Study91

If $x + \sqrt{5} = 5 + \sqrt{y}$ $x + \sqrt 5 = 5 + \sqrt y$ and x, y are positive integers, then the value of $\frac{\sqrt{x} + y}{x + \sqrt{y}}$ $\frac{{\sqrt x + y}}{{x + \sqrt y }}$ is?

11
22
35
47

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x + \sqrt{5} = 5 + \sqrt{y} \\ Put , x = 5 and y = 5 \\ 5 + \sqrt{5} = 5 + \sqrt{5} \\ L .H .S = R .H .S \\ \frac{\sqrt{x} + y}{x + \sqrt{y}} \\ = \frac{\sqrt{5} + 5}{5 + \sqrt{5}} \\ = 1 \end{aligned}

$\eqalign{ & x + \sqrt 5 = 5 + \sqrt y \cr & {\text{Put , }}x = 5{\text{ and }}y = 5 \cr & 5 + \sqrt 5 = 5 + \sqrt 5 \cr & {\text{L}}{\text{.H}}{\text{.S}} = {\text{R}}{\text{.H}}{\text{.S}} \cr & \frac{{\sqrt x + y}}{{x + \sqrt y }} \cr & = \frac{{\sqrt 5 + 5}}{{5 + \sqrt 5 }} \cr & = 1 \cr}$

If $p^{2} + \frac{1}{p^{2}} = 47,$ ${p^2} + \frac{1}{{{p^2}}} = 47{\text{,}}$ then the value of $p + \frac{1}{p}$ $p + \frac{1}{p}$ is?

15
26
37
48

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} p^{2} + \frac{1}{p^{2}} = 47 \\ On adding 2 both side \\ p^{2} + \frac{1}{p^{2}} + 2 = 47 + 2 \\ \Rightarrow {(p + \frac{1}{p})}^{2} = 49 \\ \Rightarrow (p + \frac{1}{p}) = 7 \\ \Rightarrow p + \frac{1}{p} = 7 \end{aligned}

$\eqalign{ & {p^2} + \frac{1}{{{p^2}}} = 47 \cr & {\text{On adding 2 both side}} \cr & {p^2} + \frac{1}{{{p^2}}} + 2 = 47 + 2 \cr & \Rightarrow {\left( {p + \frac{1}{p}} \right)^2} = 49 \cr & \Rightarrow \left( {p + \frac{1}{p}} \right) = 7 \cr & \Rightarrow p + \frac{1}{p} = 7 \cr}$

If $x = \sqrt[3]{x^{2} + 11} - 2,$ $x = \root 3 \of {{x^2} + 11} - 2{\text{,}}$ then the value of x³ + 5x² + 12x is?

10
23
37
411

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = \sqrt[3]{x^{2} + 11} - 2 \\ \Rightarrow x + 2 = \sqrt[3]{x^{2} + 11} \\ \Rightarrow Taking cube on both side \\ \Rightarrow {(x + 2)}^{3} = x^{2} + 11 \\ \Rightarrow x^{3} + 8 + 6 x (x + 2) = x^{2} + 11 \\ \Rightarrow x^{3} + 8 + 6 x^{2} + 12 x = x^{2} + 11 \\ \Rightarrow x^{3} + 5 x^{2} + 12 x = 3 \end{aligned}

$\eqalign{ & x = \root 3 \of {{x^2} + 11} - 2 \cr & \Rightarrow x + 2 = \root 3 \of {{x^2} + 11} \cr & \Rightarrow {\text{Taking cube on both side}} \cr & \Rightarrow {\left( {x + 2} \right)^3} = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6x\left( {x + 2} \right) = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6{x^2} + 12x = {x^2} + 11 \cr & \Rightarrow {x^3} + 5{x^2} + 12x = 3 \cr}$

If $x = \sqrt{a} + \frac{1}{\sqrt{a}},$ $x = \sqrt a + \frac{1}{{\sqrt a }}{\text{,}}$ $y = \sqrt{a} - \frac{1}{\sqrt{a}},$ $y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}}$ $(a > 0)$ $\left( {a > 0} \right)$ then the value of x⁴ + y⁴ - 2x²y² is?

