QBManager

Sets, Relation and Function

116984 If a function $F$ is such that $F (0) = 2, F (1) = 3$ , $F (n + 2) = 2 F (n) - F (n + 1)$ for $n \neq 0$ , then $F (5)$ is equal to

1 -7

2 -3

3 7

4 13

Explanation:

D Given $F (0) = 2, F (1) = 3$
$F (n + 2) = 2 F (n) - F (n + 1)$
Putting, $n = 0$
$F (0 + 2) = 2 F (0) - F (0 + 1)$
$F (2) = 2 \times 2 - 3$
$F (2) = 1$
Put $n = 1$
$F (3) = 2 F (1) - F (2)$
$F (3) = 2 \times 3 - 1$
$F (3) = 5$
Put $n = 2$
$F (2 + 2) = 2 F (2) - F (3)$
$F (4) = 2 \times 1 - 5$
$F (4) = - 3$
Put
$n = 3$
$F (5) = 2 F (3) - F (4)$
$= 2 \times 5 - (- 3)$
$= 10 + 3$
$F (5) = 13$

J and K CET-2003

Sets, Relation and Function

116986 If $f (x) = a x^{2} + b x + c$ satisfies $f (1) + 2 f (2) = 0$ and $2 f (1) + f (2) = 0$ , then $3 a + b =$

1 2

2 -1

3 0

4 1

Explanation:

C We have equation,
$f (x) = a x^{2} + b x + c$
$f (1) + 2 f (2) = 0$
$2 f (1) + f (2) = 0$
Now
$f (1) = a + b + c$
$f (2) = 4 a + 2 b + c$
$f (2) - f (1) = 3 a + b$
Now,
$f (1) + 2 f (2) = 0$
$a + b + c + 2 (4 a + 2 b + c) = 0$
$2 (a + b + c) + 4 a + 2 b + c = 0$
Adding these equation,
$3 (a + b + c) + 3 (4 a + 2 b + c) = 0$
$3 [a + b + c + 4 a + 2 b + c] = 0$
$3 [5 a + 3 b + 2 c] = 0$
It can be written as
$(a + b + c) - 2 (a + b + c) + 2 (4 a + 2 b + c)$
$- (4 a + 2 b + c)$
$As - (a + b + c) + 4 a + 2 b + c = 0$
$3 a + b = 0$
As $- (a + b + c) + 4 a + 2 b + c = 0$
$3 a + b = 0$

Shift-I

Sets, Relation and Function

116987 Let $f$ be a function defined by $f (x y) = \frac{f (x)}{y}$ for all positive real numbers $x$ and $y$ If $f (30) = 20$ , then $f (40) =$

1 10

2 15

3 25

4 17

Explanation:

B Given function
$f (xy) = \frac{f (x)}{y}$
Put
$x = 1$
$f (y) = \frac{f (1)}{y}$
Put
$y = 30$
$f (30) = \frac{f (1)}{30}$
$f (1) = f (30) \cdot 30$
$= 20 \times 30$
$f (1) = 600$
Put $y = 40$ in equation
$f (40) = \frac{f (1)}{40}$
$f (40) = \frac{600}{40}$
$f (40) = 15$

Shift-I

Sets, Relation and Function

116988 If $f (x) = \frac{1}{\sqrt{x + 2 \sqrt{2 x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2 x - 4}}}$ for $x >$ 2 , then $f (11)$ is equal to

1

\frac{7}{6}

2

\frac{5}{6}

3

\frac{6}{7}

4

\frac{5}{7}

Explanation:

