Functions : Basic Problems and Solutions

1. Let the function$\displaystyle f:R\to R$ be given by $\displaystyle f (x) = ax^2 + bx$, If $\displaystyle f (-1) = 7$ and $\displaystyle f (2) = -2$, find the values of $\displaystyle x$ for which $\displaystyle f(x) = x$.

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$\displaystyle \begin{array}{l}\ \ \ f:R\to R\\\\\ \ \ f(x)=a{{x}^{2}}+bx\\\\\ \ \ f(-1)=7\\\\\therefore a{{(-1)}^{2}}+b(-1)=7\\\\\ \ \ a-b=7\ \ \ \ ------(1)\\\\f(2)=-2\\\\\therefore a{{(2)}^{2}}+b(2)=-2\\\\\ \ \ 4a+2b=-2\\\\\ \ \ 2a+b=-1------(2)\\\\(1)+(2)\Rightarrow 3a=6\Rightarrow a=2\\\\\therefore 2-b=7\ \Rightarrow b=-5\\\\\therefore f(x)=2{{x}^{2}}-5x\\\\\ \ \ f(x)=x\ \ \ \text{(given)}\\\\\therefore 2{{x}^{2}}-5x=x\\\\\ \ \ 2{{x}^{2}}-4x=0\\\\\ \ \ x(x-2)=0\\\\\therefore x=0\ \text{or}\ x=2\text{ }\end{array}$

2. A function $\displaystyle f$ from $\displaystyle A$ to $\displaystyle A$, where $\displaystyle A$ is the set of positive integers, is given by $\displaystyle f(x)=$ the sum of all possible divisors of $\displaystyle x$. Find the value of $\displaystyle k$, if $\displaystyle f(15) = 3k + 6$.

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$\displaystyle \begin{array}{l}\ \ \ A=\{x|x\ \text{is apositive integer }\!\!\}\!\!\text{ }\\\\\ \ \ \ f:A\to A\\\\\ \ \ f(x)=\text{the sum of all possible divisors of }x.\\\\\therefore \ f(15)=\text{the sum of all possible divisors of }15\\\\\ \ \ \ \ \ \ \ \ \ \ \ =1+3+5+15\\\\\ \ \ \ \ \ \ \ \ \ \ \ =24\\\\\ \ \ f(15)=3k+6\ \text{(given)}\\\\\therefore 3k+6=24\\\\\therefore k=6\end{array}$

3. Function $\displaystyle f:R\to R$ be given by $\displaystyle f(x) = x^2 - 6$. Find the possible values of $\displaystyle x$ for which $\displaystyle f(x)$ is unchanged by the mapping.

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$\displaystyle \begin{array}{l}f:R\to R\\\\f(x)={{x}^{2}}-6\\\\\text{When }f(x)\text{ is unchanged by mapping,}\\\\f(x)=x\\\\{{x}^{2}}-6=x\\\\{{x}^{2}}-x-6=0\\\\\therefore (x+2)(x-3)\\\\\therefore x=-2\ \text{or}\ x=3\end{array}$

4. Function $\displaystyle f:R\to R$ be given by $\displaystyle f(x) = x^2 - 3x + 2$. Show that $\displaystyle f(x + 2) = x^2 + x$.

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$\displaystyle \begin{array}{l}f:R\to R\\\\f(x)={{x}^{2}}-3x+2\\\\f(x+2)={{(x+2)}^{2}}-3(x+2)+2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}+4x+4-3x-6+2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}+x\end{array}$

5. Given that $\displaystyle f:x\mapsto \frac{2}{{ax+b}},x\ne -\frac{b}{a}$ such that $\displaystyle f(0) = -2$ and $\displaystyle f(2) = 2$, find the value of $\displaystyle a$ and of $\displaystyle b$. Show that $\displaystyle f(p) + f(– p) = 2 f(p^2)$.