116
220
310
45

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = \sqrt{a} + \frac{1}{\sqrt{a}} \\ y = \sqrt{a} - \frac{1}{\sqrt{a}} \\ Put a = 4 \\ x = 2 + \frac{1}{2} = \frac{5}{2} \\ y = 2 - \frac{1}{2} = \frac{3}{2} \\ Then, x^{4} + y^{4} - 2 x^{2} y^{2} \\ = {(x^{2} - y^{2})}^{2} \\ = {(\frac{25}{4} - \frac{9}{4})}^{2} \\ = 16 \end{aligned}

$\eqalign{ & {\text{ }}x = \sqrt a + \frac{1}{{\sqrt a }} \cr & {\text{ }}y = \sqrt a - \frac{1}{{\sqrt a }} \cr & {\text{Put }}a = 4 \cr & x = 2 + \frac{1}{2} = \frac{5}{2} \cr & y = 2 - \frac{1}{2} = \frac{3}{2} \cr & {\text{Then, }}{x^4} + {y^4} - 2{x^2}{y^2}{\text{ }} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left( {\frac{{25}}{4} - \frac{9}{4}} \right)^2} \cr & = {\text{ 16}} \cr}$

05. If (x - 2)(x - p) = x² - ax + 6, then the value of (a - p) is?

10
21
32
43

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} (x - 2) (x - p) = x^{2} - a x + 6 \\ x^{2} - (2 + p) x + 2 p = x^{2} - a x + 6 \\ Comparision the cofficients \\ 2 + p = a \\ 2 p = 6 \\ \Leftrightarrow p = 3 \\ 2 + 3 = a \\ \Leftrightarrow a = 5 \\ Then , \\ p = 3, a = 5 \\ a - p = 5 - 3 \\ \Leftrightarrow a - p = 2 \end{aligned}

$\eqalign{ & \left( {x - 2} \right)\left( {x - p} \right) = {x^2} - ax + 6 \cr & {x^2} - \left( {2 + p} \right)x + 2p = {x^2} - ax + 6 \cr & {\text{Comparision the cofficients}} \cr & 2 + p = a \cr & 2p = 6 \cr & \Leftrightarrow p = 3 \cr & 2 + 3 = a \cr & \Leftrightarrow a = 5 \cr & {\text{Then , }} \cr & p = 3,{\text{ }}a = 5 \cr & a - p = 5 - 3 \cr & \Leftrightarrow a - p = 2 \cr}$

If $\frac{a}{1 - 2 a}$ $\frac{a}{{1 - 2a}}$ + $\frac{b}{1 - 2 b}$ $\frac{b}{{1 - 2b}}$ + $\frac{c}{1 - 2 c}$ $\frac{c}{{1 - 2c}}$ = $\frac{1}{2},$ $\frac{1}{2}{\text{,}}$ then the value of $\frac{1}{1 - 2 a}$ $\frac{1}{{1 - 2a}}$ + $\frac{1}{1 - 2 b}$ $\frac{1}{{1 - 2b}}$ + $\frac{1}{1 - 2 c}$ $\frac{1}{{1 - 2c}}$ is?

11
22
33
44

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{a}{1 - 2 a} + \frac{b}{1 - 2 b} + \frac{c}{1 - 2 c} = \frac{1}{2} \\ Multiply by 2 both side \\ \Rightarrow \frac{2 a}{1 - 2 a} + \frac{2 b}{1 - 2 b} + \frac{2 c}{1 - 2 c} = 1 \\ Adding 3 both side \\ \Rightarrow 1 + \frac{2 a}{1 - 2 a} + 1 + \frac{2 b}{1 - 2 b} + 1 + \frac{2 c}{1 - 2 c} = 1 + 3 \\ \Rightarrow \frac{1}{1 - 2 a} + \frac{1}{1 - 2 b} + \frac{1}{1 - 2 c} = 4 \end{aligned}

$\eqalign{ & \frac{a}{{1 - 2a}} + \frac{b}{{1 - 2b}} + \frac{c}{{1 - 2c}} = \frac{1}{2} \cr & {\text{Multiply by 2 both side}} \cr & \Rightarrow \frac{{2a}}{{1 - 2a}} + \frac{{2b}}{{1 - 2b}} + \frac{{2c}}{{1 - 2c}} = 1 \cr & {\text{Adding 3 both side}} \cr & \Rightarrow 1 + \frac{{2a}}{{1 - 2a}} + 1 + \frac{{2b}}{{1 - 2b}} + 1 + \frac{{2c}}{{1 - 2c}} = 1 + 3 \cr & \Rightarrow \frac{1}{{1 - 2a}} + \frac{1}{{1 - 2b}} + \frac{1}{{1 - 2c}} = 4 \cr}$

If x = 999, y = 1000, z = 1001 then the value of $\frac{x^{3} + y^{3} + z^{3} - 3 x y z}{x - y + z}$ $\frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{x - y + z}}$ is?

