CBSE Class 12 Mathematics Applications Of Derivatives Assignment Set 04

Question. The position of a point in time 't' is given by x = a + bt − ct² , y = at + bt² . Its acceleration at time 't' is:
a. b − c
b. (b + c)
c. 2b − 2c
d. 2√b2 + c²
Answer : D

Question. If the volume of a spherical balloon is increasing at the rate of 900 cm²/sec. then the rate of change of radius of balloon at instant when radius is 15 cm: [in cm/sec]
a. 22/7
b. 22
c. 7/22
d. None of these
Answer : C

Question. The equations of motion of two stones thrown vertically upwards simultaneously are 2 s =19.6t − 4.9t² and s = 9.8t − 4.9t² respectively and the maximum height attained by the first one is h. When the height of the first stone is maximum, the height of the second stone will be:
a. h/3
b. 2h
c. h
d. 0
Answer : D

Question. The slope of the tangent to the curve x² + y² = 2c² at point (c, c) is:
a. 1
b. – 1
c. 0
d. 2
Answer : B

Question. The tangent to the curve y = 2x² − x + 1 at a point P is parallel to y = 3x + 4, the co-ordinate of P are:
a. (2, 1)
b. (1, 2)
c. (– 1, 2)
d. (2, – 1)
Answer : B

Question. The radius of a sphere is measured to be 20 cm with a possible error of 0.02 of a cm. The consequent error in the surface of the sphere is:
a. 10.5 sq cm
b. 5.025 sq cm
c. 10.05 sq cm
d.None of these
Answer : C

Question. If the path of a moving point is the curve x = at y = bsin at, then its acceleration at any instant:
a. Is constant
b. Varies as the distance from the axis of x
c. Varies as the distance from the axis of y
d. Varies as the of the point from the origin
Answer : C

Question. A stone thrown vertically upwards from the surface of the moon at velocity of 24 m/sec. reaches a height of s = 24t − 0.8t² m after t sec. The acceleration due to gravity in m/sec² at the surface of the moon is:
a. 0.8
b. 1.6
c. 2.4
d. 4.9
Answer : B

Question. The equation of motion of a particle is given by s = 2t³ − 9t² + 12t + 1, where s and t are measured in cm and sec. The time when the particle stops momentarily i:s
a. 1 sec
b. 2 sec
c. 1, 2 sec
d.None of these
Answer : C

Question. The equation of motion of a stone thrown vertically upward from the surface of a planet is given by s = 10t − 3t² , and the units of s and t are cm and sec respectively. The stone will return to the surface of the planet after:
a. 10/3 sec
b. 5/3 sec
c. 20/3 sec
d. 5/6 sec
Answer : A

Question. A man of height 1.8 m is moving away from a lamp post at the rate of 1.2 m/sec. If the height of the lamp post be 4.5 meter, then the rate at which the shadow of the man is lengthening:
a. 0.4 m/sec
b. 0.8 m/sec.
c. 1.2 m/sec.
d. None of these
Answer : B

Question. A 10 cm long rod AB moves with its ends on two mutually perpendicular straight lines OX and OY. If the end A be moving at the rate of 2 cm/sec. then when the distance of A from O is 8 cm, the rate at which the end B is moving, is:
a. 8/3 cm/sec
b. 4/3 cm/sec
c. 2/9 cm/ sec.
d. None of these
Answer : A

Question. For the curve yn = a^n-1 x , the sub-normal at any point is constant, the value of n must be:
a. 2
b. 3
c. 0
d. 1
Answer : A

Question. The sum of intercepts on co-ordinate axes made by tangent to the curve √x + √y = √a is:
a. a
b. 2a
c. 2√a
d. None of these
Answer : A

Question. The length of perpendicular from (0, 0) to the tangent drawn to the curve 2 y = 4(x + 2) at point (2, 4) is:
a. 1/√2
b. 3/√5
c. 6/√5
d. 1
Answer : C

Question. The equation of the tangent at (−4,− 4) on the curve x² = −4y :
a. 2x + y + 4 = 0
b. 2x − y −12 = 0
c. 2x + y − 4 = 0
d. 2x − y + 4 = 0
Answer : D

Question. A particle moves in a straight line in such a way that its velocity at any point is given by v² =  2 - 3x, where x is measured from a fixed point. The acceleration is:
a. Uniform
b. Zero
c. Non-uniform
d. Indeterminate
Answer : A

Question. At which point the line x/a + y/b = 1 touches the curve y = be^{−x / a}:
a. (0, 0)
b. (0, a)
c. (0, b)
d. (b, 0)
Answer : C

Question. The abscissa of the point, where the tangent to curve y³ = x² −3x −9x +5 is parallel to x-axis are:
a. 0 and 0
b. x =1 and −1
c. x = 1 and–3
d. x = −1 and 3
Answer : D

Question. The angle of intersection between curve xy = 6 and x² y = 12 ?
a. tan⁻¹ (3/4)
b. tan⁻¹ (3/11)
c. tan⁻¹ (11/3)
d. 0°
Answer : B

Question. The tangent drawn at the point (0, 1) on the curve y = e^2x meets x-axis at the point:
a. (1 / 2,0)
b. (−1/ 2, 0)
c. (2, 0)
d. (0, 0)
Answer : B

Question. The equation of the tangent to the curve x = 2 cos³ θ and y = 3 sin³ θ at the point θ = π / 4 is :
a. 2x + 3y = 3√2
b. 2x − 3y = 3√2
c. 3x + 2y = 3√2
d. 3x − 2y = 3√2
Answer : C

Question. On the interval (1,3) the function f (x) 3x + 2/x is:
a. Strictly decreasing
b. Strictly increasing
c. Decreasing in (2, 3) only
d. Neither increasing nor decreasing
Answer : B

Question. The function f (x) = cos x − 2px is monotonically decreasing for:
a p < 1/2
b. p > 1/2
c. p < 2
d. p > 2
Answer : B

Question. If f (x) = x⁵ − 20x³ + 240x , then f (x) satisfies which of the following:
a. It is monotonically decreasing everywhere
b. It is monotonically decreasing only in (0,∞)
c. It is monotonically increasing everywhere
d. It is monotonically increasing only in (−∞,0)
Answer : C

Question. The equation of the normal to the curve sin πx/2 at (1, 1) is:
a. y = 1
b. x = 1
c. y = x
d.y - 1 = -2/π (x 1)
Answer : B

Question. The point (s) on the curve y³ + 3x² = 12y where the tangent is vertical (parallel to y-axis), is are:

Answer : D

Question. The normal of the curve x = a(cos θ + θ sinθ ) y = a(sinθ −θ cosθ ) at any θ is such that:
a. It makes a constant angle with x-axis
b. It passes through the origin
c. It is at a constant distance from the origin
d. None of these
Answer : C

Question. The value of a for which the function (a + 2)x³ − 3ax² + 9ax −1 decrease monotonically throughout for all real x, are:
a. a < −2
b. a > −2
c. −3 < a < 0
d. −∞ < a ≤ −3
Answer : D

Question. 2x³ + 18x² − 96x + 45 = 0 is an increasing function:
a. x ≤ −8, x ≥ 2
b. x < −2, x ≥ 8
c. x ≤ −2, x ≥ 8
d. 0 < x ≤ −2
Answer : A

Question. f (x) = xe^x(1-x) then f (x) is:
a. Increasing on [-1/2, 1]
b. Decreasing on R
c. Increasing on R
d. Decreasing on [-1/2, 1]
Answer : A

Question. For which value of x, the function f (x) = x² − 2x is decreasing:
a. x > 1
b. x > 2
c. x < 1
d. x < 2
Answer : C

Question. The equation of tangent to the curve y = 2 cos x at x = π/4 is:

Answer : C

Question. For the curve by² = (x + a)³ the square of subtangent is proportional to:
a. (Subnormal )^{1 / 2}
b. Subnormal
c. (Subnormal )^{3/ 2}
d. None of these
Answer : A

Question. x tends 0 to π then the given function f (x) = x sin x + cos x + cos² x is:
a. Increasing
b. Decreasing
c. Neither increasing nor decreasing
d. None of these
Answer : B

Question. The maximum and minimum values of x³ − 18x² + 96 in interval (0, 9) are:
a. 160, 0
b. 60, 0
c. 160, 128
d. 120, 28
Answer : C

Question. The minimum value of the function 2 cos 2x − cos 4x in 0 ≤ x ≤ π is:
a. 0
b. 1
c. 3/2
d. – 3
Answer : D

Question. Function f(x) λsinx + 6cos x / 2sin x + 3cos x is monotonic increasing if:
a. λ > 1
b. λ < 1
c. λ < 4
d. λ > 4
Answer : D

Question. The function f (x) = ln(π + x) / In(e + x) is:
a. Increasing on [0, ∞)
b. Decreasing on [0, ∞)
c. Decreasing on [0, π/e) and increasing on [π/e, ∞)
d. Increasing on [0, π/e) and decreasing on [π/e, ∞)
Answer : B

Question. The sum of intercepts on co-ordinate axes made by tangent to the curve √x + √y = √a is:
a. a
b. 2a
c. 2√a
d. None of these
Answer : A

Question. x and y be two variables such that x > 0 and xy = 1. Then the minimum value of x + y is:
a. 2
b. 3
c. 4
d. 0
Answer : A

Question. The real number which most exceeds its cube:
a. 1/2
b. 1/√3
c. 1/√2
d. None of these
Answer : B

Question. The adjacent sides of a rectangle with given perimeter as 100 cm and enclosing maximum area are:
a. 10 cm and 40 cm
b. 20 cm and 30 cm
c. 25 cm and 25 cm
d. 15 cm and 35 cm
Answer : C

Question. The function 4 4 sin x + cos x increase if:
a. 0 < x < π/8
b. π/4 < x < 3π/8
c. 3π/8 < x < 5π/8
d. 5π/8 < x < 3π/4
Answer : B

Question. Maximum value of (1/x) is:
a. (e)e
b. (e)1/e
c. (e)−e
d. (1/e)e
Answer : B

Question. Maximum slope of the curve y = − x³ + 3x² + 9x − 27 is:
a. 0
b. 12
c. 16
d. 32
Answer : B

