SATHEE: Application of Derivatives 4 Question 43

Application of Derivatives 4 Question 43

####45. Let $f (x)$ is a function satisfying the following conditions

(i) $f (0) = 2, f (1) = 1$

(ii) $f (x)$ has a minimum value at $x = 5 / 2$ and

(iii) For all $x, f^{'} (x) = | \begin{array}{ccc} 2 a x & 2 a x - 1 & 2 a x + b + 1 \\ b & b + 1 & - 1 \\ 2 (a x + b) & 2 a x + 2 b + 1 & 2 a x + b \end{array} |$

where, $a$ and $b$ are some constants. Determine the constants $a, b$ and the function $f (x)$ .

$(1998, 8$ M)

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Answer:

Correct Answer: 45. $\frac{4 \sqrt{3}}{9}$

Solution:

Given, $f^{'} (x) = | \begin{array}{ccc} 2 a x & 2 a x - 1 & 2 a x + b + 1 \\ b & b + 1 & - 1 \\ 2 (a x + b) & 2 a x + 2 b + 1 & 2 a x + b \end{array} |$

Applying $R_{3} \to R_{3} - R_{1} - 2 R_{2}$ , we get

$f^{'} (x) = | \begin{array}{ccc} 2 a x & 2 a x - 1 & 2 a x + b + 1 \\ b & b + 1 & - 1 \\ 0 & 0 & 1 \end{array} |$

$\Rightarrow f^{'} (x) = 2 a x + b$

On integrating both sides, we get

$f (x) = a x^{2} + b x + c$

Since, maximum at $x = 5 / 2 \Rightarrow f^{'} (5 / 2) = 0$

$\Rightarrow 5 a + b = 0$

Also, $f (0) = 2 \Rightarrow c = 2$

and $f (1) = 1 \Rightarrow a + b + c = 1$

On solving Eqs. (i), (ii) and (iii), we get

$a = \frac{1}{4}, b = - \frac{5}{4}, c = 2$

Thus,

$f (x) = \frac{1}{4} x^{2} - \frac{5}{4} x + 2$

Application of Derivatives 4 Question 43

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Application of Derivatives 4 Question 43

Answer:

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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