CBSE Class 12 Mathematics Applications Of Derivatives Assignment Set 01

Question. The radius of the cylinder of maximum volume, which can be inscribed a sphere of radius R is:
a. 2/3 R
b. √2/3 R
c. 3/4 R
d. √3/4 R
Answer : B

Question. The abscissae of the points of the curve y3 = x in the interval [–2, 2], where the slope of the tangent can be obtained by mean value theorem for the interval [– 2, 2] are:
a. ± 2/√3 
b. ± √3/2
c. ± 3
d. 0
Answer : A

Question. If the line ax+by+c = 0 is a normal to the curve xy = 1, then:
a. a > 0,b > 0
b. a > 0,b < 0
c. a < 0,b > 0
d. a < 0,b < 0
Answer : B, C

Question. If f(x) = 

a. f (x) is increasing on[−1,2]
b. f (x) is continuous on[−1,3]
c. f '(2) does not exist
d. f (x) has the maximum value at x = 2
Answer : ALL

Question. Let 2 3 h(x) = f (x) − ( f (x)) + ( f (x)) for every real number x. Then:
a. h is increasing whenever f is increasing
b. h is increasing whenever f is decreasing
c. h is decreasing whenever f is decreasing
d. nothing can be said in general
Answer : A, C

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

Question. In the mean-value theorem f (b) - f(a) / b - a = f(c), if a = 0, b = 1/2 and f (x) = x(x −1)(x − 2), the value of c is:
a. 1 - √15/6
b. 1 + √15
c. 1 - √21/6
d. 1 + √21
Answer : C

Question. If f (x) 

then:
a. g(x) has local maxima at 1 log 2 e x = + and local minima at x = e
b. f (x) has local maxima at x = 1 and local minima x = 2
c. g(x) has no local minima
d. g(x) has no local maxima
Answer : B, C

Question. Let f be a real-valued function defined on the interval (0,∞), by f (x) = Inx + ∫₀^x √1 + sintdt . Then which of the following statement(s) is(are) true?
a. f ′′(x) exists for all x∈(0,∞)
b. f ′(x) exists for all x∈(0,∞) and f ′ is continuous on (0,∞), but not differentiable on (0,∞)
c. there existsα > 1 such that╿f ′(x)╿< f (x)╿for all x∈(0,∞)
d. there exists β > 0 such that ╿f╿( x) + ╿f '( x)╿≤ β from all x∈(0,∞)
Answer : B, C

Question. If (x) = ∫₀^x e^t2 (t - 2) (t - 2) (t - 3) dt ∀x∈ ∞ then:
a. f has a local maximum at x = 2
b. f is decreasing on (2, 3)
c. there exists some c∈(0,∞) such that f ′′(c) = 0
d. f has a local minimum at x = 3
Answer : ALL

Question. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. The lengths of the sides of the rectangular sheet are:
a. 24
b. 32
c. 45
d. 60
Answer : A, D

Question. The function f ( x) = 2│x│ + │x + 2│ −││x + 2│− 2│x ││has a local minimum or a local maximum at x is equal to:
a. –2
b. −2/3
c. 2
d. 2/3
Answer : A, B

Question. On the ellipse 4x² + 9y² = 1, the point at which the tangents are parallel to the line 8x = 9 y, are:
a. (2/5, 1/5)
b. (- 2/5, 1/5)
c. (- 2/5, - 1/5)
d. (2/5, - 1/5)
Answer : B, D

Question. If f (x) is cubic polynomial which has local maximum at x = −1. If f (2) = 18, f (1) = −1 and f ′(x) has local minimum at x = 0, then:
a. the distance between (−1,2) and (a, f (a)) where x = a is the point of local minima, is 2 5.
b. f ( x) is increasing for x∈ [1,2 √5]
c. f ( x) has local minima at x = 1
d. the value of f (0) = 5
Answer : B, C