Question. The function x x is increasing, when:
a. x > 1/e
b. x < 1/e
c. x < 0
d. For all real x
Answer : A

Question. Which of the following is not a decreasing function on the interval [0, π/2)
a. cos x
b. cos2x
c. cos3x
d. cot x
Answer : C

Question. The interval of increase of the function f (x) = x - ex + tan (2π/7) is:
a. (0,∞)
b. (−∞,0)
c. (1,∞)
d. (−∞,−1)
Answer : B, D

Question. The function f (x) = ∫^x_e t(e^t − 1) (t − 1) (t − 2)3 (t −3)⁵ dt has a local minimum at x = ?
a. 0
b. 1
c. 2
d. 3
Answer : B, D

ASSIGNMENTS

1. Rate of change

LEVEL - I

Question 1. A balloon, which always remains spherical, has a variable diameter \(\frac{3}{2}(2x + 1)\). Find the rate of change of its volume with respect to x.
Answer: Radius of the spherical balloon \(r = \frac{1}{2} \times \text{diameter} = \frac{1}{2} \times \frac{3}{2}(2x + 1) = \frac{3}{4}(2x + 1)\)
Volume of the sphere \(V = \frac{4}{3}\pi r^3\)
\(V = \frac{4}{3}\pi \left[ \frac{3}{4}(2x + 1) \right]^3 = \frac{4}{3}\pi \times \frac{27}{64}(2x + 1)^3 = \frac{9}{16}\pi (2x + 1)^3\)
Rate of change of volume with respect to \(x\) is given by \(\frac{dV}{dx}\):
\(\frac{dV}{dx} = \frac{9}{16}\pi \cdot 3(2x + 1)^2 \cdot \frac{d}{dx}(2x + 1)\)
\(\frac{dV}{dx} = \frac{27}{16}\pi (2x + 1)^2 \cdot 2 = \frac{27}{8}\pi (2x + 1)^2\)
In simple words: We first find the radius in terms of \(x\), substitute it into the volume formula, and then differentiate with respect to \(x\).

Exam Tip: Always express the volume or area in terms of the single given independent variable before differentiating.

Question 2. The side of a square sheet is increasing at the rate of 4 cm per minute. At what rate is the area increasing when the side is 8 cm long?
Answer: Let \(x\) be the side length of the square and \(A\) be its area.
Given that \(\frac{dx}{dt} = 4\) cm/min and \(x = 8\) cm.
Area of the square \(A = x^2\)
Differentiating both sides with respect to time \(t\):
\(\frac{dA}{dt} = 2x \frac{dx}{dt}\)
Substitute \(x = 8\) and \(\frac{dx}{dt} = 4\):
\(\frac{dA}{dt} = 2(8)(4) = 64 \text{ cm}^2/\text{min}\)
In simple words: We use the area formula \(A = x^2\) and differentiate it with respect to time to find how fast the area grows.

Exam Tip: Pay close attention to units; area change rate uses square units per unit time.

Question 3. The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of increase of its circumference?
Answer: Let \(r\) be the radius of the circle and \(C\) be its circumference.
Given that \(\frac{dr}{dt} = 0.7\) cm/sec.
Circumference of the circle \(C = 2\pi r\)
Differentiating with respect to time \(t\):
\(\frac{dC}{dt} = 2\pi \frac{dr}{dt}\)
Substitute \(\frac{dr}{dt} = 0.7\):
\(\frac{dC}{dt} = 2\pi (0.7) = 1.4\pi \text{ cm/sec}\)
In simple words: Differentiating the formula for circumference with respect to time gives the rate of its increase.

Exam Tip: Leave the answer in terms of \(\pi\) unless a decimal approximation is specifically requested.

LEVEL - II

Question 1. Find the point on the curve \(y^2 = 8x\) for which the abscissa and ordinate change at the same rate?
Answer: Given curve: \(y^2 = 8x\)
Abscissa is \(x\) and ordinate is \(y\). They change at the same rate, meaning \(\frac{dx}{dt} = \frac{dy}{dt}\).
Differentiating both sides of the curve equation with respect to time \(t\):
\(2y \frac{dy}{dt} = 8 \frac{dx}{dt}\)
Since \(\frac{dx}{dt} = \frac{dy}{dt} \neq 0\), we can cancel it out:
\(2y = 8 \implies y = 4\)
Substitute \(y = 4\) into the curve equation \(y^2 = 8x\):
\((4)^2 = 8x \implies 16 = 8x \implies x = 2\)
Thus, the required point is \((2, 4)\).
In simple words: Equating the rates of change of \(x\) and \(y\) helps us find a relation between coordinates, which we then substitute into the curve equation.

Exam Tip: "Abscissa and ordinate change at the same rate" translates mathematically to \(\frac{dx}{dt} = \frac{dy}{dt}\).

Question 2. A man 2 metre high walks at a uniform speed of 6 km/h away from a lamp post 6 metre high. Find the rate at which the length of his shadow increases. Also find the rate at which the tip of the shadow is moving away from the lamp post.
Answer: Let AB be the lamp post of height 6 m and let CD be the man of height 2 m walking away from the lamp post at speed \(\frac{dx}{dt} = 6\) km/h.
Let \(x\) be the distance of the man from the lamp post at time \(t\), and let \(y\) be the length of his shadow.
Using similar triangles \(\triangle ABE\) and \(\triangle CDE\):
\(\frac{AB}{CD} = \frac{x + y}{y}\)
\(\frac{6}{2} = \frac{x + y}{y} \implies 3y = x + y \implies 2y = x \implies y = \frac{x}{2}\)
Differentiating with respect to time \(t\):
\(\frac{dy}{dt} = \frac{1}{2} \frac{dx}{dt} = \frac{1}{2}(6) = 3 \text{ km/h}\)
Thus, the length of the shadow increases at the rate of 3 km/h.
The distance of the tip of the shadow from the lamp post is \(x + y\).
Rate at which the tip of the shadow is moving away: \(\frac{d}{dt}(x + y) = \frac{dx}{dt} + \frac{dy}{dt} = 6 + 3 = 9 \text{ km/h}\).
In simple words: Similar triangles give a relation between the man's distance and shadow length, which we differentiate with respect to time.

Exam Tip: Draw a clear diagram showing the lamp post, man, distance \(x\), and shadow length \(y\) to apply similar triangles correctly.

Question 3. The length of a rectangle is increasing at the rate of 3.5 cm/sec and its breadth is decreasing at the rate of 3 cm/sec. Find the rate of change of the area of the rectangle when length is 12 cm and breadth is 8 cm.
Answer: Let length be \(l\) and breadth be \(b\).
Given: \(\frac{dl}{dt} = 3.5\) cm/sec, \(\frac{db}{dt} = -3\) cm/sec (negative because breadth is decreasing), \(l = 12\) cm, and \(b = 8\) cm.
Area of the rectangle \(A = l \cdot b\)
Differentiating with respect to time \(t\):
\(\frac{dA}{dt} = l \frac{db}{dt} + b \frac{dl}{dt}\)
Substitute the given values:
\(\frac{dA}{dt} = (12)(-3) + (8)(3.5)\)
\(\frac{dA}{dt} = -36 + 28 = -8 \text{ cm}^2/\text{sec}\)
Thus, the area is decreasing at the rate of \(8 \text{ cm}^2/\text{sec}\).
In simple words: Decreasing rates are treated as negative values when substituting into the derivative formula.

Exam Tip: Always assign a negative sign to rates described as "decreasing".

LEVEL - III

Question 1. A particle moves along the curve \(6y = x^3 + 2\). Find the points on the curve at which y-coordinate is changing 8 times as fast as the x-coordinate.
Answer: Given curve: \(6y = x^3 + 2\)
Given condition: \(\frac{dy}{dt} = 8 \frac{dx}{dt}\)
Differentiating the curve equation with respect to time \(t\):
\(6 \frac{dy}{dt} = 3x^2 \frac{dx}{dt}\)
Substitute \(\frac{dy}{dt} = 8 \frac{dx}{dt}\):
\(6 \left( 8 \frac{dx}{dt} \right) = 3x^2 \frac{dx}{dt}\)
\(48 \frac{dx}{dt} = 3x^2 \frac{dx}{dt}\)
Since \(\frac{dx}{dt} \neq 0\), dividing both sides by \(\frac{dx}{dt}\):
\(3x^2 = 48 \implies x^2 = 16 \implies x = \pm 4\)
When \(x = 4\):
\(6y = (4)^3 + 2 = 64 + 2 = 66 \implies y = 11\)
When \(x = -4\):
\(6y = (-4)^3 + 2 = -64 + 2 = -62 \implies y = -\frac{31}{3}\)
Thus, the required points are \((4, 11)\) and \(\left( -4, -\frac{31}{3} \right)\).
In simple words: We express the rate condition mathematically, differentiate the curve, substitute, and solve for coordinates.

Exam Tip: Do not forget to check both positive and negative roots when solving quadratic equations like \(x^2 = 16\).

Question 2. Water is leaking from a conical funnel at the rate of \(5 \text{ cm}^3/\text{sec}\). If the radius of the base of the funnel is 10 cm and altitude is 20 cm, find the rate at which water level is dropping when it is 5 cm from top.
Answer: Let radius of the water surface be \(r\) and depth be \(h\) at time \(t\).
Given base radius \(R = 10\) cm and total height \(H = 20\) cm.
By similar triangles, \(\frac{r}{h} = \frac{R}{H} = \frac{10}{20} = \frac{1}{2} \implies r = \frac{h}{2}\)
Volume of water in the cone \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi}{12} h^3\)
Given \(\frac{dV}{dt} = -5 \text{ cm}^3/\text{sec}\).
Differentiating with respect to \(t\):
\(\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}\)
When water level is 5 cm from top, the depth from vertex is \(h = 20 - 5 = 15\) cm.
Substitute \(\frac{dV}{dt} = -5\) and \(h = 15\):
\(-5 = \frac{\pi}{4} (15)^2 \frac{dh}{dt}\)
\(-5 = \frac{225\pi}{4} \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{-20}{225\pi} = -\frac{4}{45\pi} \text{ cm/sec}\)
Thus, the water level is dropping at the rate of \(\frac{4}{45\pi} \text{ cm/sec}\).
In simple words: Use similar triangles to eliminate radius \(r\) so volume is expressed in terms of height \(h\) alone before differentiating.

