Let a (a

Question

(B) For $f(x)$ to be a decreasing function,the coefficient of $x$ must be less than or equal to $0$. Let $k = [t]+1$. Then $\frac{|k|+a}{|k|+1-a} \leq

Vedclass · Accepted Answer

$[[a], [-a])$

Vedclass · Answer

(B) For $f(x)$ to be a decreasing function,the coefficient of $x$ must be less than or equal to $0$. Let $k = [t]+1$. Then $\frac{|k|+a}{|k|+1-a} \leq 0$. Since $a  0$. For the fraction to be $\leq 0$,the numerator must be $\leq 0$ (since the denominator is always positive). $|k| + a \leq 0 \implies |k| \leq -a$. Since $-a > 0$,this implies $a \leq k \leq -a$. Substituting $k = [t]+1$,we get $a \leq [t]+1 \leq -a$. Subtracting $1$ from all sides,$a-1 \leq [t] \leq -a-1$. Since $a 
otin I$,the range for $t$ is $[[a], [-a])$. Thus,the correct option is $B$.

Let $a$ $(a < 0, a \notin I)$ be a constant and $t$ be a parameter. Then the set of values of $t$ for which the function $f(x) = \left( \frac{|[t]+1|+a}{|[t]+1|+1-a} \right)x$ is a decreasing function of $x$ (where $[.]$ denotes the greatest integer function) is:

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