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CBSE Important Questions Class 12 Maths Chapter 6 – Application of Derivatives

Class 12 Maths Chapter 6 Application of Derivatives is part of the CBSE Class 12 Maths syllabus. It shows how derivatives are used to solve real problems. Students learn to find the rate of change of quantities, to decide where a function is increasing or decreasing, to find the equation of a tangent and a normal to a curve, and to find the maximum and minimum values of a function.

These CBSE important questions for Class 12 Maths Chapter 6 cover rate of change problems, increasing and decreasing functions, tangents and normals, absolute maximum and minimum on a closed interval, the first and second derivative tests, and word problems on maximising area and volume. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?

Answer: Let r be the radius and C the circumference.
Given dr/dt = 0.7 cm/s.
The circumference is C = 2πr.
Differentiating both sides with respect to time t:
dC/dt = 2π (dr/dt)
= 2π × 0.7
= 1.4π cm/s, which is about 4.4 cm/s.

Q2. Write the condition for a function to be increasing or decreasing on an interval. Find the interval in which f(x) = x² − 4x + 6 is increasing.

Answer: A function f is increasing on an interval if f′(x) > 0 there, and decreasing if f′(x) < 0.
For f(x) = x² − 4x + 6:
f′(x) = 2x − 4
For f to be increasing:
2x − 4 > 0
x > 2
So f is increasing on (2, ∞) and decreasing on (−∞, 2).

Q3. Find the equation of the tangent to the curve y = x² − 2x + 7 at the point (2, 7).

Answer: First find the slope by differentiating:
dy/dx = 2x − 2
At x = 2: slope m = 2(2) − 2 = 2
Now use the point-slope form y − y₁ = m(x − x₁):
y − 7 = 2(x − 2)
y − 7 = 2x − 4
2x − y + 3 = 0

Q4. Find the equation of the normal to the curve y = x³ at the point (1, 1).

Answer: The slope of the tangent is dy/dx = 3x².
At x = 1: slope of tangent = 3
The normal is perpendicular to the tangent, so its slope is −1/3.
Using the point-slope form at (1, 1):
y − 1 = (−1/3)(x − 1)
3y − 3 = −x + 1
x + 3y − 4 = 0

Q5. Find the absolute maximum and minimum values of f(x) = 2x³ − 15x² + 36x + 1 on the interval [1, 5].

Answer: Step 1: Find the critical points.
f′(x) = 6x² − 30x + 36 = 6(x² − 5x + 6) = 6(x − 2)(x − 3)
Setting f′(x) = 0 gives x = 2 and x = 3, both inside [1, 5].
Step 2: Evaluate f at the critical points and the endpoints.
f(1) = 2 − 15 + 36 + 1 = 24
f(2) = 16 − 60 + 72 + 1 = 29
f(3) = 54 − 135 + 108 + 1 = 28
f(5) = 250 − 375 + 180 + 1 = 56
Absolute maximum = 56 at x = 5 and absolute minimum = 24 at x = 1.

Q6. State the first derivative test and the second derivative test for local maxima and minima.

Answer: First derivative test: Let c be a critical point, that is f′(c) = 0.
• If f′ changes sign from positive to negative as x passes through c, then c is a point of local maximum.
• If f′ changes from negative to positive, then c is a point of local minimum.
• If f′ does not change sign, c is a point of inflection.
Second derivative test: Let f′(c) = 0.
• If f″(c) < 0, then c is a point of local maximum.
• If f″(c) > 0, then c is a point of local minimum.
• If f″(c) = 0, the test fails and we go back to the first derivative test.

Q7. Find the local maximum and local minimum values of f(x) = x³ − 6x² + 9x + 15.

Answer: f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3)
Setting f′(x) = 0 gives x = 1 and x = 3.
Now use the second derivative test:
f″(x) = 6x − 12
At x = 1: f″(1) = −6, which is negative, so this is a local maximum.
f(1) = 1 − 6 + 9 + 15 = 19
At x = 3: f″(3) = 6, which is positive, so this is a local minimum.
f(3) = 27 − 54 + 27 + 15 = 15

Q8. A balloon which always remains spherical is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius is increasing when the radius is 15 cm.

Answer: The volume of a sphere is V = (4/3)πr³.
Given dV/dt = 900 cm³/s.
Differentiating with respect to t:
dV/dt = 4πr² (dr/dt)
900 = 4π(15)² (dr/dt)
900 = 900π (dr/dt)
dr/dt = 1/π cm/s, which is about 0.318 cm/s.

Q9. Show that of all the rectangles with a given perimeter, the square has the largest area.

Answer: Let the perimeter be a fixed value P, and let the sides be x and y.
2(x + y) = P, so y = P/2 − x
Area: A = xy = x(P/2 − x) = Px/2 − x²
Differentiate: dA/dx = P/2 − 2x
Setting dA/dx = 0: x = P/4
Second derivative: d²A/dx² = −2, which is negative, so this gives a maximum.
When x = P/4, y = P/2 − P/4 = P/4.
Since x = y, the rectangle is a square. Hence proved.