Exam Tip: Be careful with \(h\); depth is measured from the vertex of the cone, not from the top rim.

Question 3. From a cylinder drum containing petrol and kept vertical, the petrol is leaking at the rate of 10 ml/sec. If the radius of the drum is 10 cm and height 50 cm, find the rate at which the level of the petrol is changing when petrol level is 20 cm.
Answer: Let radius of the cylindrical drum be \(r = 10\) cm (constant) and let height of the petrol be \(h\) at time \(t\).
Given \(\frac{dV}{dt} = -10 \text{ ml/sec} = -10 \text{ cm}^3/\text{sec}\).
Volume of the cylindrical petrol column \(V = \pi r^2 h = \pi (10)^2 h = 100\pi h\)
Differentiating with respect to time \(t\):
\(\frac{dV}{dt} = 100\pi \frac{dh}{dt}\)
Substitute \(\frac{dV}{dt} = -10\):
\(-10 = 100\pi \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{-10}{100\pi} = -\frac{1}{10\pi} \text{ cm/sec}\)
Thus, the petrol level is dropping at the rate of \(\frac{1}{10\pi} \text{ cm/sec}\).
In simple words: Since the radius of a cylinder is constant, only the height changes as petrol leaks out.

Exam Tip: Ensure all units are consistent (convert milliliters to cubic centimeters if necessary, though 1 ml = 1 cm³).

2. Increasing & decreasing functions

LEVEL I

Question 1. Show that \(f(x) = x^3 - 6x^2 + 18x + 5\) is an increasing function for all \(x \in R\).
Answer: Given function: \(f(x) = x^3 - 6x^2 + 18x + 5\)
Finding the first derivative \(f'(x)\):
\(f'(x) = 3x^2 - 12x + 18\)
Rewriting the quadratic expression by completing the square:
\(f'(x) = 3(x^2 - 4x + 6) = 3[(x - 2)^2 - 4 + 6] = 3[(x - 2)^2 + 2]\)
Since \((x - 2)^2 \geq 0\) for all \(x \in R\), we have:
\((x - 2)^2 + 2 > 0\)
Therefore, \(f'(x) = 3[(x - 2)^2 + 2] > 0\) for all \(x \in R\).
Hence, \(f(x)\) is an increasing function for all \(x \in R\).
In simple words: If the derivative is always positive for every real number, the function is strictly increasing everywhere.

Exam Tip: Completing the square is a standard technique to prove positivity of quadratic expressions.

Question 2. Show that the function \(x^2 - x + 1\) is neither increasing nor decreasing on \((0, 1)\).
Answer: Let \(f(x) = x^2 - x + 1\)
Differentiating with respect to \(x\):
\(f'(x) = 2x - 1\)
For \(x \in (0, 1)\), let us test values:
- If \(x = 0.3 \in (0, 1)\), then \(f'(0.3) = 2(0.3) - 1 = 0.6 - 1 = -0.4 < 0\) (decreasing)
- If \(x = 0.7 \in (0, 1)\), then \(f'(0.7) = 2(0.7) - 1 = 1.4 - 1 = 0.4 > 0\) (increasing)
Since \(f'(x) < 0\) for some values in \((0, 1)\) and \(f'(x) > 0\) for other values in \((0, 1)\), the function is neither increasing nor decreasing on \((0, 1)\).
In simple words: A function is neither if its derivative changes signs within the given interval.

Exam Tip: Pick test points inside the open interval to demonstrate sign changes for the derivative.

Question 3. Find the intervals in which the function \(f(x) = \sin x - \cos x, 0 < x < 2\pi\) is increasing or decreasing.
Answer: Given \(f(x) = \sin x - \cos x\)
Derivative: \(f'(x) = \cos x + \sin x\)
For critical points, set \(f'(x) = 0\):
\(\cos x + \sin x = 0 \implies \sin x = -\cos x \implies \tan x = -1\)
In the interval \((0, 2\pi)\), \(\tan x = -1\) when \(x = \frac{3\pi}{4}, \frac{7\pi}{4}\).
These points divide the interval \((0, 2\pi)\) into intervals: \(\left(0, \frac{3\pi}{4}\right)\), \(\left(\frac{3\pi}{4}, \frac{7\pi}{4}\right)\), and \(\left(\frac{7\pi}{4}, 2\pi\right)\).
- In \(\left(0, \frac{3\pi}{4}\right)\), let \(x = \frac{\pi}{2}\): \(f'\left(\frac{\pi}{2}\right) = \cos\frac{\pi}{2} + \sin\frac{\pi}{2} = 0 + 1 = 1 > 0\) (Increasing)
- In \(\left(\frac{3\pi}{4}, \frac{7\pi}{4}\right)\), let \(x = \pi\): \(f'(\pi) = \cos\pi + \sin\pi = -1 + 0 = -1 < 0\) (Decreasing)
- In \(\left(\frac{7\pi}{4}, 2\pi\right)\), let \(x = 2\pi\): \(f'(2\pi) = \cos(2\pi) + \sin(2\pi) = 1 + 0 = 1 > 0\) (Increasing)
Thus, \(f(x)\) is increasing in \(\left(0, \frac{3\pi}{4}\right) \cup \left(\frac{7\pi}{4}, 2\pi\right)\) and decreasing in \(\left(\frac{3\pi}{4}, \frac{7\pi}{4}\right)\).
In simple words: We find critical points using trigonometric equations, test signs in each sub-interval, and determine monotonicity.

Exam Tip: Be precise with trigonometric quadrant signs when testing values.

LEVEL II

Question 1. Indicate the interval in which the function \(f(x) = \cos x, 0 \leq x \leq 2\pi\) is decreasing.
Answer: Given \(f(x) = \cos x\)
Derivative: \(f'(x) = -\sin x\)
For \(f(x)\) to be decreasing, we must have \(f'(x) < 0\):
\(-\sin x < 0 \implies \sin x > 0\)
In the interval \([0, 2\pi]\), \(\sin x > 0\) in the first and second quadrants, which corresponds to the interval \((0, \pi)\).
Thus, the function is decreasing in \((0, \pi)\).
In simple words: Cosine is decreasing wherever its derivative \(-\sin x\) is negative.

Question 2. Show that the function \(f(x) = \frac{\sin x}{x}\) is strictly decreasing on \((0, \pi/2)\).
Answer: Given \(f(x) = \frac{\sin x}{x}\)
Using the quotient rule for differentiation:
\(f'(x) = \frac{x \cos x - \sin x}{x^2}\)
For \(x \in \left(0, \frac{\pi}{2}\right)\), we know that \(x > \sin x\) and \(\cos x < 1\).
More precisely, dividing by \(\cos x\), we have \(\tan x > x \implies x - \tan x < 0 \implies x \cos x - \sin x < 0\).
Since \(x^2 > 0\) and \(x \cos x - \sin x < 0\) for all \(x \in \left(0, \frac{\pi}{2}\right)\), we get \(f'(x) < 0\).
Therefore, \(f(x)\) is strictly decreasing on \(\left(0, \frac{\pi}{2}\right)\).
In simple words: The quotient rule derivative is negative on this interval because \(\tan x > x\).

Exam Tip: Use the well-known inequality \(\tan x > x\) for \(x \in (0, \pi/2)\) to establish the sign of the numerator.

Question 3. Find the intervals in which the function \(f(x) = \frac{\log x}{x}\) increasing or decreasing.
Answer: Domain of \(f(x)\) is \(x > 0\).
Using the quotient rule:
\(f'(x) = \frac{x \cdot \frac{1}{x} - (\log x)(1)}{x^2} = \frac{1 - \log x}{x^2}\)
For critical points, set \(f'(x) = 0\):
\(1 - \log x = 0 \implies \log x = 1 \implies x = e\)
Since \(x^2 > 0\) for all \(x > 0\), the sign of \(f'(x)\) depends entirely on \(1 - \log x\):
- If \(0 < x < e\), then \(\log x < 1 \implies 1 - \log x > 0 \implies f'(x) > 0\) (Increasing)
- If \(x > e\), then \(\log x > 1 \implies 1 - \log x < 0 \implies f'(x) < 0\) (Decreasing)
Thus, \(f(x)\) is increasing on \((0, e)\) and decreasing on \((e, \infty)\).
In simple words: We find the derivative using the quotient rule and analyze where the numerator is positive or negative.

Exam Tip: Always state the domain \(x > 0\) for logarithmic functions before proceeding.

LEVEL III

Question 1. Find the interval of monotonicity of the function \(f(x) = 2x^2 - \log x\), \(x \neq 0\).
Answer: Domain of the function is \(x > 0\) (due to \(\log x\)).
Derivative: \(f'(x) = 4x - \frac{1}{x} = \frac{4x^2 - 1}{x} = \frac{(2x - 1)(2x + 1)}{x}\)
For critical points, set \(f'(x) = 0\):
\(4x^2 - 1 = 0 \implies x^2 = \frac{1}{4} \implies x = \frac{1}{2}\) (since \(x > 0\)).
Thus, the domain \((0, \infty)\) is divided into intervals \(\left(0, \frac{1}{2}\right)\) and \(\left(\frac{1}{2}, \infty\right)\):
- In \(\left(0, \frac{1}{2}\right)\), let \(x = \frac{1}{4}\): \(f'\left(\frac{1}{4}\right) = \frac{4(1/16) - 1}{1/4} = \frac{1/4 - 1}{1/4} < 0\) (Decreasing)
- In \(\left(\frac{1}{2}, \infty\right)\), let \(x = 1\): \(f'(1) = \frac{4(1) - 1}{1} = 3 > 0\) (Increasing)
Thus, \(f(x)\) is decreasing on \(\left(0, \frac{1}{2}\right)\) and increasing on \(\left(\frac{1}{2}, \infty\right)\).
In simple words: Restrict the domain to positive numbers, find the critical point at \(x = 1/2\), and test intervals.

Exam Tip: Negative values of \(x\) are excluded automatically because \(\log x\) is undefined for \(x \leq 0\).

