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This is one of our many quizzes on Graphs and functions. Click on this to access more such competitive quizzes on Algebra.

Graphs and functions Quiz 2

mathematics

 Let the function $f:R \to R, f(x) = {x^2}$ is?

 Here $f(x)$ cannot take negative values.

  Here co-domain is given as $R$, but ${x^2}$ is always positive $ \\ \therefore f(x) \in \left( {0,\infty } \right) \\ $ Hence, $f(x)$ is INTO.

 Let the function $f:R \to R$ be defined by $f(x)= 4x + \sin x$. Then $f$ is:

 Here $f(x)$ is a monotonically increasing function.

  As per data given in the question, $ \\ $ As, $f(x) = 4x + \sin x \\ f'(x) = 4 + \cos x > 0 \\ $ Hence, $f(x)$ is increasing function $ \\ $ $\therefore f(x)$ is One – to – one $ \\ $ Now, $f(x) = 4x$ can take values in $\left( { - \infty ,\infty } \right)$ $ \\ $ $\therefore f(x) = 4x + \sin x$ is ONTO

 The domain of $f(x)\, = \sqrt {2 - 2x - {x^2}} $ is

 Expression in square root is always non-negative.

 Here, $2 - 2x - {x^2} \geqslant 0 \\ {x^2} + 2x - 2 \leqslant 0 \\ \left( {x + 1 - \sqrt 3 } \right)\left( {x + 1 + \sqrt 3 } \right) \leqslant 0 \\ \therefore \,\,x \in \left[ { - 1 - \sqrt 3 , - 1 + \sqrt 3 } \right]$

 The range of ${\text{f(x)}}\,{\text{ = 3}}{{\text{x}}^{\text{3}}} - {\text{13x}} + 4\cos {\text{x}}$ is

 Range of odd degree polynomial function is $R$.

 Let ${\text{f(x)}}\,{\text{ = g(x) + h(x)}}$ $ \\ $ Where, ${\text{g(x) = 3}}{{\text{x}}^{\text{3}}} - {\text{13x}}$ and ${\text{h(x) = 4cosx}}$ $ \\ {\text{g(x)}}$ is odd degree polynomial function, so ${\text{g(x)}} \in {\text{R}} \\ $ And ${\text{h(x)}} \in \left[ { - 4,4} \right]$ $ \\ $ Hence, ${\text{f(x)}} \in {\text{R}}$

 Let $f(x)\, = 9{x^2} + 11\cos x + \log \left( {1 + 7\left| x \right|} \right)$ is defined on $\left[ {0,1} \right]$. Then $f(x)$ is

 Replace $x$ by $ - x$ and use condition for even/odd function.

 Given, $f(x)\, = 9{x^2} + 11\cos x + \log \left( {1 + 7\left| x \right|} \right)$ $ \\ $ Now, replacing $x$ by $ - x$ we will get, $ \\ f( - x)\, = 9{\left( { - x} \right)^2} + 11\cos \left( { - x} \right) + \log \left( {1 + 7\left| { - x} \right|} \right) \\ f( - x)\, = 9{x^2} + 11\cos x + \log \left( {1 + 7\left| x \right|} \right) \\ \Rightarrow f(x) \\ $ Hence, $f(x)$ is an even function.