Let $f (x) = ⎧ ⎨ ⎩ \begin{matrix} ∣ ∣ x^{2} - 3 x ∣ ∣ + a, 0 \leq x < \frac{3}{2} - 2 x + 3 x \geq \frac{3}{2} \end{matrix}$ If f(x) has a local maximum at x =.

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If f(x)=2x3−3x2−36x+6 has a local maximum and minimum at x=a and x=b respectively, then ordered pair (a,b) is-

The correct option is C (−2,3)f′(x)=6x2−6x−36=6(x2−x−6)=0Hencex2−x−6=0(x−3)(x+2)=0x=3 and x=−2f"(x)=12x−6Now f"(3)=30 Hence minimum.f"(−2)=−30 Hence point of maximum.a=−2,b=3

If $f (x) = 2 x^{3} - 3 x^{2} - 36 x + 6$ has a local maximum and minimum at $x = a$ and $x = b$ respectively, then ordered pair $(a, b)$ is-

The correct option is C $(- 2, 3)$
$f^{'} (x)$
$= 6 x^{2} - 6 x - 36$
$= 6 (x^{2} - x - 6)$
$= 0$
Hence
$x^{2} - x - 6 = 0$
$(x - 3) (x + 2) = 0$
$x = 3$ and $x = - 2$
$f " (x)$
$= 12 x - 6$
Now
$f " (3)$
$= 30$ Hence minimum.
$f " (- 2)$
$= - 30$ Hence point of maximum.
$a = - 2, b = 3$