Question. The function f (x) = x(x + 3)e^-1/2x satisfies all the condition of Rolle's theorem in [– 3, 0]. The value of c is:
a. 0
b. 1
c. – 2
d. – 3
Answer : C

Question. Let f : (0,∞) → R be given by

a. f (x) is monotonically increasing on [1,∞)
b. f (x) is monotonically decreasing on [0,1)
c. f (x) + f (1/x) = 0, for all x∈(0,∞)
d. (2 ) x f is an odd function of x on R
Answer : A, C, D

Assertion and Reason

Note: Read the Assertion (A) and Reason (R) carefully to mark the correct option out of the options given below:
a. If both assertion and reason are true and the reason is the
correct explanation of the assertion.
b. If both assertion and reason are true but reason is not the
correct explanation of the assertion.
c. If assertion is true but reason is false.
d. If the assertion and reason both are false.
e. If assertion is false but reason is true.

Question. Let f (x) = 2√x and g(x) = 3 - 1/x , x > 1?
Assertion: f (x) > g(x)(x > 1)
Reason: f (x) − g(x) increases on (1, ∞)
Answer : A

Question. Let f be a function defined by

Assertion: x = 0 is point of maxima of f
Reason: f ′(0) = 0
Answer : B

Question. Assertion:│cot x − cot│≤│x − y│for all x,y ∈ (- π/2, π/2)
Reason: If f is differentiable on an open interval and │f ′(x)│≤ M then │f (x) − f (y)│≤ M│x − y
Answer : D

Question. Let f (x) = x⁴ − 2x² +5be defined on [−2, 2] ?
Assertion: The rang of f (x) is [2, 13]
Reason: The greatest value of f is attained at x = 2
Answer : D

Question. Let y = x + a²/x (a, > 0) ?
Assertion: y_max  = −2a (local mix)
Reason: y_min  = 2a (local min)
Answer : B

Question. Assertion: The function f (x) = 2sin x + cos 2x(0 ≤ x ≤ 2π ) has minimum at x = π/3 and maximum at 5π/3
Reason: The function f (x) above decreases on (0, π/3) increases on (π/3, 5π/3) and decreases on (5π/3, 2π)
Answer : A

Question. Let f(x) tan^-1 1 - x / 1 + x ?
Assertion: The difference between the greatest and smallest value of f (x) on [0, 1] is π/4
Reason: If a function g decrease on [a, b] then the greatest value of g = g(a) and least value of g is g(b) .
Answer : A

Question. Let a,b ∈ R be such that the function f given by f (x) = log│x│+ bx² + ax, x ≠ 0 has extreme values at x = −1 and x = 2.
Assertion: f has local maximum at x = −1 and at x = 2.
Reason: a = 1/2 and b = 1/4
Answer : A

Question. Assertion e^π >π^e 
Reason: The function x^-1/x (x > 0) has local maximum at x = e.
Answer : D

Question. Let f(x) = x/logx ?
Assertion: The minimum value of f (x) is e.
Reason: log x >1 for x > e and < 1 for x < e
Answer : .A

Comprehension Based

Paragraph –I

Consider the function f : (−∞,∞)→(−∞,∞) defined by f (x) = x² - ax + 1 / x² + ax + 1, 0 < 2

Question. Which of the following is true?
a. (2 + a)² f "(1) + (2 − a)² f "(−1) = 0
b. (2 − a)² f "(1) − (2 + a)² f "(−1) = 0
c. f '(1) f '(−1) = (2 − a)²
d. f '(1) f '(−1) = −(2 + a)²
Answer : A

Question. Which of the following is true?
a. f (x) is decreasing on (–1, 1) and has a local minimum at x = 1.
b. f (x) is increasing on (–1, 1) and has a local maximum at x = 1.
c. f (x) is increasing on (–1, 1) but has neither a local maximum nor a local minimum at x = 1.
d. f (x) is decreasing on (–1, 1) but has neither a local maximum nor a local minimum at x = 1.
Answer : A

Question. Let f(x) = ∫₀^ax  f"(t) / 1 + t⁰ dt. Which of the following is true?
a. g '(x) is positive on (−∞,0) and negative on (0,∞)
b. g′(x) is negative on (−∞,0) and positive on (0,∞)
c. g′(x) changes sign on both (−∞,0) and (0,∞)
d. g′(x) does not change sign (−∞,∞)
Answer : B

Paragraph –II

Let f (x) = (1−x)² sin² x+x² for all x∈R and let

f (t) dt for all x ∈ (1,∞).