Question 2. Prove that the function \(y = \frac{4\sin \theta}{2 + \cos \theta} - \theta\) is an increasing function of \(\theta\) in \([0, \pi/2]\) [CBSE 2011]
Answer: Given function: \(y = f(\theta) = \frac{4\sin \theta}{2 + \cos \theta} - \theta\)
Differentiating with respect to \(\theta\) using the quotient rule on the first term:
\(f'(\theta) = \frac{(2 + \cos \theta)(4\cos \theta) - (4\sin \theta)(-\sin \theta)}{(2 + \cos \theta)^2} - 1\)
\(f'(\theta) = \frac{8\cos \theta + 4\cos^2 \theta + 4\sin^2 \theta}{(2 + \cos \theta)^2} - 1 = \frac{8\cos \theta + 4(\cos^2 \theta + \sin^2 \theta)}{(2 + \cos \theta)^2} - 1\)
\(f'(\theta) = \frac{8\cos \theta + 4}{(2 + \cos \theta)^2} - 1 = \frac{8\cos \theta + 4 - (2 + \cos \theta)^2}{(2 + \cos \theta)^2}\)
\(f'(\theta) = \frac{8\cos \theta + 4 - (4 + 4\cos \theta + \cos^2 \theta)}{(2 + \cos \theta)^2} = \frac{4\cos \theta - \cos^2 \theta}{(2 + \cos \theta)^2} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos \theta)^2}\)
For \(\theta \in \left[0, \frac{\pi}{2}\right]\):
- \(\cos \theta \geq 0\)
- \(4 - \cos \theta > 0\) (since \(\cos \theta \leq 1\))
- \((2 + \cos \theta)^2 > 0\)
Therefore, \(f'(\theta) \geq 0\) for all \(\theta \in \left[0, \frac{\pi}{2}\right]\).
Hence, the function is an increasing function of \(\theta\) in \([0, \pi/2]\).
In simple words: We simplify the derivative into factored form and verify that every factor is non-negative in the given interval.

Exam Tip: Use the trigonometric identity \(\cos^2 \theta + \sin^2 \theta = 1\) to simplify the numerator.

3. Tangents & Normals

LEVEL - I

Question 1. Find the equations of the normals to the curve \(3x^2 - y^2 = 8\) which are parallel to the line \(x + 3y = 4\).
Answer: Given curve: \(3x^2 - y^2 = 8\)
Differentiating implicitly with respect to \(x\):
\(6x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{3x}{y}\)
Slope of the tangent at point \((x_1, y_1)\) is \(m_T = \frac{3x_1}{y_1}\).
Slope of the normal at point \((x_1, y_1)\) is \(m_N = -\frac{1}{m_T} = -\frac{y_1}{3x_1}\).
Given line: \(x + 3y = 4 \implies y = -\frac{1}{3}x + \frac{4}{3}\), so its slope is \(-\frac{1}{3}\).
Since the normal is parallel to the given line, their slopes must be equal:
\(-\frac{y_1}{3x_1} = -\frac{1}{3} \implies y_1 = x_1\)
Since \((x_1, y_1)\) lies on the curve \(3x^2 - y^2 = 8\):
\(3x_1^2 - x_1^2 = 8 \implies 2x_1^2 = 8 \implies x_1^2 = 4 \implies x_1 = \pm 2\)
- If \(x_1 = 2\), then \(y_1 = 2\). Point is \((2, 2)\).
- If \(x_1 = -2\), then \(y_1 = -2\). Point is \((-2, -2)\).
Slope of normal is \(-\frac{1}{3}\) for both points.
Equation of normal at \((2, 2)\):
\(y - 2 = -\frac{1}{3}(x - 2) \implies 3y - 6 = -x + 2 \implies x + 3y = 8\)
Equation of normal at \((-2, -2)\):
\(y - (-2) = -\frac{1}{3}(x - (-2)) \implies 3y + 6 = -x - 2 \implies x + 3y = -8\)
In simple words: Equate the slope of the normal to the slope of the given parallel line to find the points of contact, then write the normal equations.

Exam Tip: Remember that the slope of a normal is the negative reciprocal of the slope of the tangent.

Question 2. Find the point on the curve \(y = x^2\) where the slope of the tangent is equal to the x-coordinate of the point.
Answer: Given curve: \(y = x^2\)
Slope of the tangent: \(\frac{dy}{dx} = 2x\)
Given condition: Slope = x-coordinate
\(2x = x \implies 2x - x = 0 \implies x = 0\)
Substitute \(x = 0\) into the curve equation: \(y = (0)^2 = 0\).
Thus, the required point is \((0, 0)\).
In simple words: Set the derivative equal to \(x\) and solve for the coordinates of the point.

Exam Tip: A straightforward problem where setting \(m = x\) directly yields the solution.

Question 3. At what points on the circle \(x^2 + y^2 - 2x - 4y + 1 = 0\), the tangent is parallel to x axis?
Answer: Given circle: \(x^2 + y^2 - 2x - 4y + 1 = 0\)
Differentiating implicitly with respect to \(x\):
\(2x + 2y \frac{dy}{dx} - 2 - 4 \frac{dy}{dx} = 0\)
\((2y - 4)\frac{dy}{dx} = 2 - 2x \implies \frac{dy}{dx} = \frac{2 - 2x}{2y - 4} = \frac{1 - x}{y - 2}\)
A tangent is parallel to the x-axis if its slope is zero (\(\frac{dy}{dx} = 0\)):
\(\frac{1 - x}{y - 2} = 0 \implies 1 - x = 0 \implies x = 1\)
Substitute \(x = 1\) into the circle equation:
\((1)^2 + y^2 - 2(1) - 4y + 1 = 0\)
\(1 + y^2 - 2 - 4y + 1 = 0 \implies y^2 - 4y = 0 \implies y(y - 4) = 0 \implies y = 0 \text{ or } y = 4\)
Thus, the points are \((1, 0)\) and \((1, 4)\).
In simple words: Tangent parallel to the x-axis means the derivative is zero, which makes the numerator of \(\frac{dy}{dx}\) zero.

Exam Tip: Tangent parallel to the x-axis \(\implies \frac{dy}{dx} = 0\); tangent parallel to the y-axis \(\implies \frac{dx}{dy} = 0\).

LEVEL - II

Question 1. Find the equation of the normal to the curve \(ay^2 = x^3\) at the point \((am^2, am^3)\).
Answer: Given curve: \(ay^2 = x^3\)
Differentiating implicitly with respect to \(x\):
\(2ay \frac{dy}{dx} = 3x^2 \implies \frac{dy}{dx} = \frac{3x^2}{2ay}\)
Slope of the tangent at point \((am^2, am^3)\):
\(m_T = \frac{3(am^2)^2}{2a(am^3)} = \frac{3a^2 m^4}{2a^2 m^3} = \frac{3}{2}m\)
Slope of the normal: \(m_N = -\frac{1}{m_T} = -\frac{2}{3m}\)
Equation of the normal passing through \((am^2, am^3)\) with slope \(-\frac{2}{3m}\):
\(y - am^3 = -\frac{2}{3m}(x - am^2)\)
\(3my - 3am^4 = -2x + 2am^2\)
\(2x + 3my - a(2m^2 + 3m^4) = 0\)
In simple words: Substitute the given parametric coordinates into the derivative to find the slope of the normal, then apply the point-slope form.

Exam Tip: Take care with algebraic simplification when substituting parametric coordinates into derivatives.

Question 2. For the curve \(y = 2x^2 + 3x + 18\), find all the points at which the tangent passes through the origin.
Answer: Given curve: \(y = 2x^2 + 3x + 18\)
Let \((x_1, y_1)\) be a point on the curve where the tangent passes through the origin \((0, 0)\).
Since \((x_1, y_1)\) is on the curve, \(y_1 = 2x_1^2 + 3x_1 + 18\).
Slope of the tangent at \((x_1, y_1)\) is given by \(m = \left. \frac{dy}{dx} \right|_{(x_1, y_1)} = 4x_1 + 3\).
Since the tangent passes through \((0, 0)\) and \((x_1, y_1)\), its slope can also be expressed as:
\(m = \frac{y_1 - 0}{x_1 - 0} = \frac{y_1}{x_1}\)
Equating the two expressions for slope:
\(4x_1 + 3 = \frac{2x_1^2 + 3x_1 + 18}{x_1}\)
\(x_1(4x_1 + 3) = 2x_1^2 + 3x_1 + 18\)
\(4x_1^2 + 3x_1 = 2x_1^2 + 3x_1 + 18\)
\(2x_1^2 = 18 \implies x_1^2 = 9 \implies x_1 = \pm 3\)
- If \(x_1 = 3\): \(y_1 = 2(3)^2 + 3(3) + 18 = 18 + 9 + 18 = 45\). Point: \((3, 45)\).
- If \(x_1 = -3\): \(y_1 = 2(-3)^2 + 3(-3) + 18 = 18 - 9 + 18 = 27\). Point: \((-3, 27)\).
Thus, the points are \((3, 45)\) and \((-3, 27)\).
In simple words: Equate the derivative slope to the two-point slope formula using the origin to solve for the point of contact.

Exam Tip: Whenever a tangent passes through a specific external point, use the two-point slope formula combined with the derivative.

Question 3. Find the equation of the normals to the curve \(y = x^3 + 2x + 6\) which are parallel to the line \(x + 14y + 4 = 0\).
Answer: Given curve: \(y = x^3 + 2x + 6\)
Slope of the tangent: \(m_T = \frac{dy}{dx} = 3x^2 + 2\)
Slope of the normal: \(m_N = -\frac{1}{3x^2 + 2}\)
Given line: \(x + 14y + 4 = 0 \implies y = -\frac{1}{14}x - \frac{4}{14}\), so its slope is \(-\frac{1}{14}\).
Since the normal is parallel to the given line, their slopes are equal:
\(-\frac{1}{3x^2 + 2} = -\frac{1}{14} \implies 3x^2 + 2 = 14 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2\)
- If \(x = 2\), then \(y = (2)^3 + 2(2) + 6 = 8 + 4 + 6 = 18\). Point: \((2, 18)\).
- If \(x = -2\), then \(y = (-2)^3 + 2(-2) + 6 = -8 - 4 + 6 = -6\). Point: \((-2, -6)\).
Slope of normal is \(-\frac{1}{14}\).
Equation of normal at \((2, 18)\):
\(y - 18 = -\frac{1}{14}(x - 2) \implies 14y - 252 = -x + 2 \implies x + 14y = 254\)
Equation of normal at \((-2, -6)\):
\(y - (-6) = -\frac{1}{14}(x - (-2)) \implies 14y + 84 = -x - 2 \implies x + 14y = -86\)
In simple words: Find the normal's slope from the curve, equate it to the given line's slope, find points, and write the equations.

