If the function f given by f(x) = x3 -3(a - 2)x 2 + 3ax + 7, for some a∈ R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, (f(x)-14/(x-1)2) = 0(x ≠ 1) is:

Question

If the function f given by f(x) = x3 -3(a - 2)x 2 + 3ax + 7, for some a∈ R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, (f(x)-14/(x-1)2) = 0(x ≠ 1) is: (A) 6 (B) 5 (C) 7 (D) -7

Tardigrade Admin · Accepted Answer

f'(x) = 3x2 - 6(a - 2)x + 3a f'(x) ge 0 ∀ x ∈ (0, 1] f'(x) le 0 ∀ x ∈ [1,5) ⇒ f'(x) -14 = (x-1)2 (x-7) (f(x) -14/(x-1)2) = x-7

Q. If the function $f$ given by $f (x) = x^{3} - 3 (a - 2) x^{2} + 3 a x + 7$ , for some a $\in R$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$ , then a root of the equation, $\frac{f ( x ) - 14}{( x - 1 ) ^{2}} = 0 (x \neq = 1) i s :$

6

5

7

-7

$f^{'} (x) = 3 x^{2} - 6 (a - 2) x + 3 a$
$f^{'} (x) \geq 0\forall x \in (0, 1]$
$f^{'} (x) \leq 0\forall x \in [1, 5)$
$\Rightarrow f^{'} (x) - 14 = (x - 1)^{2} (x - 7)$
$\frac{f ( x ) - 14}{( x - 1 ) ^{2}} = x - 7$

Q. If the function f given by f(x)=x3−3(a−2)x2+3ax+7, for some a∈R is increasing in (0,1] and decreasing in [1,5), then a root of the equation, (x−1)2f(x)−14​=0(x=1)is:

6

5

7

-7

f′(x)=3x2−6(a−2)x+3a f′(x)≥0∀x∈(0,1] f′(x)≤0∀x∈[1,5) ⇒f′(x)−14=(x−1)2(x−7) (x−1)2f(x)−14​=x−7

Q. If the function $f$ given by $f (x) = x^{3} - 3 (a - 2) x^{2} + 3 a x + 7$ , for some a $\in R$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$ , then a root of the equation, $\frac{f ( x ) - 14}{( x - 1 ) ^{2}} = 0 (x \neq = 1) i s :$

$f^{'} (x) = 3 x^{2} - 6 (a - 2) x + 3 a$
$f^{'} (x) \geq 0\forall x \in (0, 1]$
$f^{'} (x) \leq 0\forall x \in [1, 5)$
$\Rightarrow f^{'} (x) - 14 = (x - 1)^{2} (x - 7)$
$\frac{f ( x ) - 14}{( x - 1 ) ^{2}} = x - 7$