Let f ( x ) = { ∣ x 2 − 3 x ∣ + a , − 2 x + 3 , 0 ≤ x < 2 3 x ≥ 2 3 . If f ( x ) has a local maxima at x = 2 3 , and the greatest value of 'a' is k , then ∣4 k ∣ is.

f ( 2 3 ) = 0 ⇒ lim x → 2 3 x 2 − 3 x + a ≤ 0 a ≤ − 4 9 Hence, ∣4 k ∣ = 9

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaMedium2 minQB

MathematicsMediuminteger

Let $f (x) = {∣ x^{2} - 3 x ∣ + a, - 2 x + 3, 0 \leq x < \frac{3}{2} x \geq \frac{3}{2}$ . If $f (x)$ has a local maxima at $x = \frac{3}{2}$ , and the greatest value of 'a' is $k$ , then $∣4 k ∣$ is.

Answer:

Solution:

$f (\frac{3}{2}) = 0 \Rightarrow lim_{x \to \frac{3}{2}} x^{2} - 3 x + a \leq 0$ $a \leq - \frac{9}{4}$ Hence, $∣4 k ∣ = 9$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima

⏱ 2mℹ️ Source: QB

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