SATHEE: Application of Derivatives 2 Question 5

Application of Derivatives 2 Question 5

####5. 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, $\frac{f (x) - 14}{(x - 1)^{2}} = 0 (x \neq 1)$ is

(a) -7

(b) 6

(d) 5

Show Answer

Answer:

(c)

Solution:

Given that function,

$f (x) = x^{3} - 3 (a - 2) x^{2} + 3 a x + 7$ , for some $a \in R$ is increasing in $(0, 1]$ and decreasing in $[1, 5)$ .

$\begin{aligned} f^{'} (1) = 0 [∵ tangent at x = 1 will be \\ \Rightarrow {(3 x^{2} - 6 (a - 2) x + 3 a)}_{x = 1} = 0 \\ \Rightarrow 3 - 6 (a - 2) + 3 a = 0 \\ \Rightarrow 3 - 6 a + 12 + 3 a = 0 \\ \Rightarrow 15 - 3 a = 0 \\ \Rightarrow a = 5 \\ So, f (x) = x^{3} - 9 x^{2} + 15 x + 7 \\ \Rightarrow f (x) - 14 = x^{3} - 9 x^{2} + 15 x - 7 \\ \Rightarrow f (x) - 14 = (x - 1) (x^{2} - 8 x + 7) = (x - 1) (x - 1) (x - 7) \\ \Rightarrow \frac{f (x) - 14}{(x - 1)^{2}} = (x - 7) \end{aligned}$

Now, $\frac{f (x) - 14}{(x - 1)^{2}} = 0, (x \neq 1)$

$\begin{array}{rrr} \Rightarrow & x - 7 = 0 \\ \Rightarrow & x = 7 \end{array}$

Application of Derivatives 2 Question 5

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Application of Derivatives 2 Question 5

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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