Let f(x)=5-|x-2| and g(x)=|x+1|, x in R. If f | vedclass

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

Question

(A) Given $f(x) = 5 - |x - 2|$. Since $|x - 2| \geq 0$,the maximum value of $f(x)$ is $5$,which occurs when $|x - 2| = 0$,i.e.,$x = 2$. Thus,$\alpha =

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Given $f(x) = 5 - |x - 2|$. Since $|x - 2| \geq 0$,the maximum value of $f(x)$ is $5$,which occurs when $|x - 2| = 0$,i.e.,$x = 2$. Thus,$\alpha = 2$.Given $g(x) = |x + 1|$. Since $|x + 1| \geq 0$,the minimum value of $g(x)$ is $0$,which occurs when $|x + 1| = 0$,i.e.,$x = -1$. Thus,$\beta = -1$.We need to evaluate $\lim _{x \rightarrow -\alpha \beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$.Since $-\alpha \beta = -(2)(-1) = 2$,the limit is $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 2)(x - 3)}{(x - 2)(x - 4)}$.Canceling the common factor $(x - 2)$,we get $\lim _{x \rightarrow 2} \frac{(x - 1)(x - 3)}{x - 4}$.Substituting $x = 2$,we get $\frac{(2 - 1)(2 - 3)}{2 - 4} = \frac{(1)(-1)}{-2} = \frac{-1}{-2} = \frac{1}{2}$.

Let $f(x)=5-|x-2|$ and $g(x)=|x+1|, x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8}$ is equal to

Similar Questions

If $a, b, c, d$ are positive,then $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{{a + bx}}} \right)^{c + dx}} = $

The value of $\lim_{x \to 0} \left( \frac{x^2 \sin^2 x}{x^2 - \sin^2 x} \right)$ is:

$\mathop {Lim}\limits_{n \to \infty } \frac{{{1^2}n + {2^2}(n - 1) + {3^2}(n - 2) + \dots + {n^2} \cdot 1}}{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}$ is equal to :

If $\lim _{x \rightarrow \infty}\left(\frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}\right)=\frac{a}{b}$,then the value of $a+b$ equals:

For each $t \in R$,let $[t]$ be the greatest integer less than or equal to $t$. Then $\lim_{x \to 0^+} x \left( [\frac{1}{x}] + [\frac{2}{x}] + \dots + [\frac{15}{x}] \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module