Let f(x)=2 x3-3(2+p) x2+12 p x+ ln (16-p2). If f(x) has exactly one local maxima and one local minima, then the number of integral values of p is (are)

Question

Let f(x)=2 x3-3(2+p) x2+12 p x+ ln (16-p2). If f(x) has exactly one local maxima and one local minima, then the number of integral values of p is (are) (A) 4 (B) 5 (C) 6 (D) 7

Tardigrade Admin · Accepted Answer

f(x)=2 x3-3(2+p) x2+12 p x+ ln (16-p2) f prime(x)=6 x2-6(2+p) x+12 p=6(x2-(2+p) x+2 p)=0 f prime(x)=0 has x=2, x=p roots. But f(x) has exactly one local maxima and one local minima so p ≠ 2 Also 16- p 2>0 ⇒ p 2-16<0 ⇒ p ∈(-4,4) Possible integral values of p are -3,-2,-1,0,1,3 ⇒ 6 integral values

Q. Let f(x)=2x3−3(2+p)x2+12px+ln(16−p2). If f(x) has exactly one local maxima and one local minima, then the number of integral values of p is (are)

4

5

6

7

f(x)=2x3−3(2+p)x2+12px+ln(16−p2) f′(x)=6x2−6(2+p)x+12p=6(x2−(2+p)x+2p)=0 f′(x)=0 has x=2,x=p roots. But f(x) has exactly one local maxima and one local minima so p=2 Also 16−p2>0⇒p2−16<0⇒p∈(−4,4) Possible integral values of p are −3,−2,−1,0,1,3 ⇒6 integral values

Q. Let $f (x) = 2 x^{3} - 3 (2 + p) x^{2} + 12 p x + ln (16 - p^{2})$ . If $f (x)$ has exactly one local maxima and one local minima, then the number of integral values of $p$ is (are)