11000
29000
31
49

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} ∴ a^{3} + b^{3} + c^{3} - 3 a b c \\ = \frac{1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}] \\ ∴ \frac{x^{3} + y^{3} + z^{3} - 3 x y z}{x - y + z} \\ = \frac{\frac{1}{2} (x + y + z) [{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2}]}{x - y + z} \\ = \frac{\frac{1}{2} (999 + 1000 + 1001) (1 + 1 + 4)}{999 - 1000 + 1001} \\ = \frac{\frac{1}{2} \times 6 \times 3000}{1000} \\ = 9 \end{aligned}

$\eqalign{ & \therefore {a^3} + {b^3} + {c^3} - 3abc \cr & = \frac{1}{2}\left( {a + b + c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & \therefore \frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{x - y + z}} \cr & = \frac{{\frac{1}{2}\left( {x + y + z} \right)\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]}}{{x - y + z}} \cr & = \frac{{\frac{1}{2}\left( {999 + 1000 + 1001} \right)\left( {1 + 1 + 4} \right)}}{{999 - 1000 + 1001}} \cr & = \frac{{\frac{1}{2} \times 6 \times 3000}}{{1000}} \cr & = 9 \cr}$

08. If (x - y)² + (y - 2)² + (z - 9)² = 0, then the value of (x + y - z) is?

116
2-1
3-2
412

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} {(x - 5)}^{2} + {(y - 2)}^{2} + {(z - 9)}^{2} = 0 \\ It is possible only when \\ x - 5 = 0 \\ x = 5 \\ y - 2 = 0 \\ y = 2 \\ z - 9 = 0 \\ z = 9 \\ ∴ x + y - z \\ = 5 + 2 - 9 \\ = 7 - 9 \\ = - 2 \end{aligned}

$\eqalign{ & {\left( {x - 5} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 9} \right)^2} = 0 \cr & {\text{It is possible only when }} \cr & x - 5 = 0 \cr & x = 5 \cr & y - 2 = 0 \cr & y = 2 \cr & z - 9 = 0 \cr & z = 9 \cr & \therefore x + y - z \cr & = 5 + 2 - 9 \cr & = 7 - 9 \cr & = - 2 \cr}$

If $\frac{1}{a} (a^{2} + 1) = 3,$ $\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}}$ then the value of $\frac{a^{6} + 1}{a^{3}}$ $\frac{{{a^6} + 1}}{{{a^3}}}$ = ?

19
218
327
41

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{1}{a} (a^{2} + 1) = 3 \\ \Rightarrow a + \frac{1}{a} = 3 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3. a . \frac{1}{a} (a + \frac{1}{a}) = 3^{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} + 3 (3) = 3^{3} \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 27 - 9 \\ \Rightarrow a^{3} + \frac{1}{a^{3}} = 18 \\ \Rightarrow \frac{a^{6} + 1}{a^{3}} = 18 \end{aligned}

$\eqalign{ & \frac{1}{a}\left( {{a^2} + 1} \right) = 3 \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( 3 \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr & \Rightarrow \frac{{{a^6} + 1}}{{{a^3}}} = 18 \cr}$

If $\frac{2 + a}{a}$ $\frac{{2 + a}}{a}$ + $\frac{2 + b}{b}$ $\frac{{2 + b}}{b}$ + $\frac{2 + c}{c}$ $\frac{{2 + c}}{c}$ = 4, then the value of $\frac{a b + b c + c a}{a b c}$ $\frac{{ab + bc + ca}}{{abc}}$ is?