:Given,
$f (x) = \frac{1}{\sqrt{x + 2 \sqrt{2 x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2 x - 4}}}$
Now find $f (11)$
Putting $x = 11$
$f (11) = \frac{1}{\sqrt{11 + 2} \sqrt{2 \times 11 - 4}} + \frac{1}{\sqrt{11 - 2 \sqrt{2 \times 11 - 4}}}$
$= \frac{1}{\sqrt{11 + 2 \sqrt{22 - 4}}} + \frac{1}{\sqrt{11 - 2 \sqrt{22 - 4}}}$
$= \frac{1}{\sqrt{11 + 2 \sqrt{18}}} + \frac{1}{\sqrt{11 - 2 \sqrt{18}}}$
$= \frac{1}{\sqrt{9 + 2 + 6 \sqrt{2}}} + \frac{1}{\sqrt{9 + 2 - 6 \sqrt{2}}}$
$= \frac{1}{\sqrt{(3)^{2} + (\sqrt{2})^{2} + 6 \sqrt{2}}} + \frac{1}{\sqrt{(3)^{2} + (\sqrt{2})^{2} - 6 \sqrt{2}}}$
$= \frac{1}{\sqrt{(\sqrt{2} + 3)^{2}}} + \frac{1}{\sqrt{(3)^{2} - (\sqrt{2})^{2}}}$
$= \frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}$
$= \frac{3 - \sqrt{2} + 3 + \sqrt{2}}{(3 + \sqrt{2}) (3 - \sqrt{2})}$
$= \frac{6}{3^{2} - (\sqrt{2})^{2}}$
$= \frac{6}{9 - 2} = \frac{6}{7}$

EAMCET-2003

Sets, Relation and Function

116984 If a function $F$ is such that $F (0) = 2, F (1) = 3$ , $F (n + 2) = 2 F (n) - F (n + 1)$ for $n \neq 0$ , then $F (5)$ is equal to

1 -7

2 -3

3 7

4 13

Explanation:

D Given $F (0) = 2, F (1) = 3$
$F (n + 2) = 2 F (n) - F (n + 1)$
Putting, $n = 0$
$F (0 + 2) = 2 F (0) - F (0 + 1)$
$F (2) = 2 \times 2 - 3$
$F (2) = 1$
Put $n = 1$
$F (3) = 2 F (1) - F (2)$
$F (3) = 2 \times 3 - 1$
$F (3) = 5$
Put $n = 2$
$F (2 + 2) = 2 F (2) - F (3)$
$F (4) = 2 \times 1 - 5$
$F (4) = - 3$
Put
$n = 3$
$F (5) = 2 F (3) - F (4)$
$= 2 \times 5 - (- 3)$
$= 10 + 3$
$F (5) = 13$

J and K CET-2003

Sets, Relation and Function

116986 If $f (x) = a x^{2} + b x + c$ satisfies $f (1) + 2 f (2) = 0$ and $2 f (1) + f (2) = 0$ , then $3 a + b =$

1 2

2 -1

3 0

4 1

Explanation:

C We have equation,
$f (x) = a x^{2} + b x + c$
$f (1) + 2 f (2) = 0$
$2 f (1) + f (2) = 0$
Now
$f (1) = a + b + c$
$f (2) = 4 a + 2 b + c$
$f (2) - f (1) = 3 a + b$
Now,
$f (1) + 2 f (2) = 0$
$a + b + c + 2 (4 a + 2 b + c) = 0$
$2 (a + b + c) + 4 a + 2 b + c = 0$
Adding these equation,
$3 (a + b + c) + 3 (4 a + 2 b + c) = 0$
$3 [a + b + c + 4 a + 2 b + c] = 0$
$3 [5 a + 3 b + 2 c] = 0$
It can be written as
$(a + b + c) - 2 (a + b + c) + 2 (4 a + 2 b + c)$
$- (4 a + 2 b + c)$
$As - (a + b + c) + 4 a + 2 b + c = 0$
$3 a + b = 0$
As $- (a + b + c) + 4 a + 2 b + c = 0$
$3 a + b = 0$

Shift-I

Sets, Relation and Function

116987 Let $f$ be a function defined by $f (x y) = \frac{f (x)}{y}$ for all positive real numbers $x$ and $y$ If $f (30) = 20$ , then $f (40) =$

1 10

2 15

3 25

4 17

Explanation:

B Given function
$f (xy) = \frac{f (x)}{y}$
Put
$x = 1$
$f (y) = \frac{f (1)}{y}$
Put
$y = 30$
$f (30) = \frac{f (1)}{30}$
$f (1) = f (30) \cdot 30$
$= 20 \times 30$
$f (1) = 600$
Put $y = 40$ in equation
$f (40) = \frac{f (1)}{40}$
$f (40) = \frac{600}{40}$
$f (40) = 15$