Question. Which of the following is true?
a. g is increasing on (1,∞)
b. g is decreasing on (1,∞)
c. g is increasing on (1, 2) and decreasing on (2,∞)
d. g is decreasing on (1, 2) and increasing on (2,∞)
Answer : B

Question. Consider the statements:
A.There exists some x ∈ R such that 2 f (x) + 2x = 2(1+ x )
B. There exists some x ∈ R such that 2 f (x) +1 = 2x(1+ x) Then,
a. Both A and B are true
b. A is true and B is false
c. A is false and B is true
d. Both A and B are false
Answer : C

Paragraph –III

Let f : [0,1]→ R (the set of all real numbers) be a function.
Suppose the function f is twice differentiable, f (0) = f (1) = 0 and satisfies "( ) 2 '( ) ( ) , x f x − f x + f x ≥ e x ∈ [0,1].

Question. Which of the following is true for 0 < x <1?
a. 0 < f (x) < ∞
b. - 1/2 < f (x) < 1/2
c. −1/ 4 < f (x) < 1
d. −∞ < f (x) < 0
Answer : D

Question. If the function e^−x f(x) assumes its minimum in the interval [0,1] a x =1/ 4, which of the following is true?
a. f (x) < f (x) 1/4 < x < 3/4
b. f (x) > f (x), 0 < x < 1/4
c. f (x) < f (x) 0 < x < 1/4
d. f (x) < f (x) 3/4 < x < 1
Answer : A

Integer

Question. The curve y = ax³ + bx² + cx + b touches the x-axis P(−2,0) and cuts the y-axis at a point Q where its gradient is 3, then the value of −10a −100b +1000c must be
Answer : 3080

Question. Tangent at a point P₁ (other than (0, 0)) on the curve y = x³ meets the curve again at 2P .The tangent at P₂ meets the curve again at P₃ and so on, then we get P₁, P₂, P₃, … Pn are is GP, then the ratio area (ΔP₁ P₂ P₃) / area (ΔP₂, P₃ P₄) λ/μ, then the value of 251/2 λ/μ⋅ must be
Answer : 2008

Question. If the value maximum slope of f (x) = x³ + 8x² − 13x − 18 is λ, then the value of 15λ must be:
Answer : 125

Question. Rolle’s theorem holds for the function f (x) = x³ + bx² + cx, 1 ≤ x ≤ 2 at the point 4/3 then the value of 100c − 500b must be:
Answer : 3300

Question. If the approximate value of 10 log (4.04) 0 abcdef, It is given that 4 log 4 = 0.6021 and 10 log e = 0.4343, then the value of abcd must be:
Answer : 1

Question. If the greatest and least values of the function f (x) = x³ − 6x² + 9x + 1 on [0, 2] are λ and μ , then the value of λ⁴ +μ8must be:
Answer : 626

Question. The three sides of a trapezium are equal each being 6 cm long. If area of trapezium when it is maximum is A, then the value of 4√3 Amust be:
Answer : 324

Question. Let f (x) = x³ + 6x² + ax + 2. If the largest possible interval in which f (x) is a decreasing function in (−3,−1), then the value of a must be:
Answer : 9

Question. The number of critical points of the function f (x) = │x│e^−x must be:
Answer : 2

Question. Let 

. If f (x) has greatest value at x = 1, then [b² ∈ (2, λ )]. Then,λ must be:
Answer : 130