Q10. Find the intervals in which f(x) = 2x³ − 3x² − 36x + 7 is increasing and decreasing.

Answer: f′(x) = 6x² − 6x − 36 = 6(x² − x − 6) = 6(x − 3)(x + 2)
The critical points are x = −2 and x = 3. These divide the number line into three intervals.
For x < −2: both factors are negative, so f′ is positive → increasing
For −2 < x < 3: (x + 2) is positive and (x − 3) is negative, so f′ is negative → decreasing
For x > 3: both factors are positive, so f′ is positive → increasing
So f is increasing on (−∞, −2) and (3, ∞), and decreasing on (−2, 3).

Q11. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?

Answer: Let x be the distance of the foot from the wall and y the height on the wall.
By the Pythagoras theorem: x² + y² = 25
Differentiating with respect to t:
2x (dx/dt) + 2y (dy/dt) = 0
When x = 4: y = √(25 − 16) = 3
Given dx/dt = 2 cm/s = 0.02 m/s.
2(4)(0.02) + 2(3)(dy/dt) = 0
0.16 + 6(dy/dt) = 0
dy/dt = −0.16/6
= −8/300 m/s, which is about −2.67 cm/s
The negative sign shows the height is decreasing.

Q12. Find two positive numbers whose sum is 24 and whose product is maximum.

Answer: Let the two numbers be x and (24 − x).
Product: P = x(24 − x) = 24x − x²
Differentiate: dP/dx = 24 − 2x
Setting dP/dx = 0:
24 − 2x = 0, so x = 12
Second derivative: d²P/dx² = −2, which is negative, confirming a maximum.
So the numbers are 12 and 12, and the maximum product is 144.

Related Study Material on Extramarks

Q1.Find the intervals in which the function f given by f(x) = 4x³ – 6x² – 72x + 30 is (a) strictly increasing (b) strictly decreasing.

Opt.

We have f(x) = 4x³ – 6x² – 72x + 30 or f'(x) = 12x² – 12x – 72

= 12(x² – x -6) = 12(x – 3) (x + 2).

Therefore, f'(x) = 0 gives x = – 2 , 3.

There are three disjoint intervals, ( 2), (? 2, 3) and (3).

f'(x) > 0 for all x ? (2) and (3),

f'(x) < 0 for all x ( 2, 3)

Ans.

We have f(x) = 4x³ – 6x² – 72x + 30 or f'(x) = 12x² – 12x – 72

= 12(x² – x -6) = 12(x – 3) (x + 2).

Therefore, f'(x) = 0 gives x = – 2 , 3.

There are three disjoint intervals, (2), ( 2, 3) and (3).

(x) > 0 for all x ( 2) and (3),

(x) < 0 for all x ( 2, 3)

Q2.

$\begin{array}{l} Show that the right circular cone of least curved surface and given \\ volume has an altitude equal to \sqrt{2} ? times the radius of the base . \end{array}$

Opt.

$\begin{array}{l} Here, volume of the cone v = \frac{1}{3} {?r}^{2} h ? r^{2} = \frac{3 V}{?h} . \dots\dots\dots (1) \\ Surface area s = rl = ?r \sqrt{h^{2} +</Ans.} \end{array}$ $\begin{array}{l} Here, volume of the cone v = \frac{1}{3} r^{2} h r^{2} = \frac{3 V}{h} . \dots\dots\dots (1) \\ Surface area s = rl = r \sqrt{h^{2} + r^{2}} . \dots\dots\dots (2) \\ Whereh = height of the cone \\ r = radius of the cone \\ l = Slant height of the cone \\ S^{2} = 2 r^{2} (h^{2} + r^{2}) by (2) \\ Let S_{1} = S^{2} then by (1) \\ S_{1} = \frac{3 V}{h} (h^{2} + \frac{3 V}{h}) = 3 Vh + \frac{9 V^{2}}{h^{2}} \\ \frac{{dS}_{1}}{dh} = 3 V + 9 V^{2} (\frac{ˆ’ 2}{h^{3}}) \\ \frac{{dS}_{1}}{dh} = 0 for maxima / minima \\ 3 V + 9 V^{2} (\frac{ˆ 2}{h^{3}}) = 0 3 V + 9 V^{2} (\frac{ˆ’ 2}{h^{3}}) = h^{3} = ‹ \frac{6 V}{} \\ As surface area is least, we have \\ \frac{{dS}_{1}}{{dh}^{2}} > 0 when h^{3} = \frac{6 V}{} \\ Therefore curved surface area is minimum when \frac{3 h^{3}}{6} = V \\ Thus, \\ \frac{h^{3}}{6} = \frac{1}{3} r^{2} h ’ h^{2} = 2 r^{2} \\ h = \sqrt{2 r} \\ Hence for least curved surface the altitude is \sqrt{2} times radius . \end{array}$

CBSE Important Questions Class 12 Maths Chapter 6 – Application of Derivatives

Related Study Material on Extramarks

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