Shift-I

Sets, Relation and Function

116988 If $f (x) = \frac{1}{\sqrt{x + 2 \sqrt{2 x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2 x - 4}}}$ for $x >$ 2 , then $f (11)$ is equal to

1

\frac{7}{6}

2

\frac{5}{6}

3

\frac{6}{7}

4

\frac{5}{7}

Explanation:

:Given,
$f (x) = \frac{1}{\sqrt{x + 2 \sqrt{2 x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2 x - 4}}}$
Now find $f (11)$
Putting $x = 11$
$f (11) = \frac{1}{\sqrt{11 + 2} \sqrt{2 \times 11 - 4}} + \frac{1}{\sqrt{11 - 2 \sqrt{2 \times 11 - 4}}}$
$= \frac{1}{\sqrt{11 + 2 \sqrt{22 - 4}}} + \frac{1}{\sqrt{11 - 2 \sqrt{22 - 4}}}$
$= \frac{1}{\sqrt{11 + 2 \sqrt{18}}} + \frac{1}{\sqrt{11 - 2 \sqrt{18}}}$
$= \frac{1}{\sqrt{9 + 2 + 6 \sqrt{2}}} + \frac{1}{\sqrt{9 + 2 - 6 \sqrt{2}}}$
$= \frac{1}{\sqrt{(3)^{2} + (\sqrt{2})^{2} + 6 \sqrt{2}}} + \frac{1}{\sqrt{(3)^{2} + (\sqrt{2})^{2} - 6 \sqrt{2}}}$
$= \frac{1}{\sqrt{(\sqrt{2} + 3)^{2}}} + \frac{1}{\sqrt{(3)^{2} - (\sqrt{2})^{2}}}$
$= \frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}$
$= \frac{3 - \sqrt{2} + 3 + \sqrt{2}}{(3 + \sqrt{2}) (3 - \sqrt{2})}$
$= \frac{6}{3^{2} - (\sqrt{2})^{2}}$
$= \frac{6}{9 - 2} = \frac{6}{7}$

EAMCET-2003

Sets, Relation and Function

116984 If a function $F$ is such that $F (0) = 2, F (1) = 3$ , $F (n + 2) = 2 F (n) - F (n + 1)$ for $n \neq 0$ , then $F (5)$ is equal to

1 -7

2 -3

3 7

4 13

Explanation:

D Given $F (0) = 2, F (1) = 3$
$F (n + 2) = 2 F (n) - F (n + 1)$
Putting, $n = 0$
$F (0 + 2) = 2 F (0) - F (0 + 1)$
$F (2) = 2 \times 2 - 3$
$F (2) = 1$
Put $n = 1$
$F (3) = 2 F (1) - F (2)$
$F (3) = 2 \times 3 - 1$
$F (3) = 5$
Put $n = 2$
$F (2 + 2) = 2 F (2) - F (3)$
$F (4) = 2 \times 1 - 5$
$F (4) = - 3$
Put
$n = 3$
$F (5) = 2 F (3) - F (4)$
$= 2 \times 5 - (- 3)$
$= 10 + 3$
$F (5) = 13$

J and K CET-2003

Sets, Relation and Function

116986 If $f (x) = a x^{2} + b x + c$ satisfies $f (1) + 2 f (2) = 0$ and $2 f (1) + f (2) = 0$ , then $3 a + b =$

1 2

2 -1

3 0

4 1

Explanation:

C We have equation,
$f (x) = a x^{2} + b x + c$
$f (1) + 2 f (2) = 0$
$2 f (1) + f (2) = 0$
Now
$f (1) = a + b + c$
$f (2) = 4 a + 2 b + c$
$f (2) - f (1) = 3 a + b$
Now,
$f (1) + 2 f (2) = 0$
$a + b + c + 2 (4 a + 2 b + c) = 0$
$2 (a + b + c) + 4 a + 2 b + c = 0$
Adding these equation,
$3 (a + b + c) + 3 (4 a + 2 b + c) = 0$
$3 [a + b + c + 4 a + 2 b + c] = 0$
$3 [5 a + 3 b + 2 c] = 0$
It can be written as
$(a + b + c) - 2 (a + b + c) + 2 (4 a + 2 b + c)$
$- (4 a + 2 b + c)$
$As - (a + b + c) + 4 a + 2 b + c = 0$
$3 a + b = 0$
As $- (a + b + c) + 4 a + 2 b + c = 0$
$3 a + b = 0$