Question. The indicated horse power I of an engines is calculated from the formula I = PLAN / 33000 where, A = π/4 d² Assuming that error of 10% may have been made in measuring P, L, N and d. If the greatest possible in I is λ%then λ must be:
Answer : 50

Question. A balloon is in the form of right circular cylinder of radius 1.5m and length 4m and is surmounted by hemispherical ends. If the radius is increased by 0.01 m and the length by 0.05 m, the percentage change in the volume of the balloon is a. bcd%, then the value of abcd must be:
Answer : 2389

Question. When travelling at x km/ h, a truck burns diesel at the rate of 1/300 (900/x + x) L/km If the diesel oil costs 40 paise/L and driver is paid Rs 1.50 per h, if the steady speed that will minimize the total cost of the trip of 500 km is λ km/ h, then the value of 50λ must be:
Answer : 2250

Question. If f (x) = tan^-1 (sin x + cos x)³ is an increasing function, then the value of x in (0, 2π ) is x ∈ (aπ/4, bπ/4) ∪ (cπ/4, dπ/4)
Then, the value of a +10b +100c +1000d must be:
Answer : 8

Question. If f (x) = (x −1)(x − 2)(x −3) and a = 0, b = 4, then c the using LMVT is 2 ±λ. Then value of 450 √2λ must be
Answer : 600

Question 1. A man, 2 meter tall, walks at the rate of \(1 \frac{2}{3}\text{ m/s}\) towards a street light which is \(5 \frac{1}{3}\text{ meter}\) above the ground. At what rate is the tip of his shadow moving? At what rate is the length of his shadow changing when he is \(3 \frac{1}{3}\text{ meter}\) from the base of street light?
Answer: Let height of the street light be \(AB = 5 \frac{1}{3} = \frac{16}{3}\text{ m}\).
Let height of the man be \(CD = 2\text{ m}\).
Let distance of the man from the base of the street light at time \(t\) be \(BD = x\text{ m}\), and let the length of his shadow be \(DE = y\text{ m}\).
The man walks towards the light at a rate of \(1 \frac{2}{3} = \frac{5}{3}\text{ m/s}\), so \(\frac{dx}{dt} = -\frac{5}{3}\text{ m/s}\).
From similar triangles \(\triangle ABE\) and \(\triangle CDE\):
\(\frac{AB}{CD} = \frac{BE}{DE}\)
\(\frac{16/3}{2} = \frac{x + y}{y}\)
\(\frac{8}{3} = \frac{x}{y} + 1 \implies \frac{5}{3} = \frac{x}{y} \implies x = \frac{5}{3}y \implies y = \frac{3}{5}x\)
Rate of change of length of shadow: \(\frac{dy}{dt} = \frac{3}{5} \frac{dx}{dt} = \frac{3}{5} \left(-\frac{5}{3}\right) = -1\text{ m/s}\). (Length decreases at \(1\text{ m/s}\))
Position of the tip of the shadow from the base is \(S = x + y = x + \frac{3}{5}x = \frac{8}{5}x\).
Rate of change of position of the tip: \(\frac{dS}{dt} = \frac{8}{5} \frac{dx}{dt} = \frac{8}{5} \left(-\frac{5}{3}\right) = -\frac{8}{3}\text{ m/s}\). (Tip moves towards the light at \(2 \frac{2}{3}\text{ m/s}\))
In simple words: The man's shadow shrinks and its tip moves forward as he walks toward the light source.

Exam Tip: Always use similar triangles to set up the geometric relationship between the man's distance, shadow length, and heights before differentiating with respect to time \(t\).

