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

$\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\frac{1}{1+{\mathrm{x}}^2}$

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)

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

$\Rightarrow \frac{dx}{dy}=\frac{12-3y^2}{6x}\dots \left(2\right)$

For vertical tangent, 

$\Rightarrow \frac{dx}{dy}=0$ 

• 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 $(\pm \frac{4}{\sqrt{3}},2)$

 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,

$\Rightarrow tan\theta =\frac{m_1-m_2}{1+m_1{m}_2}\dots \left(1\right)$

CALCULATION:

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

• For the first curve, y = x² 

⇒ $\frac{dy}{dx}=m_1=2x$

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

• For the second curve, x = y² 

⇒  1 = 2y $\frac{dy}{dx}$

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

• Using equation 1,

 $\Rightarrow tan\theta =\frac{2-\frac{1}{2}}{1+2\times \frac{1}{2}}=\frac{3}{4}$

⇒ θ = tan^-1 (3/4)

So the correct answer is option 4.

Concept:

Rate of change of 'x' is given by $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}$

Calculation:

Given that, y = 2x – x² and $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}$ = 3 units/sec

Then, the slope of the curve, $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = 2 - 2x = m

⇒$\frac{\mathrm{d}\mathrm{m}}{\mathrm{d}\mathrm{t}}$  = 0 - 2 × $\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{t}}$

= -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, 

$\Rightarrow 2y.\frac{dy}{dx}=3p{x}^2$

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

∴ ${\left|\frac{dy}{dx}\right|}_{2,3}=\frac{3p}{2}\times \frac{4}{3}=2p$

• 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,

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

CALCULATION:

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

• Using equation 1,

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

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

⇒ f'(3) = 1

So the correct answer is option 4.

Formula used:

Differentiation Formula:

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

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.