Applications of Derivatives MCQ Quiz in मल्याळम - Objective Question with Answer for Applications of Derivatives - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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.

1. f``(x) is less than 0 then the given function is said to be maxima
2. If f``(x) Is greater than 0 then the function is said to be minima

Calculation:

Let f(x) = x³ - 12x² + kx – 8

Differentiating with respect to x, we get

⇒ f’(x) = 3x² – 24x + k

It is given that function attains its maximum value of the interval [0, 3] at x = 2

∴ f’(2) = 0

⇒ 3 × 2² – (24 × 2) + k = 0

∴ k = 36

Concept:  

If f′(x) >  0 at each point in an interval, then the function is said to be strictly increasing.

Calculations:

Given , f(x) = x2 - 4x 

Differentiating, we get

f'(x) = 2x - 4

f(x) is strictly increasing function

∴ f'(x) > 0

⇒ 2x - 4 > 0

⇒ x > 2

∴ x ∈ (2, ∞)

Mistake Points

If a, b ∈ R and a < b, the following is a representation of the open and closed intervals:

• Open interval is indicated by (a, b) = {x : a < y < b}, i.e. 'a' and 'b' are not included.
• Closed interval is indicated by [a, b] = {x : a ≤ x ≤ b}, i.e. 'a' and 'b' are included.
• [a, b) = {x : a ≤ x < b} is an open interval from a to b, inclusive of 'a' but excluding 'b'.
• (a, b ] = { x : a < x ≤ b } is an open interval from a to b including 'b' but excluding 'a'.

Since '2' is not included here, Option (1) will not be correct.

Concept:

The equation of tangent at point (x1 , y1) with slope m is given by 

\(\rm (y - y_1) = m (x - x_1)\)

Calculations:

Given curve is y = 2x sin x 

Taking derivative on both side, we get

\(\rm \dfrac {dy}{dx}= 2x \;cos\;x + 2 \;sin\;x\)

Put x = \(\rm \dfrac{\pi}{2}\) to find the equation of tangent at the point \(\left(\frac \pi 2, \pi\right)\).

\(\rm \dfrac {dy}{dx}= 2 \dfrac {\pi}{2} \;cos\; \dfrac {\pi}{2} + 2 \;sin\; \dfrac {\pi}{2}\)

\(\rm \dfrac {dy}{dx}= 2\)

The equation of tangent at point (x₁ , y₁) with slope m is given by 

\(\rm (y - y_1) = m (x - x_1)\)

\(\rm (y - {\pi}) = 2 (x - \dfrac{\pi}{2})\)

y = 2x

Hence, the equation of the tangent line to the curve y = 2x sin x at the point \(\left(\frac \pi 2, \pi\right)\) is 2x.

From the question, the derivative given is:

\({\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{{\rm{sin\;x}} \:- \:{\rm{cos\;x}}}}{{\sin x \:+ \:{\rm{cos\;x}}}}} \right) \)

Now,

\( \Rightarrow {\rm{y}} = {\rm{\;ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{{\rm{sin\;x}} - {\rm{cos\;x}}}}{{\sin x + {\rm{cos\;x}}}}} \right)\)

∵ \(\left[ {\begin{array}{*{20}{c}} {\tan \left( {A - B} \right) = \frac{{{\rm{tan\;}}A - {\rm{tan\;}}B}}{{1 + {\rm{tan\;}}A\;{\rm{tan\;}}B}}}\\ {tan\left( {\frac{\pi }{4}} \right) = 1} \end{array}} \right]\)

\( \Rightarrow {\rm{y}} = {\rm{\;ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{{\rm{tan\;x}} - {\rm{1}}}}{{\tan x + {\rm{1}}}}} \right)\)

\( \Rightarrow {\rm{y}} = {\rm{\;ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{{\rm{tan\;x}} - {\rm{1}}}}{{\ {\rm{1}}+\tan x}}} \right)\)

\(\Rightarrow y = \;ta{n^{ - 1}}\left( {\frac{{{{\rm{tan\;}}x- \rm{tan}}\left( {\frac{\pi }{4}} \right) }}{{1 + {\rm{tan}}\left( {\frac{\pi }{4}} \right){\rm{tan\;}}x}}} \right)\)

\(\Rightarrow y = \;ta{n^{ - 1}}\left( {{\rm{tan}}\left( {x - \frac{\pi }{4}} \right)} \right)\)

\(\therefore y = x - \frac{\pi }{4} \)

From the question, let’s assume,

\(z = \frac{x}{2}\)

From the question, we need to find \(\frac{{dy}}{{dz}}\).

