Let f ( x )=|x 3 sin x cos x 6 -1 0 p p 2 p 3| where p is a constant. Then ( d 3(f( x ))/ dx 3) at x =0 is

Question

Let f ( x )=|x 3 sin x cos x 6 -1 0 p p 2 p 3| where p is a constant. Then ( d 3(f( x ))/ dx 3) at x =0 is (A) 6 p3 (B) p+p2 (C) p+p3 (D) independent of p

Tardigrade Admin · Accepted Answer

Q. Let $f (x) = ∣ ∣ x^{3} 6 p sin x - 1 p^{2} cos x 0 p^{3} ∣ ∣$ where $p$ is a constant. Then $\frac{d ^{3} ( f ( x ))}{d x ^{3}}$ at $x = 0$ is

$6 p^{3}$

$p + p^{2}$

$p + p^{3}$

independent of $p$

Q. Let f(x)=∣∣​x36p​sinx−1p2​cosx0p3​∣∣​ where p is a constant. Then dx3d3(f(x))​ at x=0 is

6p3

p+p2

p+p3

independent of p

f′(x)=∣∣​3x26p​cosx−1p2​−sinx0p3​∣∣​;f′′(x)=∣∣​6x6p​−sinx−1p2​−cosx0p3​∣∣​ f′′′(x)=∣∣​66p​−cosx−1p2​sinx0p3​∣∣​⇒f′′′(0)=∣∣​66p​−1−1p2​00p3​∣∣​=0 (two identical rows) ⇒ (D)

Q. Let $f (x) = ∣ ∣ x^{3} 6 p sin x - 1 p^{2} cos x 0 p^{3} ∣ ∣$ where $p$ is a constant. Then $\frac{d ^{3} ( f ( x ))}{d x ^{3}}$ at $x = 0$ is

$6 p^{3}$

$p + p^{2}$

$p + p^{3}$

independent of $p$