Chapter 2: Relations and Functions

Ans: Given: \[{{f(x)  =  }}\dfrac{1}{{1{{  -  2cosx}}}}\]

Range of function represents the value the function attains. Start with the range of functions you know.

Range of cosine function is,

  $\Rightarrow {{  -  1 }} \leqslant {{  -  cosx }} \leqslant {{ 1}}$

  $\Rightarrow {{  -  2 }} \leqslant {{  -  2cosx }} \leqslant {{ 2}}$

  $\Rightarrow {{ 1  -  2 }} \leqslant {{ 1  -  2cosx }} \leqslant {{ 1  +  2}}$ 

  $\Rightarrow {{  -  1 }} \leqslant {{ 1  -  2cosx }} \leqslant {{ 3}}$

  $\Rightarrow {{  -  1 }} \leqslant {{ }}\dfrac{1}{{1{{  -  cos2x}}}}{{ }} \leqslant {{ }}\dfrac{1}{3}$

  $\Rightarrow {{  -  1 }} \leqslant {{ f(x) }} \leqslant {{ }}\dfrac{1}{3}$

  ${{\text{ Range of }f  =  }}\left[ { - {{ 1, }}\dfrac{1}{3}} \right].$ 

Correct Answer: B

Ans: Given: Function \[{{f(x)  =  }}\sqrt {1{{  +  }}{{{x}}^2}} .\]

Replace the variables of function as \[{{x }} \to {{ xy}}\] and then compare.

We have,

  ${{f(x)  =  }}\sqrt {1{{  +  }}{{{x}}^2}}$

  ${{f(xy)  =  }}\sqrt {1{{  +  }}{{{x}}^2}{{{y}}^2}}$

  ${{f(x) }}{{. f(y)  =  }}\sqrt {1{{  +  }}{{{x}}^2}} {{ }}{{. }}\sqrt {1{{  +  }}{{{y}}^2}}$

  ${{f(x) }}{{. f(y)  =  }}\sqrt {(1{{  +  }}{{{x}}^2})(1{{  +  }}{{{y}}^2})}$

  ${{f(x) }}{{. f(y)  =  }}\sqrt {1{{  +  }}{{{x}}^2}{{  +  }}{{{y}}^2}{{  +  }}{{{x}}^2}{{{y}}^2}}$

  $\sqrt {1{{  +  }}{{{x}}^2}{{{y}}^2}} {{ }} \leqslant {{ }}\sqrt {1{{  +  }}{{{x}}^2}} {{ }}{{. }}\sqrt {1{{  +  }}{{{y}}^2}}$

  $\Rightarrow {{ f(xy) }} \leqslant {{ f(x) }}{{. f(y)}}{{.}}$ 

Correct Answer: C

Ans: Given: \[{{f(x)  =  }}\sqrt {{{{a}}^2}{{  -  }}{{{x}}^2}} \]

Domain of the function is the value of \[{{x}}\]for which the function is defined.

 Let ${{ f(x)  =  }}\sqrt {{{{a}}^2}{{  -  }}{{{x}}^2}}$

  $f(x)$ is defined, if ${{}}{{{a}}^2}\; - \;{{{x}}^2}{{ }} \geqslant \;{{0}}$

  $\Rightarrow {{ }}{{{x}}^2}\; - \;{{{a}}^2}\; \leqslant \;0$

  $\Rightarrow {{ (x  -  a)(x  +  a) }} \leqslant {{ 0}}$

  $\Rightarrow {{  -  a }} \leqslant {{ x }} \leqslant {{ a}}$

$\therefore {{ \text{ Domain of  }f  =  [ -  a, a]}}{{.}}$

Correct Answer: B

Ans: Given:

  ${{f(x)  =  ax  +  b}}$

  ${{f( -  1)  =   -  5 }}$

  ${{f(3)  =  3}}$ 

First, substitute the values in functions. Solve equations for \[{{a \text{ and } b}}{{.}}\]

  ${{for, f(x)  =  ax  +  b}}$

  $\Rightarrow {{ f( -  1)  =  a( -  1)  +  b}}$

  $\Rightarrow {{  -  5  =   -  a  +  b       - (i)}}$

  ${{\text{ and },}}$

  ${{f(3)  =  a(3)  +  b}}$

  $\Rightarrow {{ 3  =  3a  +  b              - (ii)}}$

  ${\text{from (i) and (ii), we get}}$

  ${{a  =  2 \text{ and } b  =   -  3}}$ 

Correct answer: B

Ans: Given: Function \[{{f(x)  =  }}\sqrt {4{{  -  x}}} {{  +  }}\dfrac{1}{{\sqrt {{{{x}}^2}{{  -  1}}} }}\]

Domain of the function is the value of \[{{x}}\]for which the function is defined.

  $f(x)$ is defined, if ${{}}$

  ${{4  -  x }} \geqslant {{ 0 \text{ or } }}{{{x}}^2}{{  -  1  >  0}}$

  ${{x  -  4 }} \leqslant {{ 0 \text{ or } (x  +  1)(x  -  1)  >  0}}$

  ${{x }} \leqslant {{ 4 \text{ or } x  <   -  1 \text{ and } x  >  1}}$

$\therefore {{ \text{ Domain of  }f  =  ( -  }}\infty ,{{  -  1) }} \cup {{ (1, 4]}}{{.}}$ 

Correct Answer: A

Ans: Given: \[{{f(x)  =  }}\dfrac{{4{{  -  x}}}}{{{{x  -  4}}}}\].

