Mathematics Test - 4

Concept:

If f'(x) > 0 at each point in an intervel then the function is said to be increasing.

If f'(x) < 0 at each point in an intervel then the function is said to be decreasing.

Calculation:

Here, derivative of e^x is e^x

And it is greater than zero in any interval

∴ e^x is an increasing function.

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) = tan^-1 x

Differentiating with respect to x, we get

\( \Rightarrow {\rm{\;f'}}\left( {\rm{x}} \right) = {\rm{\;}}\frac{1}{{1 + {{\rm{x}}^2}}}{\rm{\;}}\)

As we know that, x² > 0 for x ∈ R

⇒ 1 + x² > 0 for x ∈ R

So, 1 + x² gives positive values for x ∈ R

Therefore, f’(x) > 0 for x ∈ R

Hence f(x) is increasing in x ∈ R or x ∈ (-∞,∞)

CONCEPT:

• if the tangent at a point on the given curve is vertical then the normal on that same point will be horizontal with a slope of 0

​CALCULATION:

Given: The tangent is vertical

y3 + 3x2 = 12y     .....(1)

⇒ \(3y^2. \frac{dy}{dx} + 6x = 12 \frac{dy}{dx} ⇒ \frac{dy}{dx} = \frac{6x}{12 - 3y^2}\)

\(⇒ \frac{dx}{dy} = \frac{12 - 3y^2}{6x}\;\;\;\;\; \ldots \left( 2 \right)\)

For vertical tangent, 

\(⇒ \frac{dx}{dy} = 0\) 

• Using equation 2,

⇒ 12 - 3y² = 0 

⇒ y = ± 2

• Putting y = 2 in equation (i) we get,

⇒ x = ±\(\frac{4}{\sqrt 3}\) and

• On putting y = - 2 in equation (i), we get,

⇒ 3x² = -16, no real solution

• So, the final points are \(\left( \pm \frac{4}{\sqrt 3} , 2 \right) \)

 So the correct answer is option 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) is an increasing function and g(x) is a decreasing function

∴ f’(x) >  0 and g’(x) < 0

Let h(x) = gof(x) = g(f(x))

Differentiating with respect to x, we get

⇒ h’(x) = g’(f(x)) × f’(x)

We know, f’(x) >  0 and g’(x) < 0

Therefore, h’(x) = (Negative) × (Positive) = Negative

∴ h’(x) < 0

Hence, gof(x) is decreasing function.

Alternate solution:

Let x₁ < x₂

Given: f(x) is an increasing function and g(x) is a decreasing function

∴ f(x₁) < f(x₂) and g(x₁) > g(x₂)

Let h(x) = gof(x) = g(f(x))

We know that, f(x₁) < f(x₂)

⇒ g(f(x₁)) > g(f(x₂))

Hence, gof(x) is decreasing function.

CONCEPT:

• The angle of intersection between the two curves will be the same as the angle of the intersection or between the tangents on both the curves on a common point.
• If the slope of a tangent at a common point on the first curve is m1 and the slope of a tangent at a common point on the second curve is m2 then the angle of intersection,

\(⇒ tan θ = \frac{m_1 - m_2}{1 + m_1 m_2} \;\;\;\;\; \ldots \left( 1 \right)\)

CALCULATION:

Given: The common point of tangent is (1, 1)

• For the first curve, y = x² 

⇒ \(\frac{dy}{dx} = m_1 = 2x\)

 \(⇒ \left( \frac{dy}{dx} \right)_{(1, 1)} \) = 2 = m₁

• For the second curve, x = y² 

⇒  1 = 2y \(\frac{dy}{dx} \)

\(⇒ \frac{dy}{dx} = m_2 = \frac{1}{2y} ⇒ \left( \frac{dy}{dx} \right)_{(1, 1)} = \frac{1}{2}\)

• Using equation 1,

 \(⇒ tan θ = \frac{2 - \frac{1}{2}}{1 + 2 \times \frac{1}{2}}= \frac{3}{4}\)

⇒ θ = tan^-1 (3/4)

So the correct answer is option 4.

Concept:

Rate of change of 'x' is given by \(\rm \frac {dx}{dt}\)

Calculation:

Given that, y = 2x – x² and \(\rm \frac {dx}{dt}\) = 3 units/sec

Then, the slope of the curve, \(\rm \frac {dy}{dx}\) = 2 - 2x = m

⇒\(\rm \frac {dm}{dt}\)  = 0 - 2 × \(\rm \frac {dx}{dt}\)

= -2(3)

= -6 units per second

Hence, the slope of the curve is decreasing at the rate of 6 units per second when x is increasing at the rate of 3 units per second.

Hence, option (2) is correct.

CONCEPT:

• The slope of a line on a given curve, y = f(x) on a given point,

⇒ y' = dy/dx at the given point

CALCULATION:

Given: y = 4x - 5 is tangent to the curve y2 = px3 + q at (2, 3) 

• Differentiate with respect to x, 

\(⇒ 2y. \frac{dy}{dx} = 3 px^2\)

⇒ \(\frac{dy}{dx} = \frac{3p}{2} \left( \frac{x^2}{y} \right)\)

∴ \(\left| \frac{dy}{dx} \right|_{2, 3}= \frac{3p}{2} × \frac{4}{3} = 2p\)

• For given line, slope of tangent = 4 (Making comparison with y = mx + c)

⇒  2p = 4 

⇒ p = 2

• Since (2, 3) are lying on the given curve y,

⇒ 9 = 2 × 8 + q 

⇒ q = - 7

CONCEPT:

• if the tangent to a given curve y at a point is perpendicular to the given line then the multiplication of slopes of that tangent and the given line will be equal to -1 as both are perpendicular to each other.
•  The slope of tangent on a given curve y can be calculated as,

⇒ y' at given point = dy/dx at given point

CALCULATION:

Given: y = x³ + ax - b and (1, -5) lies on the curve

⇒ - 5 = 1 + a - b ⇒ a - b = -6     ...(i)

Also, y' =  dy/dx =3x² + a

⇒ y'_{(1, -5)} = 3 + a  [Slope of tangent]

• ∵ This tangent is ⊥ to - x + y + 4 = 0 which is having the slope as 1,

⇒ (3 + a)(1) = -1

⇒ a = - 4        ...(ii)

• By (i) and (ii), a = - 4, b = 2

∴ y = x³ - 4x - 2

⇒ (2, -2) lies on the curve

So the correct answer is option 2.

CONCEPT:

• If a line is making an angle θ in the anti-clockwise sense with respect to the positive X-axis then the tangent of that angle is called slope. 
• The slope of the normal at a given point on the curve,

 \(\Rightarrow m_n = \frac{-1}{dy/dx}\;\;\;\;\; \ldots \left( 1 \right)\)

CALCULATION:

Given: The angle made by normal at point (3, 4) is \(\frac{3 \pi}{4}\)

• Using equation 1,

 \(⇒\tan \frac{3 \pi}{4} = \frac{-1}{(dy/dx)_{(3, 4)}}\)

∴ \(⇒\left( \frac{dy}{dx} \right)_{(3, 4)} = 1, \) 

⇒ f'(3) = 1

So the correct answer is option 4.

Formula used:

Differentiation Formula:

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

Calculation:

f(x) = x2 - kx

⇒ f^'(x) = 2x - k

Since, f is monotonic increasing.

f'(x) > 0

⇒ 2x - k > 0

⇒ k < 2x     ----(1)

Since, we have 1 < x < ∞ 

⇒ 2 < 2x < ∞    ----(2)

From (1) and (2), we get

k < 2

∴ The correct relation is k < 2.