
2 $\frac{4}{5}$ $\frac{4}{5}$
3 $\frac{5}{3}$ $\frac{5}{3}$
4 $\frac{5}{4}$ $\frac{5}{4}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} a^{2} + b^{2} + c^{2} = 16, x^{2} + y^{2} + z^{2} = 25 \\ But b = c = 0, But y = z = 0 \\ a = 4 x = 5 \\ Now, a x + b y + c z = 20 \\ 4 \times 5 + 0 + 0 = 0 \\ 20 = 20 (Satisfy) \\ Now, \frac{a + b + c}{x + y + z} \\ = \frac{4 + 0 + 0}{5 + 0 + 0} \\ = \frac{4}{5} \end{aligned}

$\eqalign{ & {a^2} + {b^2} + {c^2} = 16,{\text{ }}{x^2} + {y^2} + {z^2} = 25{\text{ }} \cr & {\text{But }}b = c = 0,{\text{ But }}y = z = 0 \cr & {\text{ }}a = 4{\text{ }}x = 5 \cr & {\text{Now, }}ax + by + cz = 20 \cr & 4 \times 5 + 0 + 0 = 0 \cr & 20 = 20{\text{ }}\left( {{\text{Satisfy}}} \right) \cr & {\text{Now, }}\frac{{a + b + c}}{{x + y + z}} \cr & = \frac{{4 + 0 + 0}}{{5 + 0 + 0}} \cr & = \frac{4}{5} \cr}$

07. If x⁴ + 2x³ + ax² + bx + 9 is a perfect square where a and b are positive real numbers, then the value of a and b is?

1a = 5, b = 6
2a = 6, b = 7
3a = 7, b = 7
4a = 7, b = 8

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{4} + 2 x^{3} + a x^{2} + b x + 9 \\ Put x = 1 \\ = 1 + 2 \times 1 + a + b + 9 \\ = 1 + 2 + a + b + 9 \\ = 13 + a + b \end{aligned}

$\eqalign{ & {x^4} + 2{x^3} + a{x^2} + bx + 9 \cr & {\text{Put }}x = 1 \cr & = 1 + 2 \times 1 + a + b + 9 \cr & = 1 + 2 + a + b + 9 \cr & = 13 + a + b \cr}$
To make a perfect square numbers value of a + b must be either 3 or 13
Now, option (2) a = 6, b = 7

\begin{aligned} ∴ a + b = 12 \\ make perfect square \\ (25 = 5^{2}) \end{aligned}

$\eqalign{ & \therefore a + b = 12 \cr & {\text{make perfect square}} \cr & \left( {25 = {5^2}} \right) \cr}$

If $x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$ $x = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}$ and $y = \frac{\sqrt{5} - 1}{\sqrt{5} + 1},$ ${\text{y}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}{\text{,}}$ then the value of $\frac{x^{2} + x y + y^{2}}{x^{2} - x y + y^{2}}$ $\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}}$ is?

1 $\frac{3}{4}$ $\frac{3}{4}$
2 $\frac{5}{3}$ $\frac{5}{3}$
3 $\frac{4}{3}$ $\frac{4}{3}$
4 $\frac{3}{5}$ $\frac{3}{5}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} and  y = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \\ ∴ x = \frac{1}{y} \\ \Leftrightarrow x y = 1 \\ x + y = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} + \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \\ \Rightarrow x + y = \frac{5 + 1 + 2 \sqrt{5} + 5 + 1 - 2 \sqrt{5}}{5 - 1} \\ \Rightarrow x + y = \frac{12}{4} \\ \Rightarrow x + y = 3 \\ \Rightarrow x + \frac{1}{x} = 3 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} = {(3)}^{2} - 2 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} = 7 \\ Now, \frac{x^{2} + x y + y^{2}}{x^{2} - x y + y^{2}} \\ = \frac{x^{2} + y^{2} + x y}{x^{2} + y^{2} - x y} \\ = \frac{7 + 1}{7 - 1} \\ = \frac{8}{6} \\ = \frac{4}{3} \end{aligned}

$\eqalign{ & x = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ and y}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \therefore x = \frac{1}{y} \cr & \Leftrightarrow xy = 1 \cr & x + y = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \Rightarrow x + y = \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}} \cr & \Rightarrow x + y = \frac{{12}}{4} \cr & \Rightarrow x + y = 3 \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {\left( 3 \right)^2} - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr & {\text{Now,}}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} \cr & = \frac{{{x^2} + {y^2} + xy}}{{{x^2} + {y^2} - xy}} \cr & = \frac{{7 + 1}}{{7 - 1}} \cr & = \frac{8}{6} \cr & = \frac{4}{3} \cr}$

09. If x² + y² + 2x +1 = 0, then the value of x³¹ + y³⁵ is?

1-1
21
30
42

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x^{2} + y^{2} + 2 x + 1 = 0 \\ \Rightarrow {(x + 1)}^{2} + y^{2} = 0 \\ Let {(x + 1)}^{2} = 0 \\ y^{2} = 0 \\ x = - 1, y = 0 \\ Now, x^{31} + y^{35} \\ = {(- 1)}^{31} + {(0)}^{35} \\ = - 1 \end{aligned}