Shift-I

Sets, Relation and Function

116987 Let $f$ be a function defined by $f (x y) = \frac{f (x)}{y}$ for all positive real numbers $x$ and $y$ If $f (30) = 20$ , then $f (40) =$

1 10

2 15

3 25

4 17

Explanation:

B Given function
$f (xy) = \frac{f (x)}{y}$
Put
$x = 1$
$f (y) = \frac{f (1)}{y}$
Put
$y = 30$
$f (30) = \frac{f (1)}{30}$
$f (1) = f (30) \cdot 30$
$= 20 \times 30$
$f (1) = 600$
Put $y = 40$ in equation
$f (40) = \frac{f (1)}{40}$
$f (40) = \frac{600}{40}$
$f (40) = 15$

Shift-I

Sets, Relation and Function

116988 If $f (x) = \frac{1}{\sqrt{x + 2 \sqrt{2 x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2 x - 4}}}$ for $x >$ 2 , then $f (11)$ is equal to

1

\frac{7}{6}

2

\frac{5}{6}

3

\frac{6}{7}

4

\frac{5}{7}

Explanation:

:Given,
$f (x) = \frac{1}{\sqrt{x + 2 \sqrt{2 x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2 x - 4}}}$
Now find $f (11)$
Putting $x = 11$
$f (11) = \frac{1}{\sqrt{11 + 2} \sqrt{2 \times 11 - 4}} + \frac{1}{\sqrt{11 - 2 \sqrt{2 \times 11 - 4}}}$
$= \frac{1}{\sqrt{11 + 2 \sqrt{22 - 4}}} + \frac{1}{\sqrt{11 - 2 \sqrt{22 - 4}}}$
$= \frac{1}{\sqrt{11 + 2 \sqrt{18}}} + \frac{1}{\sqrt{11 - 2 \sqrt{18}}}$
$= \frac{1}{\sqrt{9 + 2 + 6 \sqrt{2}}} + \frac{1}{\sqrt{9 + 2 - 6 \sqrt{2}}}$
$= \frac{1}{\sqrt{(3)^{2} + (\sqrt{2})^{2} + 6 \sqrt{2}}} + \frac{1}{\sqrt{(3)^{2} + (\sqrt{2})^{2} - 6 \sqrt{2}}}$
$= \frac{1}{\sqrt{(\sqrt{2} + 3)^{2}}} + \frac{1}{\sqrt{(3)^{2} - (\sqrt{2})^{2}}}$
$= \frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}$
$= \frac{3 - \sqrt{2} + 3 + \sqrt{2}}{(3 + \sqrt{2}) (3 - \sqrt{2})}$
$= \frac{6}{3^{2} - (\sqrt{2})^{2}}$
$= \frac{6}{9 - 2} = \frac{6}{7}$

EAMCET-2003

Sets, Relation and Function

116984 If a function $F$ is such that $F (0) = 2, F (1) = 3$ , $F (n + 2) = 2 F (n) - F (n + 1)$ for $n \neq 0$ , then $F (5)$ is equal to

1 -7

2 -3

3 7

4 13

Explanation:

D Given $F (0) = 2, F (1) = 3$
$F (n + 2) = 2 F (n) - F (n + 1)$
Putting, $n = 0$
$F (0 + 2) = 2 F (0) - F (0 + 1)$
$F (2) = 2 \times 2 - 3$
$F (2) = 1$
Put $n = 1$
$F (3) = 2 F (1) - F (2)$
$F (3) = 2 \times 3 - 1$
$F (3) = 5$
Put $n = 2$
$F (2 + 2) = 2 F (2) - F (3)$
$F (4) = 2 \times 1 - 5$
$F (4) = - 3$
Put
$n = 3$
$F (5) = 2 F (3) - F (4)$
$= 2 \times 5 - (- 3)$
$= 10 + 3$
$F (5) = 13$