Exam Tip: Be careful to use the slope of the normal, not the tangent, when equating to the parallel line's slope.

Question 4. Show that the equation of tangent at \((x_1, y_1)\) to the parabola \(yy_1 = 2a(x + x_1)\). [CBSE 2012 Comptt.]
Answer: Standard equation of parabola is \(y^2 = 4ax\).
Differentiating implicitly with respect to \(x\):
\(2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y}\)
Slope of the tangent at \((x_1, y_1)\) is \(m = \frac{2a}{y_1}\).
Equation of the tangent using point-slope form:
\(y - y_1 = \frac{2a}{y_1}(x - x_1)\)
\(y_1(y - y_1) = 2a(x - x_1)\)
\(y_1 y - y_1^2 = 2ax - 2ax_1\)
Since \((x_1, y_1)\) lies on the parabola, \(y_1^2 = 4ax_1\):
\(y_1 y - 4ax_1 = 2ax - 2ax_1\)
\(y_1 y = 2ax - 2ax_1 + 4ax_1 = 2ax + 2ax_1 = 2a(x + x_1)\)
Hence proved: \(yy_1 = 2a(x + x_1)\).
In simple words: Substitute the derivative into the point-slope form and use the point's equation on the curve to simplify.

Exam Tip: This is a standard textbook derivation; memorize the substitution step \(y_1^2 = 4ax_1\).

LEVEL - III

Question 1. Find the equation of the tangent line to the curve \(y = \sqrt{5x - 3} - 2\) which is parallel to the line \(4x - 2y + 3 = 0\).
Answer: Given curve: \(y = \sqrt{5x - 3} - 2\)
Slope of the tangent: \(\frac{dy}{dx} = \frac{1}{2\sqrt{5x - 3}} \cdot 5 = \frac{5}{2\sqrt{5x - 3}}\)
Given line: \(4x - 2y + 3 = 0 \implies 2y = 4x + 3 \implies y = 2x + \frac{3}{2}\), so its slope is 2.
Since the tangent is parallel to the line, their slopes are equal:
\(\frac{5}{2\sqrt{5x - 3}} = 2 \implies 4\sqrt{5x - 3} = 5 \implies \sqrt{5x - 3} = \frac{5}{4}\)
Squaring both sides:
\(5x - 3 = \frac{25}{16} \implies 5x = 3 + \frac{25}{16} = \frac{73}{16} \implies x = \frac{73}{80}\)
Substitute \(x = \frac{73}{80}\) into the curve equation to find \(y\):
\(y = \sqrt{5\left(\frac{73}{80}\right) - 3} - 2 = \sqrt{\frac{73}{16} - 3} - 2 = \sqrt{\frac{25}{16}} - 2 = \frac{5}{4} - 2 = -\frac{3}{4}\)
Now, using point \(\left(\frac{73}{80}, -\frac{3}{4}\right)\) and slope \(m = 2\):
\(y - \left(-\frac{3}{4}\right) = 2\left(x - \frac{73}{80}\right)\)
\(y + \frac{3}{4} = 2x - \frac{73}{40}\)
Multiplying by 40: \(40y + 30 = 80x - 73 \implies 80x - 40y - 103 = 0\)
In simple words: Equate the derivative to the line's slope to find the point, then write the tangent equation.

Exam Tip: Take care with fractions when squaring radical equations.

Question 2. Show that the curves \(x^2 + y^2 - 2x = 0\) and \(x^2 + y^2 - 2y = 0\) cut orthogonally at the point \((0, 0)\).
Answer: Given curves:
Curve 1: \(x^2 + y^2 - 2x = 0\)
Curve 2: \(x^2 + y^2 - 2y = 0\)
Differentiating Curve 1 implicitly with respect to \(x\):
\(2x + 2y \frac{dy}{dx} - 2 = 0 \implies \frac{dy}{dx} = \frac{2 - 2x}{2y} = \frac{1 - x}{y}\)
At point \((0, 0)\), the slope of the tangent to Curve 1 is \(m_1 = \left. \frac{1 - x}{y} \right|_{(0,0)}\), which is undefined (vertical tangent, slope \(\infty\)).
Differentiating Curve 2 implicitly with respect to \(x\):
\(2x + 2y \frac{dy}{dx} - 2 \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{2x}{2 - 2y} = \frac{x}{1 - y}\)
At point \((0, 0)\), the slope of the tangent to Curve 2 is \(m_2 = \left. \frac{x}{1 - y} \right|_{(0,0)} = \frac{0}{1 - 0} = 0\) (horizontal tangent).
Since one tangent is vertical (\(m_1 = \infty\)) and the other is horizontal (\(m_2 = 0\)), they are perpendicular to each other.
Thus, the two curves intersect orthogonally at \((0, 0)\).
In simple words: Two curves are orthogonal if the product of their tangent slopes at the intersection point is \(-1\), or if one slope is zero and the other is undefined.

Exam Tip: Orthogonality means the angle between tangents at the intersection point is 90 degrees.

Question 3. Find the condition for the curves \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) and \(xy = c^2\) to intersect orthogonally.
Answer: Let \((x_1, y_1)\) be the point of intersection of the two curves.
For curve 1 (\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)), differentiating implicitly:
\(\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{b^2 x}{a^2 y}\)
Slope of tangent at \((x_1, y_1)\) is \(m_1 = \frac{b^2 x_1}{a^2 y_1}\).
For curve 2 (\(xy = c^2\)), differentiating implicitly:
\(x \frac{dy}{dx} + y = 0 \implies \frac{dy}{dx} = -\frac{y}{x}\)
Slope of tangent at \((x_1, y_1)\) is \(m_2 = -\frac{y_1}{x_1}\).
Since the curves intersect orthogonally, \(m_1 \cdot m_2 = -1\):
\(\left(\frac{b^2 x_1}{a^2 y_1}\right) \left(-\frac{y_1}{x_1}\right) = -1\)
\(-\frac{b^2}{a^2} = -1 \implies \frac{b^2}{a^2} = 1 \implies a^2 = b^2 \text{ or } \frac{1}{a^2} - \frac{1}{b^2} = 0\)
Thus, the required condition is \(\frac{1}{a^2} - \frac{1}{b^2} = 0\) or \(a^2 = b^2\).
In simple words: Multiply the derivatives of both curves at the intersection point and set their product equal to \(-1\).

Exam Tip: The condition simplifies nicely when coordinates \(x_1\) and \(y_1\) cancel out.

4. Approximations

LEVEL - I

Question Q.1. Evaluate \(\sqrt{25.3}\).
Answer: Let \(y = \sqrt{x}\). Let \(x = 25\) and \(\Delta x = 0.3\).
Then \(y + \Delta y = \sqrt{x + \Delta x} = \sqrt{25.3}\).
We know \(\Delta y \approx \frac{dy}{dx} \Delta x\).
\(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\)
At \(x = 25\), \(\frac{dy}{dx} = \frac{1}{2\sqrt{25}} = \frac{1}{10} = 0.1\).
\(\Delta y \approx 0.1 \times 0.3 = 0.03\).
\(\sqrt{25.3} = \sqrt{x} + \Delta y \approx 5 + 0.03 = 5.03\).
In simple words: Use differentials to approximate small changes: \(f(x + \Delta x) \approx f(x) + f'(x)\Delta x\).

Exam Tip: Choose a nearby perfect square or cube as the base value \(x\).

Question Q.2. Use differentials to approximate the cube root of 66.
Answer: Let \(y = x^{1/3}\). Let \(x = 64\) and \(\Delta x = 2\).
Then \(y + \Delta y = (66)^{1/3}\).
\(\frac{dy}{dx} = \frac{1}{3x^{2/3}}\)
At \(x = 64\), \(\frac{dy}{dx} = \frac{1}{3(64)^{2/3}} = \frac{1}{3(4)^2} = \frac{1}{48}\).
\(\Delta y \approx \frac{dy}{dx} \Delta x = \frac{1}{48} \times 2 = \frac{2}{48} = \frac{1}{24} \approx 0.0417\).
\((66)^{1/3} \approx 4 + 0.0417 = 4.0417\).
In simple words: Base value 64 has an exact cube root of 4, making it ideal for differential approximation.

Exam Tip: Always round off your decimal answers to the requested number of places.

Question Q.3. Evaluate \(\sqrt{0.082}\).
Answer: Let \(y = \sqrt{x}\). Let \(x = 0.09\) and \(\Delta x = -0.008\).
Then \(y + \Delta y = \sqrt{0.082}\).
\(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\)
At \(x = 0.09\), \(\frac{dy}{dx} = \frac{1}{2\sqrt{0.09}} = \frac{1}{2(0.3)} = \frac{1}{0.6} = \frac{5}{3}\).
\(\Delta y \approx \frac{5}{3} \times (-0.008) = -\frac{0.04}{3} \approx -0.0133\).
\(\sqrt{0.082} \approx 0.3 - 0.0133 = 0.2867\).
In simple words: Choose \(0.09\) as the base because its square root is \(0.3\).

Exam Tip: Keep track of decimal places when handling small numbers like \(0.008\).

Question Q.4. Evaluate \(\sqrt{49.5}\). [CBSE 2012]
Answer: Let \(y = \sqrt{x}\). Let \(x = 49\) and \(\Delta x = 0.5\).
Then \(y + \Delta y = \sqrt{49.5}\).
\(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\)
At \(x = 49\), \(\frac{dy}{dx} = \frac{1}{2\sqrt{49}} = \frac{1}{14}\).
\(\Delta y \approx \frac{1}{14} \times 0.5 = \frac{0.5}{14} = \frac{1}{28} \approx 0.0357\).
\(\sqrt{49.5} \approx 7 + 0.0357 = 7.0357\).
In simple words: The nearest square is 49, with a square root of 7.

Exam Tip: Division by 14 should be carefully computed to at least 4 decimal places before rounding.

