The least value of k for which the function f | vedclass

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

Question

(D) Let $f(x) = {x^2} + kx + 1$. For the function to be increasing on the interval $[1, 2]$,its derivative must satisfy $f'(x) \geq 0$ for all $x \in

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(D) Let $f(x) = {x^2} + kx + 1$. For the function to be increasing on the interval $[1, 2]$,its derivative must satisfy $f'(x) \geq 0$ for all $x \in [1, 2]$. Calculating the derivative,we get $f'(x) = 2x + k$. Setting the condition $f'(x) \geq 0$,we have $2x + k \geq 0$,which implies $k \geq -2x$. Since this must hold for all $x$ in the interval $[1, 2]$,$k$ must be greater than or equal to the maximum value of $-2x$ on this interval. For $x \in [1, 2]$,the function $g(x) = -2x$ is a decreasing function. Therefore,the maximum value of $g(x)$ occurs at the lower bound $x = 1$. $g(1) = -2(1) = -2$. Thus,$k \geq -2$. The least value of $k$ is $-2$.

The least value of $k$ for which the function $f(x) = {x^2} + kx + 1$ is an increasing function in the interval $1 \leq x \leq 2$ is

Similar Questions

The function $f(x) = x^2$ is increasing in the interval

Let $f(x)=e^x-x$ and $g(x)=x^2-x, \forall x \in R$. Then the set of all $x \in R$,where the function $h(x)=(fog)(x)$ is increasing is

The function $f(x) = \log(1+x) - \frac{2x}{2+x}$ is increasing on

The function $f(x) = 2x^3 - 9x^2 + 12x + 29$ is monotonically decreasing when:

The length of the longest interval,in which the function $f(x) = 3\sin x - 4\sin^3 x$ is increasing,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module