(B) Given $f(x) = x^2 + 2kx + k < 0$ for all $x \in [1, 2]$.Since the coefficient of $x^2$ is positive,the parabola opens upward.For a quadratic $f(x)

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(B) Given $f(x) = x^2 + 2kx + k Since the coefficient of $x^2$ is positive,the parabola opens upward.For a quadratic $f(x)$ to be negative on $[1, 2]$,the maximum value of $f(x)$ on the interval $[1, 2]$ must be less than $0$.Since the parabola opens upward,the maximum value on $[1, 2]$ occurs at the endpoints $x=1$ or $x=2$.$f(1) = 1^2 + 2k(1) + k = 1 + 3k $f(2) = 2^2 + 2k(2) + k = 4 + 5k For $f(x) Thus,$k Therefore,the interval for $k$ is $(-\infty, -4/5)$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Quadratic expressions and Position of roots

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If $f:[1, 2] \rightarrow R$ defined by $f(x) = x^2 + 2kx + k$ is always negative for all $x \in [1, 2]$,then the interval in which $k$ lies is:

AP EAMCET 2017 All AP EAMCET

A
$(-\infty, -1)$
B
$(-\infty, -4/5)$
C
$(-4/5, \infty)$
D
$(1, \infty)$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Quadratic expressions and Position of roots

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If $f:[1, 2] \rightarrow R$ defined by $f(x) = x^2 + 2kx + k$ is always negative for all $x \in [1, 2]$,then the interval in which $k$ lies is:

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