LEVEL - II

Question 1. If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its surface area.
Answer: Let \(r\) be the radius of the sphere and \(S\) be its surface area.
Given \(r = 9\) cm and \(dr = \Delta r = 0.03\) cm.
Surface area of the sphere \(S = 4\pi r^2\)
Differentiating with respect to \(r\):
\(\frac{dS}{dr} = 8\pi r\)
Approximate error in surface area \(dS\) is given by:
\(dS = \frac{dS}{dr} \cdot \Delta r = 8\pi r \cdot \Delta r\)
Substitute \(r = 9\) and \(\Delta r = 0.03\):
\(dS = 8\pi (9)(0.03) = 72\pi (0.03) = 2.16\pi \text{ cm}^2\)
Thus, the approximate error in calculating the surface area is \(2.16\pi \text{ cm}^2\).
In simple words: Use the differential \(dS = S'(r)\Delta r\) to calculate the error in area resulting from an error in radius.

Exam Tip: Distinguish between relative error, percentage error, and absolute error.

5. Maxima & Minima

LEVEL I

Question 1. Find the maximum and minimum value of the function \(f(x) = 3 - 2\sin x\).
Answer: Given function: \(f(x) = 3 - 2\sin x\)
We know that for any real number \(x\), the range of sine is:
\(-1 \leq \sin x \leq 1\)
Multiplying by \(-2\) reverses the inequality:
\(-2 \leq -2\sin x \leq 2\)
Adding 3 throughout:
\(3 - 2 \leq 3 - 2\sin x \leq 3 + 2\)
\(1 \leq f(x) \leq 5\)
Thus, the maximum value of \(f(x)\) is 5 and the minimum value is 1.
In simple words: Using the known bounds of the sine function directly gives the maximum and minimum values without needing derivatives.

Exam Tip: This method is much faster than calculus for bounded trigonometric functions.

Question 2. Show that the function \(f(x) = x^3 + x^2 + x + 1\) has neither a maximum value nor a minimum value.
Answer: Given function: \(f(x) = x^3 + x^2 + x + 1\)
Finding the first derivative:
\(f'(x) = 3x^2 + 2x + 1\)
Checking the discriminant of the quadratic expression \(3x^2 + 2x + 1\):
\(\Delta = b^2 - 4ac = (2)^2 - 4(3)(1) = 4 - 12 = -8 < 0\)
Since the discriminant is negative and the leading coefficient is positive (\(3 > 0\)), the quadratic expression \(3x^2 + 2x + 1 > 0\) for all \(x \in R\).
Therefore, \(f'(x) \neq 0\) for any real number \(x\), meaning there are no critical points.
Hence, the function has neither a maximum nor a minimum value.
In simple words: If the derivative never equals zero, the function has no local extrema.

Exam Tip: Always check the discriminant when analyzing the roots of quadratic derivatives.

Question 3. Find two positive numbers whose sum is 24 and whose product is maximum.
Answer: Let the two positive numbers be \(x\) and \(24 - x\).
Their product \(P(x) = x(24 - x) = 24x - x^2\).
Differentiating with respect to \(x\):
\(P'(x) = 24 - 2x\)
For critical points, set \(P'(x) = 0\):
\(24 - 2x = 0 \implies x = 12\)
Second derivative: \(P''(x) = -2 < 0\).
Since \(P''(12) < 0\), the product is maximum when \(x = 12\).
The two numbers are 12 and \(24 - 12 = 12\).
In simple words: Express the product as a function of one variable, find critical points, and confirm the maximum using the second derivative test.

Exam Tip: For problems asking for maximum product with a fixed sum, the numbers are always equal.

LEVEL II

Question 1. Prove that the area of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.
Answer: Let \(h\) be the fixed hypotenuse and let \(\theta\) be one of the acute angles of the right-angled triangle.
The base \(x = h \cos \theta\) and the perpendicular \(y = h \sin \theta\).
Area of the triangle \(A = \frac{1}{2} x y = \frac{1}{2} (h \cos \theta)(h \sin \theta) = \frac{h^2}{4} (2 \sin \theta \cos \theta) = \frac{h^2}{4} \sin 2\theta\)
Since \(h\) is constant, area \(A\) is maximum when \(\sin 2\theta\) is maximum.
\(\sin 2\theta = 1 \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}\)
When \(\theta = \frac{\pi}{4}\), base \(x = h \cos\frac{\pi}{4} = \frac{h}{\sqrt{2}}\) and perpendicular \(y = h \sin\frac{\pi}{4} = \frac{h}{\sqrt{2}}\).
Since base equals perpendicular (\(x = y\)), the right-angled triangle is isosceles.
In simple words: Using trigonometric parameters makes optimization simpler when dealing with right-angled triangles.

Question 2. A piece of wire 28 units long is cut into two pieces. One piece is bent into the shape of a circle and other into the shape of a square. How should the wire be cut so that the combined area of the two figures is as small as possible?
Answer: Let length \(x\) of the wire be used for the circle, and the remaining \(28 - x\) for the square.
Circumference of the circle = \(x \implies 2\pi r = x \implies r = \frac{x}{2\pi}\).
Area of the circle \(A_1 = \pi r^2 = \pi \left(\frac{x}{2\pi}\right)^2 = \frac{x^2}{4\pi}\).
Perimeter of the square = \(28 - x \implies \text{side } s = \frac{28 - x}{4}\).
Area of the square \(A_2 = s^2 = \left(\frac{28 - x}{4}\right)^2 = \frac{(28 - x)^2}{16}\).
Total combined area \(A(x) = \frac{x^2}{4\pi} + \frac{(28 - x)^2}{16}\).
Differentiating with respect to \(x\):
\(A'(x) = \frac{2x}{4\pi} + \frac{2(28 - x)(-1)}{16} = \frac{x}{2\pi} - \frac{28 - x}{8}\)
For critical points, set \(A'(x) = 0\):
\(\frac{x}{2\pi} = \frac{28 - x}{8} \implies \frac{x}{\pi} = \frac{28 - x}{4} \implies 4x = 28\pi - \pi x \implies x(4 + \pi) = 28\pi \implies x = \frac{28\pi}{4 + \pi}\)
Second derivative: \(A''(x) = \frac{1}{2\pi} + \frac{1}{8} > 0\).
Since \(A''(x) > 0\), the total area is minimum when \(x = \frac{28\pi}{4 + \pi}\).
In simple words: Formulate total area as a function of the cut length \(x\), differentiate, and find the minimum using the second derivative test.

Exam Tip: A positive second derivative confirms a minimum area.

Question 3. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Answer: Let the rectangular part have width \(2x\) and height \(y\). The semicircle on top has radius \(x\).
Total perimeter \(P = 2y + 2x + \pi x = 10 \implies 2y = 10 - 2x - \pi x \implies y = 5 - x - \frac{\pi}{2}x\).
Total area \(A = \text{Area of rectangle} + \text{Area of semicircle}\)
\(A = (2x)(y) + \frac{1}{2}\pi x^2 = 2x\left(5 - x - \frac{\pi}{2}x\right) + \frac{1}{2}\pi x^2\)
\(A = 10x - 2x^2 - \pi x^2 + \frac{1}{2}\pi x^2 = 10x - 2x^2 - \frac{1}{2}\pi x^2\)
Differentiating with respect to \(x\):
\(A'(x) = 10 - 4x - \pi x\)
For critical points, set \(A'(x) = 0\):
\(10 - 4x - \pi x = 0 \implies x(4 + \pi) = 10 \implies x = \frac{10}{4 + \pi}\)
Second derivative: \(A''(x) = -4 - \pi < 0\), confirming maximum area.
Width of the window \(2x = \frac{20}{4 + \pi}\) m.
Height \(y = 5 - \frac{10}{4 + \pi} - \frac{5\pi}{4 + \pi} = \frac{20 + 5\pi - 10 - 5\pi}{4 + \pi} = \frac{10}{4 + \pi}\) m.
In simple words: Express total area in terms of \(x\), find the critical point by setting the first derivative to zero, and verify with the second derivative.

Exam Tip: Pay close attention to perimeter components; do not include the diameter dividing the rectangle and semicircle.

LEVEL III

Question 1. Find the area of the greatest isosceles triangle that can be inscribed in a given ellipse having its vertex coincident with one extremity of major axis.
Answer: Let the ellipse be \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
Let the vertex of the isosceles triangle be at the extremity of the major axis, i.e., \((a, 0)\).
Let the other two vertices on the ellipse be \((-x_1, y_1)\) and \((-x_1, -y_1)\).
The base of the isosceles triangle is \(2y_1\) and its height is \(a + x_1\).
Area of the triangle \(A = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2}(2y_1)(a + x_1) = y_1(a + x_1)\)
Since \((-x_1, y_1)\) lies on the ellipse, \(\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1 \implies y_1^2 = \frac{b^2}{a^2}(a^2 - x_1^2) \implies y_1 = \frac{b}{a}\sqrt{(a - x_1)(a + x_1)}\)
Substitute \(y_1\) into the area formula:
\(A = \frac{b}{a}(a + x_1)^{3/2}(a - x_1)^{1/2}\)
Differentiating with respect to \(x_1\) and setting \(\frac{dA}{dx_1} = 0\) gives \(x_1 = \frac{a}{2}\).
Substituting \(x_1 = \frac{a}{2}\) back into the area formula yields maximum area \(A = \frac{3\sqrt{3}}{4} ab\).
In simple words: Express the triangle's area in terms of coordinates on the ellipse, differentiate, and solve for the maximum area.

Exam Tip: Product rule differentiation is required when optimizing the combined radical terms.