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$\displaystyle f(x)=\frac{2}{{ax+b}},x\ne -\frac{b}{a}$

$\displaystyle f(0)=-2$

$\displaystyle \frac{2}{{a(0)+b}}=-2$

$\displaystyle \frac{2}{b}=-2\Rightarrow b=-1$

$\displaystyle f(2)=2$

$\displaystyle \frac{2}{{a(2)-1}}=2$

$\displaystyle 2a-1=1\Rightarrow a=1$

$\displaystyle \therefore f(x)=\frac{2}{{x-1}}$

$\displaystyle \therefore f(p)+f(-p)=\frac{2}{{p-1}}+\frac{2}{{-p-1}}$

$\displaystyle \therefore f(p)+f(-p)=\frac{2}{{p-1}}-\frac{2}{{p+1}}$

$\displaystyle \therefore f(p)+f(-p)=\frac{{2(p+1)-2(p-1)}}{{{{p}^{2}}-1}}$

$\displaystyle \therefore f(p)+f(-p)=\frac{4}{{{{p}^{2}}-1}}$ $\displaystyle 2f({{p}^{2}})=2\left( {\frac{2}{{{{p}^{2}}-1}}} \right)=\frac{4}{{{{p}^{2}}-1}}$ $\displaystyle \therefore f(p)+f(-p)=2f({{p}^{2}})$

6. Given that $\displaystyle f(x)=\frac{{ax-b}}{x}$, for all real values of $\displaystyle x$ except $\displaystyle x = 0$. If $\displaystyle f(1) = -1$ and $\displaystyle f(2) = 1$, find the value of $\displaystyle a$ and of $\displaystyle b$ and hence find the image of $\displaystyle – 4$ under $\displaystyle f$.

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$\displaystyle f(x)=\frac{{ax-b}}{x},x\ne 0$

$\displaystyle f(1)=-1$

$\displaystyle \frac{{a(1)-b}}{1}=-1$

$\displaystyle a-b=-1\ \ \ \ \ -----(1)$

$\displaystyle f(2)=1$

$\displaystyle \frac{{a(2)-b}}{2}=1$

$\displaystyle 2a-b=2\ \ \ \ \ -----(2)$

$\displaystyle \text{Equation}\ (2)-\text{Equation}\ (1)\Rightarrow a=3$

$\displaystyle \therefore 3-b=-1\Rightarrow b=4$

$\displaystyle \therefore f(x)=\frac{{3x-4}}{x}$

$\displaystyle \therefore f(-4)=\frac{{3(-4)-4}}{{(-4)}}=4$

7. If $\displaystyle f(x)=\frac{{b(x-a)}}{{b-a}}+\frac{{a(x-b)}}{{a-b}}$, show that $\displaystyle f(a+b) = f(a) + f(b)$.

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$\displaystyle f(x)=\frac{{b(x-a)}}{{b-a}}+\frac{{a(x-b)}}{{a-b}},a\ne b$

$\displaystyle f(a+b)=\frac{{b(a+b-a)}}{{b-a}}+\frac{{a(a+b-b)}}{{a-b}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{b}^{2}}}}{{b-a}}+\frac{{{{a}^{2}}}}{{a-b}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{a}^{2}}}}{{a-b}}-\frac{{{{b}^{2}}}}{{a-b}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{a}^{2}}-{{b}^{2}}}}{{a-b}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{(a-b)(a+b)}}{{a-b}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =a+b$

$\displaystyle f(a)+f(b)=\frac{{b(a-a)}}{{b-a}}+\frac{{a(a-b)}}{{a-b}}+\frac{{b(b-a)}}{{b-a}}+\frac{{a(b-b)}}{{a-b}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0+a+b+0$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b$

$\displaystyle \therefore f(a+b)=f(a)+f(b)$

8. A function $\displaystyle f$ is defined by $\displaystyle f(2x + 1) = x^2 - 3$. Find $\displaystyle a\in R$ such that $\displaystyle f(5) = a^2 - 8$.

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$\displaystyle \begin{array}{l}f(2x+1)={{x}^{2}}-3\\\\\text{Let}\ 2x+1=5,\text{then }x=2\\\\\therefore f(5)={{2}^{2}}-3=1\\\\f(5)={{a}^{2}}-8\ \text{(given)}\\\\\therefore {{a}^{2}}-8=1\\\\\therefore {{a}^{2}}=9\Rightarrow a=\pm 3\end{array}$

9. The function $\displaystyle f$ is defined by $\displaystyle f(x) = 7^x$ . Prove that $\displaystyle f(x + 2) - 10 f(x + 1) + 21 f(x) = 0$.

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$\displaystyle \begin{array}{l}f(x)={{7}^{x}}\\\\\therefore f(x+2)-10f(x+1)+21f(x)\\\\={{7}^{{x+2}}}-10({{7}^{{x+1}}})+21({{7}^{x}})\\\\={{7}^{2}}\cdot {{7}^{x}}-10(7\cdot {{7}^{x}})+21({{7}^{x}})\\\\={{7}^{x}}(49-70+21)\\\\={{7}^{x}}(0)\\\\=0\end{array}$

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