The correct answer is C.

EXPLANATION

$|2 x+23|=|x+11| \Rightarrow 2 x+23=x+11$ or $2 x+23=-(x+11)$.

$\Rightarrow x=11-23$ or $3 x=-11-23$.

$\Rightarrow x=-12$ or $x=-\frac{34}{3}$.

The correct answer is C.

EXPLANATION

Given that $f(x)=4(x-3)$ and $g(x)=x^{2}-7 x+12=(x-3)(x-4)$. It follows that $f(0)=g(0)=0$. Therefore, $\frac{g(x)}{f(x)}$ is undefined when $x=3$.

The correct answer is B.

EXPLANATION

Given that $f(x)=2 x^{2}+5 x-24$ and $g(x)=3 x+2(x$ real $)$,

$f(x)+g(x)=2 x^{2}+8 x-22$.

The discriminant $b^{2}-4 a c=64-4(2)(-22)=240>0$. Hence, the roots are real and distinct.

The correct answer is B.

EXPLANATION

$$
f(x)= \begin{cases}7 & \text { if } x \leq 3 \\ m x+n & \text { if } 3<x<12 \\ 18 & \text { if } x \geq 12\end{cases}
$$

In order that the function is continuous at $x=3, f(3-)=f(3+)$, that is, $7=3 m+n$.

Similarly, for $x=7, f(7-)=f(7+)$. This gives $12 m+n=18$.

The solution of these two equations is $m=\frac{11}{9}$ and $n=\frac{10}{3}$.

The correct answer is C.

EXPLANATION

$f(x)=-2 x^{3}-9 x^{2}-12 x+1, f^{\prime}(x)=-6 x^{2}-18 x-12$.

$f^{\prime}(x)=0 \Rightarrow x^{2}+3 x+2=0 \Rightarrow(x \mp 1)(x+2)=0 \Rightarrow x=-1$ or $x=-2$.

The possible intervals are $(-\infty,-2),(-2,-1),(-1, \infty)$.

It is clear that $f^{\prime}(x)0$ for $x \in(-2,-1)$ and $f^{\prime}(x)<$ 0 for $x \in(-1, \infty)$. Hence $f(x)$ is increasing in $(-2,-1)$.

The correct answer is A.

EXPLANATION

Obviously, choices B, C and D are not true.

Let $A \cup B=A \cap B$. To prove $A=B$, let $x \in A \Rightarrow x \in A \cup B$

$\Rightarrow x \in A \cap B \Rightarrow x \in A$ and $B \Rightarrow x \in B$. Hence, $A \subset B$.

Similarly it can be proved that $B \subset A$. Hence, $A=B$.

The correct answer is A.

EXPLANATION

$$
\begin{aligned}
&(\sqrt{3}+1)^{4}-(\sqrt{3}-1)^{4} \\
&= {\left[(\sqrt{3})^{4}+\left(\begin{array}{l}
4 \\
3
\end{array}\right)(\sqrt{3})^{3}+\left(\begin{array}{l}
4 \\
2
\end{array}\right)(\sqrt{3})^{2}+\left(\begin{array}{l}
4 \\
1
\end{array}\right)(\sqrt{3})^{1}+1\right] } \\
&-\left[(\sqrt{3})^{4}-\left(\begin{array}{l}
4 \\
3
\end{array}\right)(\sqrt{3})^{3}+\left(\begin{array}{l}
4 \\
2
\end{array}\right)(\sqrt{3})^{2}-\left(\begin{array}{l}
4 \\
1
\end{array}\right)(\sqrt{3})^{1}+1\right] \\
&=2\left[\left(\begin{array}{l}
4 \\
3
\end{array}\right)(\sqrt{3})^{3}++\left(\begin{array}{l}
4 \\
1
\end{array}\right)(\sqrt{3})^{1}\right]=2\left[4(\sqrt{3})^{3}++4 \sqrt{3}\right]=32 \sqrt{3} .
\end{aligned}
$$

The correct answer is C.

EXPLANATION

$\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}=\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \times \frac{1}{7}}\right)=\tan ^{-1}\left(\frac{21+4}{28-3}\right)$. $=\tan ^{-1}(1)=\frac{\pi}{4}$.

The correct answer is D.

EXPLANATION

$\sum_{j=1}^{n}(3 j-1)^{2}=\sum_{j=1}^{n}\left(9 j^{2}-6 j+1\right)=9 \sum_{j=1}^{n} j^{2}-6 \sum_{j=1}^{n} j+\sum_{j=1}^{n} 1=$ $9\left(\frac{n(n+1)(2 n+1)}{6}\right)-6\left(\frac{n(n+1)}{2}\right)+n=\frac{n}{2}\left[6 n^{2}+3 n-1\right]$.

The correct answer is A.

EXPLANATION

$\frac{(1+2 i)(2-i)}{(3-2 i)(2+3 i)}=\frac{2-i+4 i-2 i^{2}}{6+9 i-4 i-6 i^{2}}=\frac{2+3 i+2}{6+5 i+6}=\frac{3 i+4}{5 i+12}=\frac{3 i+4}{5 i+12} \times \frac{5 i-12}{5 i-12}$

$$
=\frac{15 i^{2}-36 i+20 i-48}{(5 i)^{2}-(12)^{2}}=\frac{-15-16 i-48}{-25-144}=\frac{63+16 i}{169} .
$$

Hence the conjugate of $\frac{(1+2 i)(2-i)}{(3-2 i)(2+3 i)}$ is $\frac{63}{169}-\frac{16 i}{169}$.

The correct answer is B.

EXPLANATION

The area bounded by the curve $y=\frac{x^{2}}{2}$, the lines $y=0, x=1$ and $x=3$ is $\int_{1}^{3} \frac{x^{2}}{2} d x=$ $4 \frac{1}{3}$.

The correct answer is A.

EXPLANATION

Given that $f(x)=x^{2}-2 \cdot f(1)=-1$ (-ve) and $f(2)=2(+\mathrm{ve})$. Hence, there exists a positive root between 1 and 2. $f^{\prime}(x)=2 x$.

The $n^{\text {th }}$ iteration: $x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$.

We start with the initial value $x_{0}=1 . x_{1}=1-\frac{-1}{2}=1.5$;

$x_{2}=1.5-\frac{\frac{9}{4}-2}{3}=\frac{17}{12}=1.41667 ; x_{3}=\frac{17}{12}-\frac{\left(\frac{17}{12}\right)^{2}-2}{\frac{17}{6}}=\frac{577}{408}=1.41426$.