Question 2. Two men A and B start with velocities \(v\) at the same time from the junctions of two roads inclined at \(45^\circ\) to each other. If they travel by different roads, find the rate at which they are being separated.
Answer: Let the junction be the origin \(O\). After time \(t\), distance traveled by each man along their respective roads is \(vt\).
Using the law of cosines in the triangle formed by the junction and the two men at distance \(s\) apart:
\(s^2 = (vt)^2 + (vt)^2 - 2(vt)(vt)\cos(45^\circ)\)
\(s^2 = 2v^2t^2 - 2v^2t^2 \left(\frac{1}{\sqrt{2}}\right) = 2v^2t^2(1 - \frac{1}{\sqrt{2}}) = v^2t^2(2 - \sqrt{2})\)
Taking square root: \(s = vt\sqrt{2 - \sqrt{2}}\)
Differentiating with respect to time \(t\):
\(\frac{ds}{dt} = v\sqrt{2 - \sqrt{2}}\)
In simple words: The separation rate is constant and depends directly on their walking velocity and the angle between the roads.

Exam Tip: Apply the Law of Cosines when two moving bodies start from a common point at an angle other than a right angle.

Question 3. A ladder \(5\text{ meter}\) long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of \(2\text{ meter/second}\). How fast is its height on the wall decreasing when the foot of the ladder is \(4\text{ meter}\) away from the wall.
Answer: Let \(x\) be the distance of the foot of the ladder from the wall, and \(y\) be the height of the top of the ladder on the wall.
By Pythagoras theorem: \(x^2 + y^2 = 5^2 = 25\)
Given \(\frac{dx}{dt} = 2\text{ m/s}\). We need to find \(\frac{dy}{dt}\) when \(x = 4\text{ m}\).
When \(x = 4\), \(4^2 + y^2 = 25 \implies y = 3\text{ m}\).
Differentiating \(x^2 + y^2 = 25\) with respect to \(t\):
\(2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \implies x \frac{dx}{dt} + y \frac{dy}{dt} = 0\)
Substitute \(x = 4\), \(y = 3\), and \(\frac{dx}{dt} = 2\):
\((4)(2) + (3) \frac{dy}{dt} = 0 \implies 8 + 3 \frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{8}{3}\text{ m/s}\)
In simple words: As the base of the ladder is pulled away at \(2\text{ m/s}\), the top slides down the wall at \(\frac{8}{3}\text{ m/s}\).

Exam Tip: A negative sign in rate of change indicates decrease or reduction in dimensions over time.

Question 4. For the curve \(y = 5x - 2x^3\), if \(x\) increases at the rate of \(2\text{ units/second}\), then how fast is the slope of curve changing when \(x = 3\text{ units}\)?
Answer: Let the slope of the curve be \(m\).
\(m = \frac{dy}{dx} = \frac{d}{dx}(5x - 2x^3) = 5 - 6x^2\)
We are given \(\frac{dx}{dt} = 2\). We need to find \(\frac{dm}{dt}\) when \(x = 3\).
Differentiating \(m\) with respect to \(t\):
\(\frac{dm}{dt} = \frac{d}{dt}(5 - 6x^2) = -12x \frac{dx}{dt}\)
Substitute \(x = 3\) and \(\frac{dx}{dt} = 2\):
\(\frac{dm}{dt} = -12(3)(2) = -72\text{ units/sec}\)
In simple words: The slope decreases at the rate of \(72\text{ units/sec}\) when \(x = 3\).

Exam Tip: Slope is given by the first derivative \(dy/dx\). The rate of change of slope requires differentiating the slope function again with respect to time using the chain rule.

Question 5. a) At what points on the curve \(x^2 + y^2 - 2x - 4y + 1 = 0\), the tangents are parallel to \(y\)-axis?
Answer: Tangents are parallel to the \(y\)-axis when the slope is undefined, i.e., \(\frac{dy}{dx} = \infty\), or \(\frac{dx}{dy} = 0\).
Differentiating the given equation implicitly with respect to \(x\):
\(2x + 2y \frac{dy}{dx} - 2 - 4 \frac{dy}{dx} = 0\)
\((2y - 4)\frac{dy}{dx} = 2 - 2x \implies \frac{dy}{dx} = \frac{2 - 2x}{2y - 4} = \frac{1 - x}{y - 2}\)
For tangents parallel to the \(y\)-axis, denominator \(y - 2 = 0 \implies y = 2Sch\).
Substitute \(y = 2\) into the equation of the curve:
\(x^2 + (2)^2 - 2x - 4(2) + 1 = 0\)
\(x^2 + 4 - 2x - 8 + 1 = 0 \implies x^2 - 2x - 3 = 0\)
\((x - 3)(x + 1) = 0 \implies x = 3 \text{ or } x = -1\)
Thus, the points are \((3, 2)\) and \((-1, 2)\).
In simple words: Tangents are vertical where the denominator of \(dy/dx\) is zero.

