The correct answer is C.

EXPLANATION

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

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

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

The correct answer is C.

EXPLANATION

Given that $f(x)=4(x-3)$ and $g(x)=x^2-7x+12=(x-3)(x-4)$. 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 $f(x)=2x^2+5x-24$ and $g(x)=3x+2(x$ real $)$,

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

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

The correct answer is B.

EXPLANATION

$$f(x)=\{\begin{array}{ll}7& \text{ if }x\le 3\\ mx+n& \text{ if }3<x<12\\ 18& \text{ if }x\ge 12\end{array}$$

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

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

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

The correct answer is C.

EXPLANATION

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

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

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

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

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

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

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

The correct answer is A.

EXPLANATION

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

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

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

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 ${\int }_1^3\frac{x^2}{2}dx=$ $4\frac{1}{3}$.

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. $f^{\prime }(x)=2x$.

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 $x_0=1.x_1=1-\frac{-1}{2}=1.5$;

$x_2=1.5-\frac{\frac{9}{4}-2}{3}=\frac{17}{12}=1.41667;x_3=\frac{17}{12}-\frac{{\left(\frac{17}{12}\right)}^2-2}{\frac{17}{6}}=\frac{577}{408}=1.41426$.