Let
\(f(x) = \begin{vmatrix} a & -1 & 0\\ ax & a & -1\\ ax^2 & ax & a \end{vmatrix}\)
a ∈ R. Then the sum of the square of all the values of a, for which 2f′(10) –f′(5) + 100 = 0, is
Let
\(f(x) = \begin{vmatrix} a & -1 & 0\\ ax & a & -1\\ ax^2 & ax & a \end{vmatrix}\)
a ∈ R. Then the sum of the square of all the values of a, for which 2f′(10) –f′(5) + 100 = 0, is
117
106
125
136
The Correct Option is C
Solution and Explanation
The correct answer is (C) : 125
\(f(x) = \begin{vmatrix} a & -1 & 0\\ ax & a & -1\\ ax^2 & ax & a \end{vmatrix},\) \(a∈R\)
f(x) = a(a2 + ax) + 1(a2x + ax2) = a(x + a)2
f′(x) = 2a(x + a)
Now, 2f′(10) – f′(5) + 100 = 0
⇒ 2·2a(10 + a) – 2a(5 + a) + 100 = 0
⇒ 2a(a + 15) + 100 = 0
⇒ a2 + 15a + 50 = 0
⇒ a = –10, –5
Therefore,
Sum of squares of values of a is 125.
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
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