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Let $f(x) = 3x^2 + x$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2001 - Paper 1

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Let-$f(x)-=-3x^2-+-x$-HSC-SSCE Mathematics Extension 1-Question 2-2001-Paper 1.png

Let $f(x) = 3x^2 + x$. Use the definition $$f'(a) = ext{lim}_{h \to 0} \frac{f(a + h) - f(a)}{h}$$ to find the derivative of $f(x)$ at the point $x = a$. (i) Fin... show full transcript

Worked Solution & Example Answer:Let $f(x) = 3x^2 + x$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2001 - Paper 1

Step 1

Let $f(x) = 3x^2 + x$. Use the definition to find the derivative of $f(x)$ at the point $x = a$.

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Answer

To find the derivative using the limit definition, we substitute into the formula:

f(a)=limh0f(a+h)f(a)hf'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}

Calculating f(a+h)f(a + h):

f(a+h)=3(a+h)2+(a+h)=3(a2+2ah+h2)+a+h=3a2+6ah+3h2+a+hf(a + h) = 3(a + h)^2 + (a + h) = 3(a^2 + 2ah + h^2) + a + h = 3a^2 + 6ah + 3h^2 + a + h

Now compute f(a+h)f(a)f(a + h) - f(a):

f(a+h)f(a)=(3a2+6ah+3h2+a+h)(3a2+a)=6ah+3h2+hf(a + h) - f(a) = (3a^2 + 6ah + 3h^2 + a + h) - (3a^2 + a) = 6ah + 3h^2 + h

Thus, substituting back into the limit gives us:

f(a)=limh06ah+3h2+hh=limh0(6a+3h+1)=6a+1f'(a) = \lim_{h \to 0} \frac{6ah + 3h^2 + h}{h} = \lim_{h \to 0} (6a + 3h + 1) = 6a + 1

Step 2

Find $\int \frac{e^x}{1 + e^x} dx$

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Answer

To solve the integral, we can use substitution. Let:

u=1+exdu=exdxu = 1 + e^x \Rightarrow du = e^x dx

The limits change accordingly: When x=0x = 0, u=2u = 2. Therefore:

ex1+exdx=duu=lnu+C=ln1+ex+C\int \frac{e^x}{1 + e^x} dx = \int \frac{du}{u} = \ln |u| + C = \ln |1 + e^x| + C

Step 3

Find $\int_0^2 3 \cos x \, dx$

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Answer

Calculating the definite integral:

3cosxdx=3sinx+C\int 3 \cos x \, dx = 3 \sin x + C

Evaluating from 0 to 2:

[3sin(2)3sin(0)]=3sin(2)0=3sin(2)[3 \sin(2) - 3 \sin(0)] = 3\sin(2) - 0 = 3\sin(2)

Step 4

How many arrangements of the letters in the word ALGBRAIC are possible?

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Answer

The word ALGBRAIC consists of 9 letters with the letters A and I repeating once. The number of distinct arrangements is given by:

9!2!=3628802=181440\frac{9!}{2!} = \frac{362880}{2} = 181440

Step 5

How many arrangements of the letters in the word ALGBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th and 8th positions?

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Answer

The vowels A, A, I occupy 4 slots: 2nd, 3rd, 5th, and 8th. The arrangements of the vowels:

4!2!=12\frac{4!}{2!} = 12

The remaining 5 letters (L, G, B, R, C) can occupy the remaining positions, giving:

5!=1205! = 120

Thus, the total arrangements are:

12×120=144012 \times 120 = 1440

Step 6

Find the term independent of $x$ in the binomial expansion of $(x^2 - 1)^9$.

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Answer

The general term in the binomial expansion of (x21)9(x^2 - 1)^9 is given by:

Tk=(9k)(x2)9k(1)kT_k = \binom{9}{k} (x^2)^{9-k} (-1)^k

For the term independent of xx, we require the power of xx to be zero, i.e., 2(9k)=02(9-k) = 0. Thus:

9k=0k=99 - k = 0 \Rightarrow k = 9

The term is:

T9=(99)(x2)0(1)9=1T_9 = \binom{9}{9} (x^2)^0 (-1)^9 = -1

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