Mathematics Test - 17

Concept:

Following steps to finding maxima and minima using derivatives.

Find the derivative of the function.

Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.

Now we have to find the second derivative.

• f"(x) is less than 0 then the given function is said to be maxima
• If f"(x) Is greater than 0 then the function is said to be minima

We know,

sin 2x = 2 sinx.cosx

Calculation:

Let, f(x) = sinx .cos x

= $\frac{1}{2}\times \mathrm{s}\mathrm{i}\mathrm{n}2\mathrm{x}$            (∵ $\mathrm{s}\mathrm{i}\mathrm{n}2\mathrm{x}=2\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}.\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}$) 

f'(x) = $\frac{1}{2}\times (2\mathrm{c}\mathrm{o}\mathrm{s}2\mathrm{x})$

= cos 2x

Now, f'(x) = 0 ⇒ cos 2x = 0

We know, $\mathrm{c}\mathrm{o}\mathrm{s}(\frac{\mathrm{\pi }}{2})=0$

∴ cos 2x = $\mathrm{c}\mathrm{o}\mathrm{s}(\frac{\mathrm{\pi }}{2})$

⇒ 2x = π/2 

⇒ x = π/4

Also, f''(x) = $-(4\mathrm{s}\mathrm{i}\mathrm{n}2\mathrm{x})$

⇒ f''($\frac{\pi }{4}$) = - 4 < 0

At x = $\frac{\pi }{4}$ , f(x) is maximum.

∴ f($\frac{\pi }{4}$) = $\frac{1}{2}\times \mathrm{s}\mathrm{i}\mathrm{n}(\frac{\mathrm{\pi }}{2})$ 

= 1/2

Hence, option (4) is correct.

Concept:

|-a| = a,  where a = constant

Calculation:

We have, f(x) = |x - 4|

At, x = 0, f(0) = |0 - 4| = 4

At, x = 2, f(2) = |2 - 4| = |-2| = 2

At, x = 4, f(4) = |4 - 4| = 0

At, x = -4, f(-4) = |-4 - 4| = 8

So , at x = 4, we get minimum value 

Hence, option (3) is correct.

Formula used:

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}{\mathrm{x}}^{\mathrm{n}}=\mathrm{n}{\mathrm{x}}^{\mathrm{n}-1}$

Concept Used:

If f''(a) > 0 then x = a is a point of local minima

If f''(a) < 0 then x = a is a point of local maxima

Calculation:

Let f(x) = x + $\frac{1}{x}$​

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}=1-\frac{1}{{\mathrm{x}}^2}$

$\frac{\mathrm{d}\mathrm{f}(\mathrm{x})}{\mathrm{d}\mathrm{x}}=0$

x = ±1

$\frac{{\mathrm{d}}^2\mathrm{f}(\mathrm{x})}{\mathrm{d}{\mathrm{x}}^2}=\frac{2}{{\mathrm{x}}^3}$

At x = 1, $\frac{{\mathrm{d}}^2\mathrm{f}(\mathrm{x})}{\mathrm{d}{\mathrm{x}}^2}=2$ 

$\frac{{\mathrm{d}}^2\mathrm{f}(\mathrm{x})}{\mathrm{d}{\mathrm{x}}^2}>0$

So, the function has minima and its value is f(1) = 2

At x = -1, $\frac{{\mathrm{d}}^2\mathrm{f}(\mathrm{x})}{\mathrm{d}{\mathrm{x}}^2}=-2$

$\frac{{\mathrm{d}}^2\mathrm{f}(\mathrm{x})}{\mathrm{d}{\mathrm{x}}^2}<0$

So, the function has maxima and its value is f(-1) = -2

⇒ fmax < fmin 

∴ Statement 1 is only correct.

Concept:

Trigonometric formulas used:

• sin 3x = 3sin x – 4sin³ x

Calculation:

Given: f(x) = 3sin x – 4sin³ x

We know that, sin 3x = 3sin x – 4sin³ x

So, f(x) = 3 sin x – 4 sin³ x = sin 3x

We know that, sin 3x is a periodic function.