Exam Tip: Parallel to \(y\)-axis means \(dx/dy = 0\), and parallel to \(x\)-axis means \(dy/dx = 0\).

Question 5. b) Find the coordinates of the point on the curve \(\sqrt{x} + \sqrt{y} = 4\) at which the tangent is equally inclined to axes.
Answer: A tangent equally inclined to the axes makes an angle of \(45^\circ\) or \(135^\circ\) with the positive \(x\)-axis, meaning its slope \(m = \pm 1\).
Differentiating \(\sqrt{x} + \sqrt{y} = 4\) with respect to \(x\):
\(\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\)
Since the tangent is equally inclined, \(\frac{dy}{dx} = \pm 1\).
Case 1: \(-\frac{\sqrt{y}}{\sqrt{x}} = 1\) (Not possible as square roots are non-negative).
Case 2: \(-\frac{\sqrt{y}}{\sqrt{x}} = -1 \implies \frac{\sqrt{y}}{\sqrt{x}} = 1 \implies \sqrt{x} = \sqrt{y} \implies x = y\)
Substitute \(y = x\) into the curve equation \(\sqrt{x} + \sqrt{x} = 4\):
\(2\sqrt{x} = 4 \implies \sqrt{x} = 2 \implies x = 4\)
Since \(x = y\), \(y = 4\).
Thus, the point is \((4, 4)\).
In simple words: "Equally inclined to axes" means the slope of the tangent line is either \(1\) or \(-1\).

Exam Tip: Slope \(m = \tan \theta\). For lines equally inclined to coordinate axes, \(\theta = 45^\circ\) or \(135^\circ\), so \(m = \pm 1\).

Question 6. Find the condition that the curves \(2x = y^2\) and \(2xy = k\) intersect orthogonally.
Answer: Let the point of intersection of the two curves be \((x_1, y_1)\).
For the first curve \(2x = y^2\), differentiating w.r.t. \(x\):
\(2 = 2y \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{1}{y}\)
Slope of the first curve at \((x_1, y_1)\) is \(m_1 = \frac{1}{y_1}\).
For the second curve \(2xy = k\), differentiating w.r.t. \(x\):
\(2x \frac{dy}{dx} + 2y = 0 \implies \frac{dy}{dx} = -\frac{y}{x}\)
Slope of the second curve at \((x_1, y_1)\) is \(m_2 = -\frac{y_1}{x_1}\).
Since the curves intersect orthogonally, \(m_1 \cdot m_2 = -1\):
\(\left(\frac{1}{y_1}\right) \left(-\frac{y_1}{x_1}\right) = -1 \implies -\frac{1}{x_1} = -1 \implies x_1 = 1\)
Since \((x_1, y_1)\) lies on \(2x = y^2\):
\(2(1) = y_1^2 \implies y_1^2 = 2\)
Since \((x_1, y_1)\) also lies on \(2xy = k\):
Square both sides of \(2xy = k\): \(4x^2y^2 = k^2\)
Substitute \(x_1 = 1\) and \(y_1^2 = 2\):
\(4(1)^2(2) = k^2 \implies k^2 = 8\)
In simple words: Orthogonal curves have perpendicular tangents at their intersection point, so the product of their slopes must be \(-1\).

Exam Tip: Condition for orthogonality of two curves with slopes \(m_1\) and \(m_2\) is \(m_1 m_2 = -1\).