$\eqalign{ & {x^2} + {y^2} + 2x + 1 = 0 \cr & \Rightarrow {\left( {x + 1} \right)^2} + {y^2} = 0 \cr & {\text{Let }}{\left( {x + 1} \right)^2} = 0 \cr & {y^2} = 0 \cr & x = - 1,y = 0 \cr & {\text{Now, }}{x^{31}} + {y^{35}} \cr & = {\left( { - 1} \right)^{31}} + {\left( 0 \right)^{35}} \cr & = - 1 \cr}$

10. A complete factorisation of x⁴ + 64 is?

1(x² + 8)²
2(x² + 8)(x² - 8)
3(x² - 4x + 8)(x² - 4x - 8)
4(x² + 4x + 8)(x² - 4x + 8)

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} (x^{4} + 64) \\ = x^{4} + 8^{2} + 2. x^{2} .8 - 2. x^{2} .8 \\ = {(x^{2} + 8)}^{2} - (16 x^{2}) \\ = {(x^{2} + 8)}^{2} - {(4 x)}^{2} \\ = (x^{2} + 8 + 4 x) (x^{2} + 8 - 4 x) \\ = (x^{2} + 4 x + 8) (x^{2} - 4 x + 8) \end{aligned}

$\eqalign{ & \left( {{x^4} + 64} \right) \cr & = {x^4} + {8^2} + 2.{x^2}.8 - 2.{x^2}.8 \cr & = {\left( {{x^2} + 8} \right)^2} - \left( {16{x^2}} \right) \cr & = {\left( {{x^2} + 8} \right)^2} - {\left( {4x} \right)^2} \cr & = \left( {{x^2} + 8 + 4x} \right)\left( {{x^2} + 8 - 4x} \right) \cr & = \left( {{x^2} + 4x + 8} \right)\left( {{x^2} - 4x + 8} \right) \cr}$

Hello Guest

Algebra - 01

01. If x, y, z are the three factors of a³ - 7a - 6, then value of x + y + z will be?

02. If x - 11, then the value of x⁵ - 12x⁴ + 12x³ - 12x² + 12x - 1 is?

04. If x = y + z then x³ - y³ - z³ is?

If $x^{2} + \frac{1}{x^{2}} = 98,$ ${x^2} + \frac{1}{{{x^2}}} = 98{\text{,}}$ $(x > 0),$ $\left( {x > 0} \right){\text{,}}$ then the value of $^{3} + \frac{1}{x^{3}}$ $^3 + \frac{1}{{{x^3}}}$ is?

06. If a² + b² + c² = 16, x² + y² + z² = 25 and ax + by + cz = 20 then the value of $\frac{a + b + c}{x + y + z}$ $\frac{{a + b + c}}{{x + y + z}}$ = ?

07. If x⁴ + 2x³ + ax² + bx + 9 is a perfect square where a and b are positive real numbers, then the value of a and b is?

09. If x² + y² + 2x +1 = 0, then the value of x³¹ + y³⁵ is?

10. A complete factorisation of x⁴ + 64 is?

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NCERT
6 to 12

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Practice Batch

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Hello Guest

Algebra - 01

01. If x, y, z are the three factors of a3 - 7a - 6, then value of x + y + z will be?

02. If x - 11, then the value of x5 - 12x4 + 12x3 - 12x2 + 12x - 1 is?

04. If x = y + z then x3 - y3 - z3 is?

05. If x2+1x2=98,x2+1x2=98,{x^2} + \frac{1}{{{x^2}}} = 98{\text{,}} (x>0),(x>0),\left( {x > 0} \right){\text{,}} then the value of 3+1x33+1x3^3 + \frac{1}{{{x^3}}} is?

06. If a2 + b2 + c2 = 16, x2 + y2 + z2 = 25 and ax + by + cz = 20 then the value of a+b+cx+y+za+b+cx+y+z\frac{{a + b + c}}{{x + y + z}} = ?

07. If x4 + 2x3 + ax2 + bx + 9 is a perfect square where a and b are positive real numbers, then the value of a and b is?

08. If x=5–√+15–√−1x=5+15−1x = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} and y=5–√−15–√+1,y=5−15+1,{\text{y}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}{\text{,}} then the value of x2+xy+y2x2−xy+y2x2+xy+y2x2−xy+y2\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} is?

09. If x2 + y2 + 2x +1 = 0, then the value of x31 + y35 is?

10. A complete factorisation of x4 + 64 is?

Subject Wise

Exam Wise

NCERT 6 to 12

Paid Course

Practice Batch

01. If x, y, z are the three factors of a³ - 7a - 6, then value of x + y + z will be?

02. If x - 11, then the value of x⁵ - 12x⁴ + 12x³ - 12x² + 12x - 1 is?

04. If x = y + z then x³ - y³ - z³ is?

If $x^{2} + \frac{1}{x^{2}} = 98,$ ${x^2} + \frac{1}{{{x^2}}} = 98{\text{,}}$ $(x > 0),$ $\left( {x > 0} \right){\text{,}}$ then the value of $^{3} + \frac{1}{x^{3}}$ $^3 + \frac{1}{{{x^3}}}$ is?

06. If a² + b² + c² = 16, x² + y² + z² = 25 and ax + by + cz = 20 then the value of $\frac{a + b + c}{x + y + z}$ $\frac{{a + b + c}}{{x + y + z}}$ = ?

07. If x⁴ + 2x³ + ax² + bx + 9 is a perfect square where a and b are positive real numbers, then the value of a and b is?

09. If x² + y² + 2x +1 = 0, then the value of x³¹ + y³⁵ is?

10. A complete factorisation of x⁴ + 64 is?

NCERT
6 to 12