As we know Sin x is a periodic function [-π/2 to π/2]  

Since here the function is Sin 3x the range becomes [-π/6 to π/6]

So, sin 3x increases from –π/6 to π/6.

∴ The length of the increasing interval = π/6 – (-π/6) = π/3 

Concept:

Following steps to finding maxima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have find second derivative.
  • f"(x) is less than 0 then the given function is said to be maxima

Calculation:

Let f(x) = xy + 5,

2x + y = 4 

⇒ y = 4 - 2x

So, f(x) =  x(4 - 2x) + 5 

f(x) = 4x - 2x² + 5

Differentiating with respect to x, we get

f'(x) = 4 - 4x 

Set f(x) = 0 

⇒ 4 - 4x = 0

⇒ x = 1

Again differentiating with respect to x

f''(x) = -4 < 0

Hence function is said to be maxima

So, maximum value will obtain at x = 1, and y = 4  - 2(1) = 2

∴ The maximum value of xy + 5 = 1(2) + 5 = 7

Hence, option (4) is correct. 

Concept:

A function f(x) is maximum at a point, where,

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

Calculation:

Given, s = 64t - 16t2,

So, the function s(t) is maximum when, 

s'(t) = 0 and s"(t) < 0

⇒ s'(t) = 64 - 16(2t) = 0 

⇒ 64 = 32 t 

⇒ t = 2

And at t = 2, s"(t) = -32 < 0

So, s(t) is maximum at t = 2.

So, The time required to reach the maximum height by the stone is 2s.

∴ The correct answer is an option (2).

Concept:

Differentiation formula

Quotient rule:

$\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}(\frac{\mathrm{u}}{\mathrm{v}})=\frac{\mathrm{v}\frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}-\mathrm{u}\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}}{{\mathrm{v}}^2}$

Calculation:

f(x) = $\frac{(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x})}{\mathrm{x}}$

⇒ f^'(x) = $\frac{1}{{\mathrm{x}}^2}-\frac{\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}}{{\mathrm{x}}^2}$

⇒ f′(x) = 0 at x = e

⇒ f′(x) > 0 when x < e

⇒ f′(x) < 0 when x > e

So, f(x) is maxima at x = e

⇒  f(e) = $\frac{1}{\mathrm{e}}$

∴ The maximum value of $\frac{(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x})}{\mathrm{x}}$ is $\frac{1}{\mathrm{e}}$

Concept:

 Decreasing rate always represent by negative sign and increasing rate by positive sign.

Area of rectangle is given by A = x × y where x is length and y is breadth

Calculations:

Here $\frac{dx}{dt}=-7and\frac{dy}{dt}=6$ and x = 10, y = 8

Area of Rectangle A = xy

The rate of change of area of rectangle

$\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}$

$\frac{dA}{dt}=\left(10\right)\left(6\right)+8\left(-7\right)$  

$\frac{dA}{dt}=4$

Concept:

Formula:

• cos-1 (cos x) = x
• sin^-1 (sin x) = x
• Slope of the curve = tan θ = $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$

Calculation:

Given: Slope = tan θ

y = sin-1 (cos x)

⇒ y = sin-1 (sin (π/2 -x))

⇒ y = π/2 – x                                        (∵ sin-1 (sin x) = x)

Differentiating both sides with respect to x, we get

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = -1

As we know, Slope = $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$

⇒ tan θ  = -1

∴ θ = 3π/4

Concept:

• If f′(x) > 0 then the function is said to be increasing.
• If f′(x) < 0 then the function is said to be decreasing.

Calculation:

Given:

f(x) = x - e^x

Differentiating with respect to x, we get

⇒ f’(x) = 1 - e^x

For increasing function,

f'(x) > 0

⇒ 1 – e^x > 0

⇒ e^x < 1

⇒ e^x < e⁰

∴ x < 0

So, x ∈ (-∞, 0)