Question

Which of the following statement(s) is/are CORRECT?

• A. If a + b <1, then $\log_b a > 0$ wherever defined
• B. Equation $3[x]^2 - x - 4 = 0$ has a unique solution ([.] denotes greater integer function)
• C. If $x, y \in R$, then there exists exactly one ordered pair (x,y) which satisfies the equation $2^x + 3^y = 0$
• D. equation $|x-1| + 2|x+1| = |x+3|$ has 3 integral solutions

Answer 1

Answer: (A), (B), (D)

Statements 1, 2, and 4 are correct.

Answer 2

Solution:

1. Statement 1:

For the logarithm $\log_b a$ to be defined we require $a>0$ and $b>0, b\neq1$. Notice that if $a+b<1$ and $a>0$, then $b<1$ (because if $b\ge1$, then $a+b\ge1+a>1$). When $0<b<1$, the logarithm function is decreasing so:

$\log_b a>0 \iff a<1$.

Since $a+b<1$ implies $a<1-b<1$, the inequality holds wherever defined.

⇒ Statement 1 is correct.

2. Statement 2:

Let $[x]=n$ (the greatest integer $\le x$). Then the equation

$3n^2 - x - 4 = 0$

becomes

$x = 3n^2 - 4$.

For consistency with the floor function we must have:

$n \le x < n+1 \quad \Longrightarrow \quad n \le 3n^2-4 < n+1$.

Check for integer values $n$:

• For $n=-1$:

$x=3(-1)^2-4=3-4=-1$.

Verify: $[-1]= -1$ and $-1 \le -1 < 0$. (Valid)

• For other $n$ the inequality fails.

Hence, there is a unique solution $x=-1$.

⇒ Statement 2 is correct.

3. Statement 3:

The equation

$2^x+3^y=0$

involves two exponential terms. Since $2^x>0$ and $3^y>0$ for all real $x,y$, their sum is always positive. Thus, no solution exists.

⇒ Statement 3 is false.

4. Statement 4:

Consider the equation

$|x-1|+2|x+1|=|x+3|$.

Identify the critical points: $x=-3,\,-1,\,1$. Solve piecewise:

• For $x\ge1$:

$|x-1|=x-1$ and $|x+1|=x+1$ while $|x+3|=x+3$.

Equation: $(x-1)+2(x+1)=x+3 \implies 3x+1=x+3 \implies 2x=2 \implies x=1.$

• For $-1\le x<1$:

$|x-1|=1-x$, $|x+1|=x+1$, and $|x+3|=x+3$ (since $x+3>0$).

Equation: $(1-x)+2(x+1)=x+3,$

which simplifies to $x+3=x+3$ (an identity).

So every $x$ in this interval is a solution.

• For $x<-1$:

(A quick check shows no additional integer solution arises.)

Now, we want the integral solutions.

The solutions in $-1\le x<1$ include $x=-1$ and $x=0$, and from the first case, $x=1$ is also a solution. No other integers satisfy the equation.

⇒ Statement 4 is correct.