J and K CET-2003

Sets, Relation and Function

116986 If $f (x) = a x^{2} + b x + c$ satisfies $f (1) + 2 f (2) = 0$ and $2 f (1) + f (2) = 0$ , then $3 a + b =$

1 2

2 -1

3 0

4 1

Explanation:

C We have equation,
$f (x) = a x^{2} + b x + c$
$f (1) + 2 f (2) = 0$
$2 f (1) + f (2) = 0$
Now
$f (1) = a + b + c$
$f (2) = 4 a + 2 b + c$
$f (2) - f (1) = 3 a + b$
Now,
$f (1) + 2 f (2) = 0$
$a + b + c + 2 (4 a + 2 b + c) = 0$
$2 (a + b + c) + 4 a + 2 b + c = 0$
Adding these equation,
$3 (a + b + c) + 3 (4 a + 2 b + c) = 0$
$3 [a + b + c + 4 a + 2 b + c] = 0$
$3 [5 a + 3 b + 2 c] = 0$
It can be written as
$(a + b + c) - 2 (a + b + c) + 2 (4 a + 2 b + c)$
$- (4 a + 2 b + c)$
$As - (a + b + c) + 4 a + 2 b + c = 0$
$3 a + b = 0$
As $- (a + b + c) + 4 a + 2 b + c = 0$
$3 a + b = 0$

Shift-I

Sets, Relation and Function

116987 Let $f$ be a function defined by $f (x y) = \frac{f (x)}{y}$ for all positive real numbers $x$ and $y$ If $f (30) = 20$ , then $f (40) =$

1 10

2 15

3 25

4 17

Explanation:

B Given function
$f (xy) = \frac{f (x)}{y}$
Put
$x = 1$
$f (y) = \frac{f (1)}{y}$
Put
$y = 30$
$f (30) = \frac{f (1)}{30}$
$f (1) = f (30) \cdot 30$
$= 20 \times 30$
$f (1) = 600$
Put $y = 40$ in equation
$f (40) = \frac{f (1)}{40}$
$f (40) = \frac{600}{40}$
$f (40) = 15$

Shift-I

Sets, Relation and Function

116988 If $f (x) = \frac{1}{\sqrt{x + 2 \sqrt{2 x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2 x - 4}}}$ for $x >$ 2 , then $f (11)$ is equal to

1

\frac{7}{6}

2

\frac{5}{6}

3

\frac{6}{7}

4

\frac{5}{7}

Explanation:

:Given,
$f (x) = \frac{1}{\sqrt{x + 2 \sqrt{2 x - 4}}} + \frac{1}{\sqrt{x - 2 \sqrt{2 x - 4}}}$
Now find $f (11)$
Putting $x = 11$
$f (11) = \frac{1}{\sqrt{11 + 2} \sqrt{2 \times 11 - 4}} + \frac{1}{\sqrt{11 - 2 \sqrt{2 \times 11 - 4}}}$
$= \frac{1}{\sqrt{11 + 2 \sqrt{22 - 4}}} + \frac{1}{\sqrt{11 - 2 \sqrt{22 - 4}}}$
$= \frac{1}{\sqrt{11 + 2 \sqrt{18}}} + \frac{1}{\sqrt{11 - 2 \sqrt{18}}}$
$= \frac{1}{\sqrt{9 + 2 + 6 \sqrt{2}}} + \frac{1}{\sqrt{9 + 2 - 6 \sqrt{2}}}$
$= \frac{1}{\sqrt{(3)^{2} + (\sqrt{2})^{2} + 6 \sqrt{2}}} + \frac{1}{\sqrt{(3)^{2} + (\sqrt{2})^{2} - 6 \sqrt{2}}}$
$= \frac{1}{\sqrt{(\sqrt{2} + 3)^{2}}} + \frac{1}{\sqrt{(3)^{2} - (\sqrt{2})^{2}}}$
$= \frac{1}{3 + \sqrt{2}} + \frac{1}{3 - \sqrt{2}}$
$= \frac{3 - \sqrt{2} + 3 + \sqrt{2}}{(3 + \sqrt{2}) (3 - \sqrt{2})}$
$= \frac{6}{3^{2} - (\sqrt{2})^{2}}$
$= \frac{6}{9 - 2} = \frac{6}{7}$

EAMCET-2003

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