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 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.
Find the Derivative \( \frac{dy}{dx} \)
Given:\[ y = \cos(x^2) + \cos(2x) + \cos^2(x^2) + \cos(x^x) \]
Find the intervals in which the function\[ f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11 \]
is:
If \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6 \], then f(1) is equal to:
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.
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
Read More: Application of Derivatives