12
21
30
4 $\frac{1}{2}$ $\frac{1}{2}$

View Answer Discuss in forum Workspace Report

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{2 + a}{a} + \frac{2 + b}{b} + \frac{2 + c}{c} = 4 \\ \Rightarrow \frac{2 b c + a b c + 2 a c + a b c + 2 a b + a b c}{a b c} = 4 \\ \Rightarrow \frac{2 (b c + a b + c a)}{a b c} + \frac{3 a b c}{a b c} = 4 \\ \Rightarrow \frac{2 (b c + a b + c a)}{a b c} + 3 = 4 \\ \Rightarrow \frac{a b + b c + c a}{a b c} = \frac{1}{2} \end{aligned}

$\eqalign{ & \frac{{2 + a}}{a} + \frac{{2 + b}}{b} + \frac{{2 + c}}{c} = 4 \cr & \Rightarrow \frac{{2bc + abc + 2ac + abc + 2ab + abc}}{{abc}} = 4 \cr & \Rightarrow \frac{{2\left( {bc + ab + ca} \right)}}{{abc}} + \frac{{3abc}}{{abc}} = 4 \cr & \Rightarrow \frac{{2\left( {bc + ab + ca} \right)}}{{abc}} + 3 = 4 \cr & \Rightarrow \boxed{\frac{{ab + bc + ca}}{{abc}} = \frac{1}{2}} \cr}$

Hello Guest

algebra - 03

If $x + \sqrt{5} = 5 + \sqrt{y}$ $x + \sqrt 5 = 5 + \sqrt y$ and x, y are positive integers, then the value of $\frac{\sqrt{x} + y}{x + \sqrt{y}}$ $\frac{{\sqrt x + y}}{{x + \sqrt y }}$ is?

If $p^{2} + \frac{1}{p^{2}} = 47,$ ${p^2} + \frac{1}{{{p^2}}} = 47{\text{,}}$ then the value of $p + \frac{1}{p}$ $p + \frac{1}{p}$ is?

If $x = \sqrt[3]{x^{2} + 11} - 2,$ $x = \root 3 \of {{x^2} + 11} - 2{\text{,}}$ then the value of x³ + 5x² + 12x is?

If $x = \sqrt{a} + \frac{1}{\sqrt{a}},$ $x = \sqrt a + \frac{1}{{\sqrt a }}{\text{,}}$ $y = \sqrt{a} - \frac{1}{\sqrt{a}},$ $y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}}$ $(a > 0)$ $\left( {a > 0} \right)$ then the value of x⁴ + y⁴ - 2x²y² is?

05. If (x - 2)(x - p) = x² - ax + 6, then the value of (a - p) is?

If x = 999, y = 1000, z = 1001 then the value of $\frac{x^{3} + y^{3} + z^{3} - 3 x y z}{x - y + z}$ $\frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{x - y + z}}$ is?

08. If (x - y)² + (y - 2)² + (z - 9)² = 0, then the value of (x + y - z) is?

If $\frac{1}{a} (a^{2} + 1) = 3,$ $\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}}$ then the value of $\frac{a^{6} + 1}{a^{3}}$ $\frac{{{a^6} + 1}}{{{a^3}}}$ = ?

If $\frac{2 + a}{a}$ $\frac{{2 + a}}{a}$ + $\frac{2 + b}{b}$ $\frac{{2 + b}}{b}$ + $\frac{2 + c}{c}$ $\frac{{2 + c}}{c}$ = 4, then the value of $\frac{a b + b c + c a}{a b c}$ $\frac{{ab + bc + ca}}{{abc}}$ is?

Subject Wise

Exam Wise

NCERT
6 to 12

Paid Course

Practice Batch

Daily Current Affairs

Monthly Current Affairs

Topic Wise Current Affairs

State Wise Current Affairs

Prelims Facts

Hello Guest

algebra - 03

01. If x+5–√=5+y√x+5=5+yx + \sqrt 5 = 5 + \sqrt y and x, y are positive integers, then the value of x−−√+yx+y√x+yx+y\frac{{\sqrt x + y}}{{x + \sqrt y }} is?

