Solve the Following Question.(2 Marks)

Question 12 Marks

If $\mathrm{f}(\mathrm{x})=\frac{x+3}{4 x-5}, \mathrm{~g}(\mathrm{x})=\frac{3+5 x}{4 x-1}$, then verify that $(\mathrm{fog})(\mathrm{x})=\mathrm{x}$

Answer

$\begin{aligned} & f(x)=\frac{x+3}{4 x-5}, g(x)=\frac{3+5 x}{4 x-1} \\ & \begin{aligned}(f \circ g)(x) & =f(g(x)) \\ & =f\left(\frac{3+5 x}{4 x-1}\right) \\ & =\frac{\frac{3+5 x}{4 x-1}+3}{4\left(\frac{3+5 x}{4 x-1}\right)-5} \\ & =\frac{3+5 x+12 x-3}{12+20 x-20 x+5}=\frac{17 x}{17}=x\end{aligned}\end{aligned}$

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Question 22 Marks

If $\mathrm{f}(\mathrm{x})=\frac{2 x-1}{5 x-2}, x \neq \frac{2}{5}$, verify whether (fof) $(\mathrm{x})=\mathrm{x}$

Answer

$\begin{aligned} & (f \circ f)(x)=f(f(x)) \\ & =f\left(\frac{2 x-1}{5 x-2}\right) \\ & =\frac{2\left(\frac{2 x-1}{5 x-2}\right)-1}{5\left(\frac{2 x-1}{5 x-2}\right)-2} \\ & =\frac{4 x-2-5 x+2}{10 x-5-10 x+4}=\frac{-x}{-1}=x\end{aligned}$

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Question 32 Marks

If $f(x)=a x^2+b x+2$ and $f(1)=3, f(4)=42$, find $a$ and $b$.

Answer

$
\begin{aligned}
& f(x)=a x^2+b x+2 \\
& f(1)=3 \\
& \therefore a(1)^2+b(1)+2=3 \\
& \therefore a+b=1 \ldots \ldots .(i) \\
& f(4)=42 \\
& \therefore a(4)^2+b(4)+2=42 \\
& \therefore 16 a+4 b=40
\end{aligned}
$
Dividing by 4 , we get
$
4 a+b=10
$
Solving (i) and (ii), we get $a=3, b=-2$

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Question 42 Marks

If f(x) = 3x + a and f(1) = 7, find a and f(4).

Answer

f(x) = 3x + a,
f(1) = 7
∴ 3(1) + a = 7
∴ a = 7 – 3 = 4
∴ f(x) = 3x + 4
∴ f(4) = 3(4) + 4
= 12 + 4
= 16

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Question 52 Marks

If $f(x)=3 x^2-5 x+7$, find $f(x-1)$

Answer

$\begin{aligned} & f(x)=3 x^2-5 x+7 \\ & \therefore f(x-1)=3(x-1)^2-5(x-1)+7 \\ & =3\left(x^2-2 x+1\right)-5(x-1)+7 \\ & =3 x^2-6 x+3-5 x+5+7 \\ & =3 x^2-11 x+15\end{aligned}$

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Question 62 Marks

A function f is defined as follows:
f(x) = 4x + 5, for -4 ≤ x < 0. Find the values of f(-1), f(-2), f(0), if they exist.

Answer

f(x) = 4x + 5, -4 ≤ x < 0
f(-1) = 4(-1) + 5 = -4 + 5 = 1
f(-2) = 4(-2) + 5 = -8 + 5 = -3
x = 0 ∉ domain of f
∴ f(0) does not exist.

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Question 72 Marks

Find $x$, if $f(x)=g(x)$ where $f(x)=x^4+2 x^2, g(x)=11 x^2$

Answer

$\begin{aligned} & f(x)=x^4+2 x^2, g(x)=11 x^2 \\ & f(x)=g(x) \\ & \therefore x^4+2 x^2=11 x^2 \\ & \therefore x^4-9 x^2=0 \\ & \therefore x^2\left(x^2-9\right)=0 \\ & \therefore x^2=0 \text { or } x^2-9=0 \\ & \therefore x=0 \text { or } x^2=9 \\ & \therefore x=0 \text { or } x= \pm 3\end{aligned}$

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Maths (commerce) Solve the Following Question.(2 Marks)

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