የሒሳብ መግቢያ ፈተና፣ 2005 ዓ.ም. (2013)

The rule \(a_{n+1} = a_n + 4\) means every term is 4 more than the one before it, so this is an arithmetic sequence with \(a_1 = 2\) and \(d = 4\). From the 1st term to the 35th there are 34 steps, so \(a_{35} = 2 + 34(4) = 138\). The sum of an arithmetic sequence is \(S_n = \frac{n}{2}(a_1 + a_n)\), so \(S_{35} = \frac{35}{2}(2 + 138) = 35 \times 70 = 2450\). The common slip is to count 35 steps and write \(a_{35} = 142\), which gives a total of 2520. A total that is not on the list is your signal to recount the steps.

That limit is the definition of the derivative of \(f\) at \(x = 2\), so find \(f'\) and put 2 into it. For \(f(x) = x^2 + 2\ln x\), the derivative is \(f'(x) = 2x + \frac{2}{x}\). At \(x = 2\) this gives \(4 + 1 = 5\). Option B, 4, is what you get if you differentiate only the \(x^2\) part and forget that \(2\ln x\) contributes \(\frac{2}{x}\).

Differentiate twice. The first derivative is \(f'(x) = 2e^{2x} + 1 + 3\sin x\): the chain rule pulls a factor of 2 out of \(e^{2x}\), and the derivative of \(-3\cos x\) is \(+3\sin x\). Differentiating again, the constant 1 disappears and \(3\sin x\) becomes \(3\cos x\), so \(f''(x) = 4e^{2x} + 3\cos x\). Option C keeps a minus sign in front of the cosine. That minus sign was already used up at the first step, when \(-3\cos x\) turned into \(+3\sin x\).

A function increases on the interval where its derivative is not negative. Here \(f'(x) = 4x^3 + 4 = 4(x^3 + 1)\), and \(x^3 + 1 \geq 0\) exactly when \(x^3 \geq -1\), which means \(x \geq -1\). So \(f\) increases on \([-1, \infty)\). Option A is the mirror image of the answer: on \((-\infty, -1]\) the derivative is negative, so the function is falling there.

Handle the two pieces separately. The modulus is \(|3+4i| = \sqrt{9+16} = 5\). For the fraction, multiply top and bottom by the conjugate \(3-4i\): \(\frac{25i(3-4i)}{(3+4i)(3-4i)} = \frac{75i - 100i^2}{25} = \frac{100 + 75i}{25} = 4 + 3i\). So the expression is \(5 - (4+3i) = 1 - 3i\). Option C, \(1 + 3i\), comes from subtracting the real part but leaving the imaginary part with the sign it had before the subtraction.

The number is already in the form \(\cos\theta + i\sin\theta\) with \(\theta = \frac{\pi}{10}\), so use de Moivre's rule: raising to the power 5 multiplies the angle by 5. That gives \(Z^5 = \cos\frac{5\pi}{10} + i\sin\frac{5\pi}{10} = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\). Since \(\cos\frac{\pi}{2} = 0\) and \(\sin\frac{\pi}{2} = 1\), the value is \(i\). Option B comes from raising the two numbers \(\cos\frac{\pi}{10}\) and \(\sin\frac{\pi}{10}\) to the power 5 one by one, instead of multiplying the angle by 5.