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The graph of $f(x) = ext{log}_k x$ is drawn below - NSC Mathematics - Question 6 - 2021 - Paper 1

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The graph of $f(x) = ext{log}_k x$ is drawn below. B(k; 2) is a point on $f$. 6.1 Calculate the value of $k$. 6.2 Determine the values of $x$ for which $-1 ext{ ≤... show full transcript

Worked Solution & Example Answer:The graph of $f(x) = ext{log}_k x$ is drawn below - NSC Mathematics - Question 6 - 2021 - Paper 1

Step 1

6.1 Calculate the value of $k$.

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Answer

Given that point B(k; 2) is on the graph of f(x)=extlogkxf(x) = ext{log}_k x, we substitute the coordinates into the function:

2=extlogkk2 = ext{log}_k k

Using the property of logarithms, extlogkk=1 ext{log}_k k = 1. Therefore, we have:

2=12 = 1

This doesn't hold. Instead, we set up the equation as:

2=extlogkk22 = ext{log}_k k^2

This implies:

k2=k2k^2 = k^2

Squaring both sides and equating gives:

4=k.4 = k. Therefore, k=16k=16.

Step 2

6.2 Determine the values of $x$ for which $-1 ext{ ≤ } f(x) ext{ ≤ } 2$.

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Answer

We start by analyzing the inequalities one at a time.

  1. For the left inequality: 1extextlogkx-1 ext{ ≤ } ext{log}_k x This is equivalent to: x ext{ ≥ } k^{-1} = rac{1}{4}.

  2. For the right inequality: extlogkxext2 ext{log}_k x ext{ ≤ } 2 This can be rewritten as: xextk2=16.x ext{ ≤ } k^2 = 16.

Combining these gives: rac{1}{4} ext{ ≤ } x ext{ ≤ } 16.

Step 3

6.3 Write down the equation of $f^{-1}$, the inverse of $f$, in the form $y = ...$

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Answer

To find the inverse function, we swap xx and yy in the equation:

y=extlogkx.y = ext{log}_k x.

Swapping gives:

x=extlogky.x = ext{log}_k y.

Now, solving for yy, we rewrite the equation in exponential form:

y=kx.y = k^x.

Step 4

6.4 For which values of $x$ will $x f'^{-1}(x) < 0$?

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Answer

To determine where the product is negative, we need to examine the behavior of f1(x)f'^{-1}(x):

  1. The function f(x)f'(x) indicates the rate of change of f(x)f(x): Since f(x)=extlogkxf(x) = ext{log}_k x, we have: f'(x) = rac{1}{x ext{ln} k}.

The inverse function's derivative: f'^{-1}(x) = rac{1}{k^x ext{ln} k}.

  1. Now for the condition: x rac{1}{f'^{-1}(x)} < 0. The product will be negative if:
    • Either x<0x < 0 (where f1f'^{-1} is positive)
    • Or when both terms are negative or positive, leading to: xext(ext;0).x ext{ ∈ } (- ext{∞}; 0).

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