02. If p2+1p2=47,p2+1p2=47,{p^2} + \frac{1}{{{p^2}}} = 47{\text{,}} then the value of p+1pp+1pp + \frac{1}{p} is?

03. If x=x2+11−−−−−−√3−2,x=x2+113−2,x = \root 3 \of {{x^2} + 11} - 2{\text{,}} then the value of x3 + 5x2 + 12x is?

04. If x=a−−√+1a−−√,x=a+1a,x = \sqrt a + \frac{1}{{\sqrt a }}{\text{,}} y=a−−√−1a−−√,y=a−1a,y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}} (a>0)(a>0)\left( {a > 0} \right) then the value of x4 + y4 - 2x2y2 is?

05. If (x - 2)(x - p) = x2 - ax + 6, then the value of (a - p) is?

06. If a1−2aa1−2a\frac{a}{{1 - 2a}} + b1−2bb1−2b\frac{b}{{1 - 2b}} + c1−2cc1−2c\frac{c}{{1 - 2c}} = 12,12,\frac{1}{2}{\text{,}} then the value of 11−2a11−2a\frac{1}{{1 - 2a}} + 11−2b11−2b\frac{1}{{1 - 2b}} + 11−2c11−2c\frac{1}{{1 - 2c}} is?

07. If x = 999, y = 1000, z = 1001 then the value of x3+y3+z3−3xyzx−y+zx3+y3+z3−3xyzx−y+z\frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{x - y + z}} is?

08. If (x - y)2 + (y - 2)2 + (z - 9)2 = 0, then the value of (x + y - z) is?

09. If 1a(a2+1)=3,1a(a2+1)=3,\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}} then the value of a6+1a3a6+1a3\frac{{{a^6} + 1}}{{{a^3}}} = ?

10. If 2+aa2+aa\frac{{2 + a}}{a} + 2+bb2+bb\frac{{2 + b}}{b} + 2+cc2+cc\frac{{2 + c}}{c} = 4, then the value of ab+bc+caabcab+bc+caabc\frac{{ab + bc + ca}}{{abc}} is?

Subject Wise

Exam Wise

NCERT 6 to 12

Paid Course

Practice Batch

If $x + \sqrt{5} = 5 + \sqrt{y}$ $x + \sqrt 5 = 5 + \sqrt y$ and x, y are positive integers, then the value of $\frac{\sqrt{x} + y}{x + \sqrt{y}}$ $\frac{{\sqrt x + y}}{{x + \sqrt y }}$ is?

If $p^{2} + \frac{1}{p^{2}} = 47,$ ${p^2} + \frac{1}{{{p^2}}} = 47{\text{,}}$ then the value of $p + \frac{1}{p}$ $p + \frac{1}{p}$ is?

If $x = \sqrt[3]{x^{2} + 11} - 2,$ $x = \root 3 \of {{x^2} + 11} - 2{\text{,}}$ then the value of x³ + 5x² + 12x is?

If $x = \sqrt{a} + \frac{1}{\sqrt{a}},$ $x = \sqrt a + \frac{1}{{\sqrt a }}{\text{,}}$ $y = \sqrt{a} - \frac{1}{\sqrt{a}},$ $y = \sqrt a - \frac{1}{{\sqrt a }}{\text{,}}$ $(a > 0)$ $\left( {a > 0} \right)$ then the value of x⁴ + y⁴ - 2x²y² is?

05. If (x - 2)(x - p) = x² - ax + 6, then the value of (a - p) is?

If x = 999, y = 1000, z = 1001 then the value of $\frac{x^{3} + y^{3} + z^{3} - 3 x y z}{x - y + z}$ $\frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{x - y + z}}$ is?

08. If (x - y)² + (y - 2)² + (z - 9)² = 0, then the value of (x + y - z) is?

If $\frac{1}{a} (a^{2} + 1) = 3,$ $\frac{1}{a}\left( {{a^2} + 1} \right) = 3{\text{,}}$ then the value of $\frac{a^{6} + 1}{a^{3}}$ $\frac{{{a^6} + 1}}{{{a^3}}}$ = ?

If $\frac{2 + a}{a}$ $\frac{{2 + a}}{a}$ + $\frac{2 + b}{b}$ $\frac{{2 + b}}{b}$ + $\frac{2 + c}{c}$ $\frac{{2 + c}}{c}$ = 4, then the value of $\frac{a b + b c + c a}{a b c}$ $\frac{{ab + bc + ca}}{{abc}}$ is?

NCERT
6 to 12