\(\Rightarrow \frac{{dy}}{{dz}} = \frac{{dy/dx}}{{dz/dx}}\)

Now,

\(\Rightarrow \frac{{dy}}{{dx}} = 1\)

\(\Rightarrow \frac{{dz}}{{dx}} = \frac{1}{2}\)

Now,

\(\Rightarrow \frac{{dy}}{{dz}} = \frac{1}{{1/2}}\)

Concept:

Differentiation formula

Quotient rule:

\(\rm \frac{d}{dx} (\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2}}\)

Calculation:

f(x) = \(\rm \frac{(log x)}{x}\)

⇒ f^'(x) = \(\rm \frac{1}{x^{2}} - \frac{log x}{x^{2}}\)

⇒ 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) = \(\rm \frac{1}{e}\)

∴ The maximum value of \(\rm \frac{(log x)}{x}\) is \(\rm \frac{1}{e}\)

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) 

Concept:

Monotonic function: If a function is differentiable on the interval (a, b) and if the function is increasing/strictly increasing or decreasing/strictly decreasing, then the function is known as monotonic function.

First derivative test:

• If \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} \ge 0\) for all x in (a, b) then, f(x) is an increasing function in (a, b).
• If \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} \le 0\) for all x in (a, b) then, f(x) is an decreasing function in (a, b).

For strictly increasing or decreasing:

• \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} > 0\) for strictly increasing.
• \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} < 0\) for strictly decreasing.

Calculation:

Given: \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + \frac{1}{{\rm{x}}}\)

Differentiating w.r.t x

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

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

So, for x ∈ (0, 1), f’(x) < 0.

Hence, f(x) is decreases for x ∈ (0, 1).

Concept:

Circumference of circle = 2πr, where r is the radius of the circle.

Calculation:

Given: \(\frac{{dr}}{{dt}} = 0.8\;cm/sec\) .

Circumference of circle (C) = 2πr

Differentiating both the sides with respect to t, we get:

⇒  \(\frac{{dC}}{{dt}} = 2π \times \frac{{dr}}{{dt}}\)        ----(1)

By, substituting the values of \(\frac{{dr}}{{dt}}\) the equation(1), we get:

⇒ \(\frac{{dC}}{{dt}} = 2π \times 0.8 = \frac{{8\pi}}{{5}}\)

CONCEPT:

• If y = f(x), then dy/dx denotes the rate of change of y with respect to x its value at x = a is denoted by: \({\left[ {\frac{{dy}}{{dx}}} \right]_{x = a}}\)
• Decreasing rate is represented by negative sign whereas increasing rate is represented by positive sign.
• Volume of cylinder is given by: πr²h where r is the radius and h is the altitude.

CALCULATION:

Given: dr/dt = 3m/s and dh/dt = - 4m/s

As we know that, volume of cylinder is given by: πr2h where r is the radius and h is the altitude.

Let V = πr2h

Now by differentiating v with respect to t we get,

\(\Rightarrow \frac{{dV}}{{dt}} = 2\pi rh \cdot \frac{{dr}}{{dt}} + \pi {r^2} \cdot \frac{{dh}}{{dt}}\)

Now by substituting r = 4m, h = 6m, dr/dt = 3m/s and dh/dt = - 4m/s we get,

\(\Rightarrow {\left( {\frac{{dV}}{{dt}}} \right)_{\left( {r = 4,\;\;\;h = 6} \right)}} = 2\pi \cdot \left( 4 \right) \cdot \left( 6 \right) \cdot \left( 3 \right) + \pi \cdot {\left( 4 \right)^2} \cdot \left( { - \;4} \right) = 80\pi \)

Hence, option D is the correct answer.

Concept: 

Let small charge in x be Δx and the corresponding change in y is Δy.

\(\rm Δ y = \rm \dfrac{dy}{dx}Δ x = f'(x) Δ x \)

Now that Δy = f(x + Δx) - f(x)

Therefore, f(x + Δx) = f(x) + Δy

Calculation: 

Given:f(x) = 3x2 +  3

Let x + Δx = 3.01 = 3 + 0.01

Therefore, x = 3 and Δx = 0.01

f(x + Δx) = f(x) + Δy

= f(x + Δx) = f(x) + f'(x)Δx

= f(3.01) = 3x2 +  3 + (6x)Δx

= f(3.01) = 3(3)² + 3 + (6⋅3)(0.01)

= f(3.01) = 30 + 0.18

= f(3.01) = 30.18