Differentiation (Part-38) | S N De

Differentiation (Part-31) | S N Dey

5. If the function \( f(x) = x^3 + ax^2 - bx + 4 \) defined in \( -2 \le x \le 2 \) satisfies Rolle's theorem when \( -2 < c < 2 \) where \( c = \frac{1}{3}(1+\sqrt{13}) \), then find the values of \( a \) and \( b \).

Solution.

We know that if a real-valued function \( f(x) \) defined on \( [a, b] \) satisfies:

(i) \( f(x) \) is continuous in \( [a, b] \)
(ii) \( f(x) \) is differentiable in \( (a, b) \)
(iii) \( f(a) = f(b) \)

then there exists at least one point \( c \in (a, b) \) such that \( f'(c) = 0 \).

For \( f(x) = x^3 + ax^2 - bx + 4 \) on \( [-2, 2] \):

$$ f'(x) = 3x^2 + 2ax - b $$

$$ \therefore f'(c) = 3c^2 + 2ac - b $$

Substituting \( c = \frac{1+\sqrt{13}}{3} \):

$$ \left[f'(c)\right]_{c=\frac{1+\sqrt{13}}{3}} = 3 \times \frac{1}{9}(1+\sqrt{13})^2 + \frac{2a}{3}(1+\sqrt{13}) - b = 0 $$

$$ \implies \frac{1}{3}(1 + 13 + 2\sqrt{13}) + \frac{2a}{3}(1+\sqrt{13}) - b = 0 $$

$$ \implies b = \frac{14+2\sqrt{13}}{3} + \frac{2a}{3}(1+\sqrt{13}) \quad \rightarrow (1) $$

Now, applying condition (iii), \( f(2) = f(-2) \):

$$ 2^3 + a(2)^2 - b(2) + 4 = (-2)^3 + a(-2)^2 - b(-2) + 4 $$

$$ \implies 8 + 4a - 2b = -8 + 4a + 2b $$

$$ \implies 16 = 4b \implies b = 4 \quad \rightarrow (2) $$

Substituting \( b = 4 \) into equation (1):

$$ 4 = \frac{14+2\sqrt{13}}{3} + \frac{2a}{3}(1+\sqrt{13}) $$

$$ \implies 4 - \frac{14+2\sqrt{13}}{3} = \frac{2a}{3}(1+\sqrt{13}) $$

$$ \implies \frac{12 - 14 - 2\sqrt{13}}{3} = \frac{2a}{3}(1+\sqrt{13}) $$

$$ \implies -2(1+\sqrt{13}) = 2a(1+\sqrt{13}) $$

$$ \implies a = -1 \quad \rightarrow (3) $$

\( \therefore a = -1 \) and \( b = 4 \).

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6. Using Rolle's theorem, find the point on the curve where the tangent is parallel to the x-axis:
(a) \( y = x(x-4), \quad x \in [0, 4] \)
(b) \( f(x) = 1 - \cos x \quad \text{in } 0 \le x \le 2\pi \)

Solution (a).

We have \( f(x) = y = x(x-4) = x^2 - 4x \).

(i) \( f(x) \) is continuous on \( [0, 4] \)
(ii) \( f'(x) = 2x - 4 \) exists for all \( x \in (0, 4) \)
(iii) \( f(0) = 0 = f(4) \)

By Rolle's theorem, there exists a point \( c \in (0, 4) \) such that \( f'(c) = 0 \):

$$ 2c - 4 = 0 \implies 2c = 4 \implies c = 2 $$

At \( c = 2 \):

$$ f(2) = 2(2 - 4) = -4 $$

So, the point on the curve is \( (2, -4) \).

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Solution (b).

We have \( f(x) = y = 1 - \cos x \) in \( 0 \le x \le 2\pi \).

(i) \( f(x) \) is continuous on \( [0, 2\pi] \)
(ii) \( f'(x) = \sin x \) exists for all \( x \in (0, 2\pi) \)
(iii) \( f(0) = 0 = f(2\pi) \)

By Rolle's theorem, there exists a point \( c \in (0, 2\pi) \) such that \( f'(c) = 0 \):

$$ \sin c = 0 \implies c = \pi \quad [\because c \in (0, 2\pi)] $$

At \( c = \pi \):

$$ f(\pi) = 1 - \cos\pi = 1 - (-1) = 2 $$

So, the point on the curve is \( (\pi, 2) \).

7. At what point is the tangent to the curve \( f(x) = \log x \) parallel to the chord joining the points \( A(1, 0) \) and \( B(e, 1) \)?

Solution.

We have \( f(x) = \log x \):

$$ f'(x) = \frac{1}{x} \quad \rightarrow (1) $$

The slope of the chord joining \( A(1, 0) \) and \( B(e, 1) \) is:

$$ m = \frac{1 - 0}{e - 1} = \frac{1}{e - 1} $$

Let \( x = c \) be the abscissa of the point where the tangent is parallel to the chord. Equating the slopes:

$$ \frac{1}{c} = \frac{1}{e - 1} \implies c = e - 1 $$

Hence, the tangent is parallel to the chord at the point with abscissa \( c = e - 1 \).

8. In the mean value theorem, \( f(b) - f(a) = (b-a)f'(c) \), \( (a < c < b) \); if \( f(x) = x^3 - 3x - 1 \) and \( a = -\frac{11}{7}, \, b = \frac{13}{7} \), find the value of \( c \).

Solution.

First, compute \( f(a) \) and \( f(b) \):

$$ f(a) = f\left(-\frac{11}{7}\right) = \left(-\frac{11}{7}\right)^3 - 3\left(-\frac{11}{7}\right) - 1 $$

$$ = -\frac{1331}{343} + \frac{33}{7} - 1 = \frac{-1331 + 1617 - 343}{343} = -\frac{57}{343} \quad \rightarrow (1) $$

$$ f(b) = f\left(\frac{13}{7}\right) = \left(\frac{13}{7}\right)^3 - 3\left(\frac{13}{7}\right) - 1 $$

$$ = \frac{2197}{343} - \frac{39}{7} - 1 = \frac{2197 - 1911 - 343}{343} = -\frac{57}{343} \quad \rightarrow (2) $$

Now, finding \( f'(x) \):

$$ f'(x) = 3x^2 - 3 = 3(x^2 - 1) \implies f'(c) = 3(c^2 - 1) \quad \rightarrow (3) $$

Applying the Mean Value Theorem formula:

$$ f(b) - f(a) = (b - a)f'(c) $$

$$ \implies -\frac{57}{343} - \left(-\frac{57}{343}\right) = \left[\frac{13}{7} - \left(-\frac{11}{7}\right)\right] f'(c) $$

$$ \implies 0 = \left(\frac{24}{7}\right) f'(c) \implies f'(c) = 0 $$

$$ \implies 3(c^2 - 1) = 0 \implies c^2 = 1 \implies c = \pm 1 $$

\( \therefore c = \pm 1 \).