A rational function  \[\dfrac{{{{f(x)}}}}{{{{g(x)}}}}\] is defined if \[{{g(x) }} \ne {{ 0}}{{.}}\]

  $f(x)$ is defined, if ${{x  -  4 }} \ne {{ 0 \text{ or } x }} \ne {{ 4}}$

$\therefore {{ \text{ Domain of  }f  =  R  -  \{ 4\} }}{{.}}$

  ${{Let f(x)  =  y}}$

  $\Rightarrow {{ y  =  }}\dfrac{{4{{  -  x}}}}{{{{x  -  4}}}}$

  $\Rightarrow {{ xy  -  4y  =  4  -  x}}$

  $\Rightarrow {{ xy  +  x  =  4  +  4y}}$

  $\Rightarrow {{ x(y  +  1)  =  4(1  +  y)}}$ 

\[\Rightarrow {{ x  =  }}\dfrac{{4(1{{  +  y)}}}}{{{{y  +  1}}}}\]

${\text{x is defined, if y  +  1 }} \ne {{ 0, y }} \ne {{  -  1}}$

$\therefore$  Range of ${{f  =  R  -  \{  -  1\} }}{{.}}$ 

Correct Answer: C

Ans: Given: \[{{f(x)  =  }}\sqrt {{{x  -  1}}} \].

A square root function \[{{f(x)  =  }}\sqrt {{x}} \] is defined by \[{{x }} \geqslant {{ 0}}{{.}}\]

  $f(x)$ is defined, if ${{x  -  1 }} \geqslant {{ 0}}$

  $\Rightarrow {{ x }} \geqslant {{ 1}}$

$\therefore {{ \text{ Domain of  }f  =  [1, }}\infty {{)}}{{.}}$

  ${{Let f(x)  =  y}}$

  $\Rightarrow {{ y  =  }}\sqrt {{{x  -  1}}}$

  $\Rightarrow {{ }}{{{y}}^2}{{  =  x  -  1}}$

  $\Rightarrow {{ x  =  }}{{{y}}^2}{{  +  1}}$

  $x$ is defined for ${{y }} \in {{ R}}{{.}}$

  ${{but, y }} \geqslant {{ 0}}$

$\therefore$  Range of ${{f  =  [0, }}\infty ).$ 

Correct Answer: D

Ans: Given: \[{{f(x)  =  }}\dfrac{{{{{x}}^2}{{  +  2x  +  1}}}}{{{{{x}}^2}{{  -  x  -  6}}}}.\]

The domain is the value of \[{{x}}\] for which function is defined.

  ${{f(x) \text{ is defined, if } }}{{{x}}^2}{{  -  x  -  6  =  0}}$

  $\Rightarrow {{ }}{{{x}}^2}{{  -  3x  +  2x  -  6  =  0}}$

  $\Rightarrow {{ x(x  -  3)  +  2(x  -  3)  =  0}}$ 

  $\Rightarrow {{ (x  -  3)(x  +  2)  =  0}}$

  $\Rightarrow {{ x  =  3,  -  2}}$

$\therefore$ Domain of ${{f  =  R  -  \{ 3,  -  2\} }}{{.}}$ 

Correct Answer: A

Ans: Given: \[{{f(x)  =  2  -  | x  -  5 |}}{{.}}\]

The absolute modulus function \[|{{x | }} \geqslant {{ 0}}{{.}}\]

  ${\text{f(x) is defined for all x }} \in {{ R}}$

$\therefore {{ \text{ Domain of  }f  =  R}}{{.}}$

  ${{Now, | x  -  5 | }} \geqslant {{ 0}}$ 

  $\Rightarrow {{  -  | x  -  5 | }} \leqslant {{ 0}}$

  $\Rightarrow {{ 2  -  | x  -  5 | }} \leqslant {{ 2}}$

  $\Rightarrow {{ f(x)}}{{ }} \leqslant {{ 2}}$

$\therefore$  Range of ${{f  =  ( -  }}\infty {{, 2]}}{{.}}$ 

Correct Answer: B

Ans: Given: \[{{f(x)  =  3}}{{{x}}^2}{{  -  1 \text{ and } g(x)  =  3  +  x}}{{.}}\]

Substitute the functions equal and then find \[{{x}}{{.}}\]

  ${{\text{For}, f(x)  =  g(x)}}$

  $\Rightarrow {{ 3}}{{{x}}^2}{{  -  1  =  3  +  x}}$

  $\Rightarrow {{ 3}}{{{x}}^2}{{  -  x  -  4  =  0}}$

  $\Rightarrow {{ 3}}{{{x}}^2}{{  -  4x  +  3x  -  4  =  0}}$

  $\Rightarrow {{ x(3x  -  4)  +  1(x  +  1)  =  0}}$

  $\Rightarrow {{ (3x  -  4)(x  +  1)  =  0}}$

  $\Rightarrow {{ x  =   - 1, }}\dfrac{4}{3}$

$\therefore {{ \text{Domain for } f(x)  =  g(x) \text{ is } }}\left[ { - {{ 1, }}\dfrac{4}{3}} \right].$ 

Correct Answer: A

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