Mathematics Test - 5

CONCEPT:

• Referring to the curve with stationary point is shown below it can be seen from the figure that before the slope becomes zero it was negative, after it gets zero it becomes positive.
• It can be said that dy/dx is negative before a stationary point dy/dx is positive after a stationary point. So the slope is increasing at that point.

\(\Rightarrow \frac{d^2y}{dx^2}>0\)

• So at this stationary point, the function will have a minimum and the shape of the curve will be concave down.

EXPLANATION:

• At the stationary point, the slope of the tangent is 0 so the tangent will be horizontal. So, statement 1 is correct and statement 3 is incorrect.
• Let f be a function defined on an open interval I. Let f be continuous at a critical point c in I.
• Referring to the curve, If f'(x) does not change significantly as x increases through C, then C is neither a point of local maxima nor a point of local minima. Such a point is called a point of inflection. So, statement 2 is correct.
• To calculate the points of inflection,

\(\Rightarrow \frac{d^2y}{dx^2} = 0\)

• So the correct answer is option 1.

Concept:

We know that if f(x) = |x|, then

f(x) = - x, x < 0

f(x) = x, x ≥ 0

Calculation:

Given function is f(x) =  |x|.

We know that,

f(x) = - x, x < 0

f(x) = x, x ≥ 0

The value of |x| is a real positive set.

The minimum value in real positive set is zero.

∴ The minimum value of |x| is zero.

Formula used:

\(\rm \frac{d}{dx} x^{n} = nx^{n - 1}\)

Calculation:

f(x) = x3 + x2 + kx

f^'(x) = 3x² + 2x + k

For no extremum, f should have zero turning points.

There will be no slope existing either it is +ve or -ve

For that above quadratic have no real roots.

⇒ f'(x) ≠ 0 which is only possible when,

2² - 4 × 3 × k < 0

⇒ 4 - 12k < 0

⇒ 4 < 12k

⇒ 1 < 3k

⇒ 3k > 1

∴ The condition that f(x) = x3 + x2 + kx has no local extremum is 3k > 1

Concept:

For maximum value of A, dA/dx = 0;

Calculation:

Given rectangle perimeter = 48 cm;

Let the sides of rectangle be x, y;

⇒ 2x + 2y = 48 cm ⇒ x + y = 24 cm;

Area of the rectangle will be A = xy;

For area to be maximum, dA/dx = 0

We have y = A/x ⇒ x + A/x = 24;

⇒ x² – 24x = A

dA/dx = 0 ⇒ 2x – 24 = 0

⇒ x = 12 cm ⇒ y = 12 cm;

Now the maximum area will be 12 × 12 = 144 cm²;

Given:

x2 + y2 = 1.

x2 + 4xy - y2

Concept Used:

The maximum value of a sin θ + b cos θ is \(\rm\sqrt{a^2+b^2}\).

Calculation:

Since, x2 + y2 = 1, we can suppose x = cos θ and y = sin θ.

Now, x2 + 4xy - y2

= cos2 θ + 4 sin θ cos θ - sin2 θ

= (cos2 θ - sin2 θ) + 4 sin θ cos θ

= cos 2θ + 2 sin 2θ

Its maximum value is \(\rm\sqrt{1^2+2^2}\) = √5.

Concept:

Steps to find the maxima and minima

1.find \(\rm \dfrac {dy}{dx}\) and equate to zero

2. find the value of x i. e. critical point

3. for the maximum value we put x in f(x) to get the value of f(x)  at the point.

Calculations:

Given, f(x) = \(\rm \left(\dfrac{1}{x}\right)^x\)

Consider, y = \(\rm \left(\dfrac{1}{x}\right)^x\)

Taking log on both side, we get

ln y = x ln(\(\rm \dfrac 1 x\))

⇒ln y = -x ln x

⇒(1/y) × \(\rm \dfrac {dy}{dx}= -ln\; x -1\)

Equating \(\rm \dfrac {dy}{dx} = 0\)

⇒\(\rm -ln\; x -1 = 0\)

⇒ x = \(\dfrac 1 e\)

So, for the maximum value we put x = \(\dfrac 1 e\) in f(x) to get the value of f(x)  at the point.

f(x) = e1/e

Hence, The maximum value of \(\rm \left(\dfrac{1}{x}\right)^x\) is e1/e

Concept:

The area of a triangle with a fixed perimeter is maximum when it is equilateral.

Calculation:

Using the property mentioned above, the area is maximum when the triangle is equilateral.

Height of the triangle = \(\rm \frac{\sqrt3}{2}a=\frac{\sqrt3}{2}8=4\sqrt3\) cm.

