The number of positive integral values of K f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $f(x) = |x + |2x - 1|| - |x - |2x - 1||$. We know that $|a+b| - |a-b| = 2 	ext{sgn}(b) \min(|a|, |b|)$ is not quite right,but specifically $|

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $f(x) = |x + |2x - 1|| - |x - |2x - 1||$. We know that $|a+b| - |a-b| = 2 	ext{sgn}(b) \min(|a|, |b|)$ is not quite right,but specifically $|x+y| - |x-y| = 2x$ if $x \ge |y|$ and $2y$ if $|y| > x$. Here,let $y = |2x-1|$. The expression is $f(x) = |x+y| - |x-y|$. If $x \ge |2x-1|$,then $f(x) = 2|2x-1|$. If $x Solving $x = |2x-1|$: $x = 2x-1 \implies x=1$ or $x = -(2x-1) \implies 3x=1 \implies x=1/3$. For $x \in [1/3, 1]$,$x \ge |2x-1|$ is false,so $f(x) = 2x$. For $x For $x > 1$,$f(x) = 2|2x-1| = 4x-2$. Analyzing the function $f(x)$: $1$) If $x $2$) If $1/3 \le x \le 1$,$f(x) = 2x$. $3$) If $x > 1$,$f(x) = 4x-2$. Wait,let's re-evaluate: If $x If $1/3 \le x \le 1$,$|2x-1| = 2x-1$. Then $f(x) = |x+2x-1| - |x-(2x-1)| = |3x-1| - |1-x| = (3x-1) - (1-x) = 4x-2$. If $x > 1$,$|2x-1| = 2x-1$. Then $f(x) = |x+2x-1| - |x-(2x-1)| = |3x-1| - |1-x| = (3x-1) - (x-1) = 2x$. The function $f(x)$ is $2x$ for $x  1$. At $x=1/3$,$f(x) = 2/3$. At $x=1$,$f(x) = 2$. The graph increases from $-\infty$ to $2/3$,then increases faster to $2$,then increases to $\infty$. Since the function is strictly increasing,it can have at most one solution for any $K$. Thus,there are no values of $K$ for which it has three solutions. The answer is $0$.

The number of positive integral values of $K$ for which the equation $K = |x + |2x - 1|| - |x - |2x - 1||$ has exactly three real solutions is

Similar Questions

Which of the following functions is an even function?

Which of the following functions are odd?
$I. f(x)=x\left(\frac{e^x-1}{e^x+1}\right)$
$II. f(x)=k^x+k^{-x}+\cos x$
$III. f(x)=\log \left(x+\sqrt{x^2+1}\right)$

Statement $-1$: Any function $f(x)$ is an even function if $f(-x) = f(x)$ for all $x$ in its domain.
Statement $-2$: The function $f(x) = \frac{1}{\sqrt{1 - x^2}} + \left[ \frac{x^2 + x + 1}{4} \right]$,where $[.]$ denotes the greatest integer function,is an even function.

If $f(x) = \cos([\pi^2]x) + \cos([- \pi^2]x)$,then which of the following is true?

If $f(x) = \log \left( \frac{1 + x}{1 - x} \right)$,then $f(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module