Question 2. An open box with a square base is to be made out of a given quantity of cardboard of area \(c^2\) square units. Show that the maximum volume of the box is \(\frac{c^3}{6\sqrt{3}}\) cubic units. [CBSE 2012 Comptt.]
Answer: Let the side of the square base be \(x\) and the height of the box be \(y\).
Total surface area of the open box \(S = x^2 + 4xy = c^2\) (given constant).
From this, express height \(y\) in terms of \(x\):
\(4xy = c^2 - x^2 \implies y = \frac{c^2 - x^2}{4x}\)
Volume of the box \(V = x^2 y = x^2 \left(\frac{c^2 - x^2}{4x}\right) = \frac{x(c^2 - x^2)}{4} = \frac{c^2 x - x^3}{4}\)
Differentiating with respect to \(x\):
\(\frac{dV}{dx} = \frac{c^2 - 3x^2}{4}\)
For critical points, set \(\frac{dV}{dx} = 0\):
\(c^2 - 3x^2 = 0 \implies 3x^2 = c^2 \implies x = \frac{c}{\sqrt{3}}\)
Second derivative: \(\frac{d^2V}{dx^2} = -\frac{6x}{4} < 0\), confirming maximum volume.
Substitute \(x = \frac{c}{\sqrt{3}}\) into the volume formula:
\(V = \frac{c^2\left(\frac{c}{\sqrt{3}}\right) - \left(\frac{c}{\sqrt{3}}\right)^3}{4} = \frac{\frac{c^3}{\sqrt{3}} - \frac{c^3}{3\sqrt{3}}}{4} = \frac{\frac{2c^3}{3\sqrt{3}}}{4} = \frac{2c^3}{12\sqrt{3}} = \frac{c^3}{6\sqrt{3}}\)
Hence proved that the maximum volume is \(\frac{c^3}{6\sqrt{3}}\).
In simple words: Express volume in terms of base side \(x\), use surface area constraint to eliminate height \(y\), and apply the second derivative test.

Exam Tip: This is a classic Board exam optimization question; practice the algebraic simplification of radicals at the end.

Question 3. A window is in the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 m, find the dimensions of the rectangle that will produce the largest area of the window. [CBSE 2011]
Answer: Let the width of the rectangle be \(x\) and height be \(y\).
The equilateral triangle surmounting the rectangle has side length \(x\).
Perimeter of the window \(P = 2y + 3x = 12 \implies 2y = 12 - 3x \implies y = 6 - \frac{3}{2}x\).
Total area \(A = \text{Area of rectangle} + \text{Area of equilateral triangle}\)
\(A = xy + \frac{\sqrt{3}}{4}x^2 = x\left(6 - \frac{3}{2}x\right) + \frac{\sqrt{3}}{4}x^2 = 6x - \frac{3}{2}x^2 + \frac{\sqrt{3}}{4}x^2\)
Differentiating with respect to \(x\):
\(\frac{dA}{dx} = 6 - 3x + \frac{\sqrt{3}}{2}x = 6 - x\left(3 - \frac{\sqrt{3}}{2}\right)\)
For critical points, set \(\frac{dA}{dx} = 0\):
\(6 - x\left(\frac{6 - \sqrt{3}}{2}\right) = 0 \implies x\left(\frac{6 - \sqrt{3}}{2}\right) = 6 \implies x = \frac{12}{6 - \sqrt{3}}\)
Rationalizing the denominator:
\(x = \frac{12(6 + \sqrt{3})}{36 - 3} = \frac{12(6 + \sqrt{3})}{33} = \frac{4(6 + \sqrt{3})}{11} \text{ m}\)
Height \(y = 6 - \frac{3}{2}\left[\frac{4(6 + \sqrt{3})}{11}\right] = 6 - \frac{6(6 + \sqrt{3})}{11} = \frac{66 - 36 - 6\sqrt{3}}{11} = \frac{30 - 6\sqrt{3}}{11} \text{ m}\).
In simple words: Combine the rectangular and triangular areas into a single function of width \(x\), differentiate, and solve for dimensions.

Exam Tip: Remember that an equilateral triangle's area formula is \(\frac{\sqrt{3}}{4}(\text{side})^2\).

Questions for self evaluation

Question 1. Sand is pouring from a pipe at the rate of \(12 \text{ cm}^3/\text{s}\). The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Answer: Given \(\frac{dV}{dt} = 12 \text{ cm}^3/\text{s}\) and \(h = \frac{1}{6}r \implies r = 6h\).
Volume of the cone \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (6h)^2 h = \frac{1}{3}\pi (36h^2) h = 12\pi h^3\)
Differentiating with respect to time \(t\):
\(\frac{dV}{dt} = 12\pi \cdot 3h^2 \frac{dh}{dt} = 36\pi h^2 \frac{dh}{dt}\)
Substitute \(\frac{dV}{dt} = 12\) and \(h = 4\):
\(12 = 36\pi (4)^2 \frac{dh}{dt} \implies 12 = 36\pi (16) \frac{dh}{dt} \implies 12 = 576\pi \frac{dh}{dt}\)
\(\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi} \text{ cm/s}\)
In simple words: Express base radius in terms of height using the given ratio, substitute into the volume formula, and differentiate.

Exam Tip: Substituting the relation \(r = 6h\) early eliminates extra variables before differentiation.

Question 2. The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?
Answer: Let the equal sides be \(a\) and the fixed base be \(b\).
Given \(\frac{da}{dt} = -3\) cm/s.
Height of the isosceles triangle from the vertex to the base: \(h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2}\).
Area of the triangle \(A = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} b \sqrt{a^2 - \frac{b^2}{4}}\).
Differentiating with respect to time \(t\):
\(\frac{dA}{dt} = \frac{b}{2} \cdot \frac{1}{2\sqrt{a^2 - \frac{b^2}{4}}} \cdot 2a \frac{da}{dt} = \frac{ab}{\sqrt{4a^2 - b^2}} \frac{da}{dt}\)
Given that the equal sides are equal to the base (\(a = b\)):
\(\frac{dA}{dt} = \frac{(b)(b)}{\sqrt{4b^2 - b^2}} (-3) = \frac{b^2}{\sqrt{3b^2}} (-3) = \frac{b^2}{\sqrt{3}b} (-3) = -\sqrt{3}b \text{ cm}^2/\text{s}\).
Thus, the area is decreasing at the rate of \(\sqrt{3}b \text{ cm}^2/\text{s}\).
In simple words: Express the area using the height formula for an isosceles triangle, then differentiate with respect to time using the chain rule.

Exam Tip: Keep the base \(b\) as a constant parameter throughout the differentiation.

Question 3. Find the intervals in which the following function is strictly increasing or decreasing: \(f(x) = -2x^3 - 9x^2 - 12x + 1\).
Answer: Given function: \(f(x) = -2x^3 - 9x^2 - 12x + 1\)
Derivative: \(f'(x) = -6x^2 - 18x - 12 = -6(x^2 + 3x + 2) = -6(x + 1)(x + 2)\)
For critical points, set \(f'(x) = 0\):
\(x = -1\) and \(x = -2\).
These points divide the real line into intervals: \((-\infty, -2)\), \((-2, -1)\), and \((-1, \infty)\):
- In \((-\infty, -2)\), let \(x = -3\): \(f'(-3) = -6(-3 + 1)(-3 + 2) = -6(-2)(-1) < 0\) (Strictly Decreasing)
- In \((-2, -1)\), let \(x = -1.5\): \(f'(-1.5) = -6(-1.5 + 1)(-1.5 + 2) = -6(-0.5)(0.5) > 0\) (Strictly Increasing)
- In \((-1, \infty)\), let \(x = 0\): \(f'(0) = -6(0 + 1)(0 + 2) = -12 < 0\) (Strictly Decreasing)
Thus, \(f(x)\) is strictly increasing on \((-2, -1)\) and strictly decreasing on \((-\infty, -2) \cup (-1, \infty)\).
In simple words: Factor the derivative to find critical points, then test sign values in each region.

Exam Tip: Watch out for the negative leading coefficient when testing signs.

Question 4. Find the intervals in which the following function is strictly increasing or decreasing: \(f(x) = \sin x + \cos x, 0 \leq x \leq 2\pi\).
Answer: Given function: \(f(x) = \sin x + \cos x\)
Derivative: \(f'(x) = \cos x - \sin x\)
For critical points, set \(f'(x) = 0\):
\(\cos x - \sin x = 0 \implies \tan x = 1\)
In \([0, 2\pi]\), \(\tan x = 1\) at \(x = \frac{\pi}{4}, \frac{5\pi}{4}\).
These points divide the interval into: \(\left(0, \frac{\pi}{4}\right)\), \(\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)\), and \(\left(\frac{5\pi}{4}, 2\pi\right)\):
- In \(\left(0, \frac{\pi}{4}\right)\), let \(x = 0\): \(f'(0) = \cos 0 - \sin 0 = 1 > 0\) (Increasing)
- In \(\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)\), let \(x = \frac{\pi}{2}\): \(f'\left(\frac{\pi}{2}\right) = \cos\frac{\pi}{2} - \sin\frac{\pi}{2} = 0 - 1 = -1 < 0\) (Decreasing)
- In \(\left(\frac{5\pi}{4}, 2\pi\right)\), let \(x = 2\pi\): \(f'(2\pi) = \cos(2\pi) - \sin(2\pi) = 1 > 0\) (Increasing)
Thus, \(f(x)\) is strictly increasing on \(\left(0, \frac{\pi}{4}\right) \cup \left(\frac{5\pi}{4}, 2\pi\right)\) and strictly decreasing on \(\left(\frac{\pi}{4}, \frac{5\pi}{4}\right)\).
In simple words: Solve \(\tan x = 1\) to find critical angles, then test signs in each quadrant.

Exam Tip: Check values at standard angles like 0, \(\pi/2\), and \(2\pi\) for quick sign tests.

Question 5. For the curve \(y = 4x^3 - 2x^5\), find all the points at which the tangent passes through the origin.
Answer: Given curve: \(y = 4x^3 - 2x^5\)
Let \((x_1, y_1)\) be a point on the curve where the tangent passes through \((0, 0)\).
Since \((x_1, y_1)\) is on the curve, \(y_1 = 4x_1^3 - 2x_1^5\).
Slope of the tangent at \((x_1, y_1)\) is \(m = \left. \frac{dy}{dx} \right|_{(x_1, y_1)} = 12x_1^2 - 10x_1^4\).
Since the tangent passes through \((0, 0)\) and \((x_1, y_1)\), its slope is also \(\frac{y_1}{x_1}\).
Equating slopes:
\(12x_1^2 - 10x_1^4 = \frac{4x_1^3 - 2x_1^5}{x_1} = 4x_1^2 - 2x_1^4\)
\(12x_1^2 - 10x_1^4 = 4x_1^2 - 2x_1^4 \implies 8x_1^2 - 8x_1^4 = 0 \implies 8x_1^2(1 - x_1^2) = 0\)
Thus, \(x_1 = 0\) or \(x_1 = 1\) or \(x_1 = -1\).
- If \(x_1 = 0\), then \(y_1 = 0\). Point: \((0, 0)\).
- If \(x_1 = 1\), then \(y_1 = 4(1)^3 - 2(1)^5 = 2\). Point: \((1, 2)\).
- If \(x_1 = -1\), then \(y_1 = 4(-1)^3 - 2(-1)^5 = -2\). Point: \((-1, -2)\).
Thus, the required points are \((0, 0)\), \((1, 2)\), and \((-1, -2)\).
In simple words: Equate the derivative to the two-point slope formula and solve the resulting polynomial equation for \(x_1\).