Additional Information

Area of a triangle is \(\rm\sqrt{s(s-a)(s-b)(s-c)}\) (Heron's formula), where s = (a + b + c)/2.

Since we are trying to maximize the Area, we can say that we are trying to maximize s(s - a)(s - b)(s - c).

Since s is a constant (because the perimeter is a fixed quantity), this is equivalent to maximizing (s - a)(s - b)(s - c).

Note that (s - a) + (s - b) + (s - c) = 3s - (a + b + c) = 3s - 2s = s, which is a constant.

So, by AM-GM inequality, a product xyz is maximum, if sum x + y + z is a constant, when x = y = z.

∴ (s - a) = (s - b) = (s - c)

⇒ a = b = c, or the triangle is equilateral.

Concept:

A function f(x) has extremum when f ' (x) = 0

And the extremum is maximum point, when,

f ' (x) = 0 and f " (x) < 0

And the extremum is minimum point, when,

f ' (x) = 0 and f " (x) > 0

Calculation:

Given, \(f(x)=\frac{x}{1+x \tan x}\)

⇒ f ' (x) = \({1(1+ x. tan\ x) - x(0 + tan \ x + x sec^2 x)\over (1+ xtan x)^2 }\)

⇒ f ' (x) = \({(1+ x. tan\ x) - x tan \ x - x ^2sec^2 x\over (1+ xtan x)^2 }\)

⇒ f ' (x) = \({1 - x ^2sec^2 x\over (1+ xtan x)^2 }\)           ___(i)

And f ' (x) = 0 

⇒ 1 - x²sec²x = 0

⇒ x² = cos²x 

⇒ x = cos x in (0, π/2).

Which has only one root in (0, π/2).

⇒ f(x) has only one extremum point.

And, differentiating equation (i),

f " (x) = \({( -2 xsec^2 x -x ^2 2sec^2 x \tan x)(1+xtan x)^2 - (1 - x ^2sec^2 x)2(1+x tan x)(tanx + x sec ^2 x)\over (1+ xtan x)^4 }\)

f "(x) = \((1+x tan x){ -2 xsec^2 x(1+x \tan x)(1+xtan x) - (1 - x ^2sec^2 x)(2tanx + x sec ^2 x)\over (1+ xtan x)^4 }\)

⇒ f "(x) = \({ -2 xsec^2 x(1+x \tan x)^2- (1 - x ^2sec^2 x)(2tanx + x sec ^2 x)\over (1+ xtan x)^3 }\)

when x = cos x

⇒ f "(x) = \({ -2 sec x(1+\sin x)^2- (1 - 1)(2tanx + x sec ^2 x)\over (1+ sin x)^3 }\)

⇒ f "(x) = \({ -2 sec x(1+\sin x)^2\over (1+ sin x)^3 }\) < 0 in (0, π/2).

⇒ f(x) has one maximum point.

∴ The correct answer is option (2).

Calculation:

f(x) = x - 6  and g(x) = x2 - 2x + 3

g(f(x)) = (x - 6)² - (2 × (x - 6)) + 3

⇒ x² - 12x + 36 - 2x + 12 + 3

⇒ x² - 14x + 51

Similarly f(g(x)) = (x² - 2x + 3) - 6

⇒ f(g(x)) = x² - 2x -3

Let k(x) = g(f(x)) +  f(g(x))

k(x) = (x2 - 14x + 51) + (x2 - 2x -3)

⇒ 2x2 - 16x + 48

To find the point at which function is minimum or maximum, differentiating k with respect to x and equating with zero.

k^| = 2x - 8 = 0

⇒ x = 4

Additional Information

• A quadratic equation, ax² + bx + c can be represented as a parabola.
• 'a' is positive for an upward parabola and negative for a downward parabola.
• For an upward parabola, differentiation of ax² + bx + c will give us the minima.
• For a downward parabola,  differentiation of ax2 + bx + c will give us the maxima.

Concept:

Differentiation formula

\(\rm \frac{d}{dx} \sin x = \cos x\)

\(\rm \frac{d}{dx} \sin ax = a\cos ax\)

Calculation:

f(x) = k sin x + \(\rm \frac{1}{3}\)sin 3x

⇒ f’(x) = k cos x + cos 3x 

⇒ f’(\(\rm \frac{\pi}{3}\)) = k cos (\(\rm \frac{\pi}{3}\)) + cos (π) = 0

⇒ \(\rm \frac{k}{2}\) – 1 = 0

⇒ k – 2 = 0

⇒ k = 2

∴ The value of k is 2.