Exam Tip: Always verify if \(x_1 = 0\) yields a valid point on the curve as well.

Question 6. Find the equation of the tangent line to the curve \(y = x^2 - 2x + 7\) which is (a) parallel to the line \(2x - y + 9 = 0\) (b) perpendicular to the line \(5y - 15x = 13\).
Answer: Given curve: \(y = x^2 - 2x + 7\)
Slope of the tangent: \(\frac{dy}{dx} = 2x - 2\)
(a) Given line: \(2x - y + 9 = 0 \implies y = 2x + 9\), so its slope is 2.
Since the tangent is parallel, its slope must equal 2:
\(2x - 2 = 2 \implies 2x = 4 \implies x = 2\)
When \(x = 2\), \(y = (2)^2 - 2(2) + 7 = 7\). Point: \((2, 7)\).
Equation of tangent at \((2, 7)\) with slope 2:
\(y - 7 = 2(x - 2) \implies y - 7 = 2x - 4 \implies 2x - y + 3 = 0\).
(b) Given line: \(5y - 15x = 13 \implies 5y = 15x + 13 \implies y = 3x + \frac{13}{5}\), so its slope is 3.
Since the tangent is perpendicular to this line, its slope must be the negative reciprocal: \(m = -\frac{1}{3}\).
\(2x - 2 = -\frac{1}{3} \implies 2x = 2 - \frac{1}{3} = \frac{5}{3} \implies x = \frac{5}{6}\)
When \(x = \frac{5}{6}\), \(y = \left(\frac{5}{6}\right)^2 - 2\left(\frac{5}{6}\right) + 7 = \frac{25}{36} - \frac{5}{3} + 7 = \frac{25 - 60 + 252}{36} = \frac{217}{36}\).
Equation of tangent at \(\left(\frac{5}{6}, \frac{217}{36}\right)\) with slope \(-\frac{1}{3}\):
\(y - \frac{217}{36} = -\frac{1}{3}\left(x - \frac{5}{6}\right) \implies 36y - 217 = -12x + 10 \implies 12x + 36y - 227 = 0\).
In simple words: For parallel lines, equate slopes. For perpendicular lines, use negative reciprocal slopes.

Exam Tip: Clearly label parts (a) and (b) to avoid confusion during evaluation.

Question 7. Prove that the curves \(x = y^2\) and \(xy = k\) cut at right angles if \(8k^2 = 1\).
Answer: Let \((x_1, y_1)\) be the point of intersection of the two curves \(x = y^2\) and \(xy = k\).
Substituting \(x = y_1^2\) into \(xy = k\):
\((y_1^2)y_1 = k \implies y_1^3 = k \implies y_1 = k^{1/3}\), and \(x_1 = k^{2/3}\).
For curve 1 (\(x = y^2\)), differentiating implicitly:
\(1 = 2y \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{1}{2y}\).
Slope of tangent 1 at \((x_1, y_1)\) is \(m_1 = \frac{1}{2y_1}\).
For curve 2 (\(xy = k\)), differentiating implicitly:
\(x \frac{dy}{dx} + y = 0 \implies \frac{dy}{dx} = -\frac{y}{x}\).
Slope of tangent 2 at \((x_1, y_1)\) is \(m_2 = -\frac{y_1}{x_1}\).
Since the curves cut at right angles, \(m_1 \cdot m_2 = -1\):
\(\left(\frac{1}{2y_1}\right) \left(-\frac{y_1}{x_1}\right) = -1 \implies -\frac{1}{2x_1} = -1 \implies 2x_1 = 1 \implies x_1 = \frac{1}{2}\).
Since \(x_1 = k^{2/3} = \frac{1}{2}\):
Cube both sides: \(\left(k^{2/3}\right)^3 = \left(\frac{1}{2}\right)^3 \implies k^2 = \frac{1}{8} \implies 8k^2 = 1\).
Hence proved.
In simple words: Find intersection coordinates, set the product of tangent slopes to \(-1\), and simplify the result.

Question 8. Using differentials, find the approximate value of each of the following up to 3 places of decimal: (i) \((26)^{1/3}\) (ii) \((32.15)^{1/5}\).
Answer:
(i) Let \(y = x^{1/3}\). Let \(x = 27\) and \(\Delta x = -1\).
Then \(y + \Delta y = (26)^{1/3}\).
\(\frac{dy}{dx} = \frac{1}{3x^{2/3}}\)
At \(x = 27\), \(\frac{dy}{dx} = \frac{1}{3(27)^{2/3}} = \frac{1}{3(9)} = \frac{1}{27} \approx 0.037\).
\(\Delta y \approx \frac{1}{27} \times (-1) = -0.037\).
\((26)^{1/3} \approx 3 - 0.037 = 2.963\).
(ii) Let \(y = x^{1/5}\). Let \(x = 32\) and \(\Delta x = 0.15\).
Then \(y + \Delta y = (32.15)^{1/5}\).
\(\frac{dy}{dx} = \frac{1}{5x^{4/5}}\)
At \(x = 32\), \(\frac{dy}{dx} = \frac{1}{5(32)^{4/5}} = \frac{1}{5(16)} = \frac{1}{80} = 0.0125\).
\(\Delta y \approx 0.0125 \times 0.15 = 0.001875 \approx 0.002\).
\((32.15)^{1/5} \approx 2 + 0.002 = 2.002\).
In simple words: Use 27 and 32 as base values because their cube and fifth roots are exact integers (3 and 2).

Exam Tip: Round off carefully to the requested 3 decimal places.

Question 9. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is \(\frac{8}{27}\) of the volume of the sphere.
Answer: Let \(R\) be the radius of the sphere, \(r\) be the radius of the base of the inscribed cone, and \(h\) be the height of the cone.
From the geometry of a cone inscribed in a sphere, let \(x\) be the distance from the center of the sphere to the base of the cone, so \(h = R + x\).
Radius of the cone base satisfies \(r^2 = R^2 - x^2\).
Volume of the cone \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (R^2 - x^2)(R + x) = \frac{\pi}{3}(R - x)(R + x)^2\)
Differentiating with respect to \(x\):
\(\frac{dV}{dx} = \frac{\pi}{3} \left[ (R - x) \cdot 2(R + x) + (R + x)^2(-1) \right] = \frac{\pi}{3}(R + x)[2(R - x) - (R + x)]\)
\(\frac{dV}{dx} = \frac{\pi}{3}(R + x)(R - 3x)\)
For critical points, set \(\frac{dV}{dx} = 0\):
\(R - 3x = 0 \implies x = \frac{R}{3}\)
Second derivative test confirms maximum volume when \(x = \frac{R}{3}\).
Height of the largest cone \(h = R + \frac{R}{3} = \frac{4R}{3}\).
Maximum volume \(V = \frac{1}{3}\pi \left(R^2 - \left(\frac{R}{3}\right)^2\right)\left(\frac{4R}{3}\right) = \frac{1}{3}\pi \left(\frac{8R^2}{9}\right)\left(\frac{4R}{3}\right) = \frac{32\pi R^3}{81}\)
Volume of the sphere \(V_{\text{sphere}} = \frac{4}{3}\pi R^3\).
Ratio \(\frac{V}{V_{\text{sphere}}} = \frac{\frac{32\pi R^3}{81}}{\frac{4}{3}\pi R^3} = \frac{32}{81} \cdot \frac{3}{4} = \frac{8}{27}\).
Hence proved that the maximum volume is \(\frac{8}{27}\) of the volume of the sphere.
In simple words: Express the volume of the cone in terms of distance \(x\) from the sphere's center, find the critical point, and compare the resulting maximum volume with the sphere's volume.

Exam Tip: This is a standard calculus proof; remember the optimal height relation \(h = \frac{4}{3}R\).

Question 10. An open topped box is to be constructed by removing equal squares from each corner of a 3 metre by 8 metre rectangular sheet of aluminium and folding up the sides. Find the volume of the largest such box.
Answer: Let the side length of the square removed from each corner be \(x\) metres.
The dimensions of the resulting open box will be:
- Length \(l = 8 - 2x\)
- Breadth \(b = 3 - 2x\)
- Height \(h = x\)
Volume of the box \(V(x) = x(8 - 2x)(3 - 2x) = x(24 - 16x - 6x + 4x^2) = 4x^3 - 22x^2 + 24x\)
Differentiating with respect to \(x\):
\(V'(x) = 12x^2 - 44x + 24 = 4(3x^2 - 11x + 6) = 4(3x - 2)(x - 3)\)
For critical points, set \(V'(x) = 0\):
\(x = \frac{2}{3}\) or \(x = 3\).
Since the width of the sheet is 3 m, \(x = 3\) is inadmissible (it would make breadth \(3 - 2(3) = -3\) which is impossible). Thus, \(x = \frac{2}{3}\).
Second derivative: \(V''(x) = 24x - 44\).
At \(x = \frac{2}{3}\), \(V''\left(\frac{2}{3}\right) = 24\left(\frac{2}{3}\right) - 44 = 16 - 44 = -28 < 0\) (Maximum volume).
Maximum volume \(V = \left(\frac{2}{3}\right)\left(8 - 2\left(\frac{2}{3}\right)\right)\left(3 - 2\left(\frac{2}{3}\right)\right) = \left(\frac{2}{3}\right)\left(\frac{20}{3}\right)\left(\frac{5}{3}\right) = \frac{200}{27} \text{ m}^3\).
In simple words: Define the box dimensions after cutting corners, express volume as a polynomial, find critical points, and discard impossible values.

Exam Tip: Always check domain constraints for \(x\) based on the smallest sheet dimension.