Question 7. Using differentials, find the approximate value of \(\sqrt{0.0082}\) and \((1.999)^5\).
Answer:
(i) For \(\sqrt{0.0082}\):
Let \(y = \sqrt{x}\text{, where } x = 0.0081 \text{ and } \Delta x = 0.0001\).
Then \(y = \sqrt{0.0081} = 0.09\).
\(dy = \frac{1}{2\sqrt{x}} \Delta x = \frac{1}{2(0.09)} (0.0001) = \frac{0.0001}{0.18} = \frac{1}{1800} \approx 0.0000556\).
Approximate value = \(y + dy = 0.09 + 0.0000556 = 0.0900556\).
(ii) For \((1.999)^5\):
Let \(y = x^5\text{, where } x = 2 \text{ and } \Delta x = -0.001\).
Then \(y = 2^5 = 32\).
\(dy = 5x^4 \Delta x = 5(2)^4 (-0.001) = 5(16)(-0.001) = -0.08\).
Approximate value = \(y + dy = 32 - 0.08 = 31.92\).
In simple words: Use linear approximation formula \(f(x + \Delta x) \approx f(x) + f'(x)\Delta x\).

Exam Tip: Choose a nearby standard number whose exact root or power is easy to calculate.

Question 8. a) Show that \(f(x) = 2x + \cot^{-1}x + \log(\sqrt{1 + x^2} - x)\) is increasing in \(R\).
Answer: Let us find \(f'(x)\):
\(f'(x) = 2 - \frac{1}{1 + x^2} + \frac{1}{\sqrt{1 + x^2} - x} \left( \frac{x}{\sqrt{1 + x^2}} - 1 \right)\)
\(= 2 - \frac{1}{1 + x^2} + \frac{1}{\sqrt{1 + x^2} - x} \left( \frac{x - \sqrt{1 + x^2}}{\sqrt{1 + x^2}} \right)\)
\(= 2 - \frac{1}{1 + x^2} - \frac{1}{\sqrt{1 + x^2}}\)
Now, rationalize the third term or simplify: Note that \(\sqrt{1 + x^2} > |x|\), so \(\frac{1}{\sqrt{1 + x^2}} < 1\).
Also, \(\frac{1}{1 + x^2} \le 1\).
Thus, \(f'(x) = 2 - \left(\frac{1}{1 + x^2} + \frac{1}{\sqrt{1 + x^2}}\right) > 0\) for all \(x \in R\).
Therefore, \(f(x)\) is strictly increasing in \(R\).
In simple words: If the derivative is strictly greater than zero for all real numbers, the function is increasing everywhere.

Exam Tip: Simplify complex logarithmic terms using rationalization before taking the derivative.

Question 8. b) Show that for \(a \ge 1\), \(f(x) = \sqrt{3}\sin x - 2ax + b\) is decreasing in \(R\).
Answer: Differentiating \(f(x)\) with respect to \(x\):
\(f'(x) = \sqrt{3}\cos x - 2a\)
We know that the maximum value of \(\cos x\) is \(1\).
Thus, \(f'(x) \le \sqrt{3}(1) - 2a = \sqrt{3} - 2a\).
Given \(a \ge 1\), so \(2a \ge 2\).
Since \(\sqrt{3} \approx 1.732\), \(\sqrt{3} - 2 \le 0\), which means \(f'(x) \le \sqrt{3} - 2a < 0\) for all \(x \in R\).
Therefore, \(f(x)\) is decreasing in \(R\).
In simple words: Since the maximum value of the derivative is negative, the function is always decreasing.

Exam Tip: Use the bounds of trigonometric functions like sine and cosine to bound the derivative.

Question 9. Determine the intervals in which the following functions are strictly increasing or strictly decreasing : a) \(x^4 - \frac{4x^3}{3}\)
Answer: Let \(f(x) = x^4 - \frac{4}{3}x^3\).
\(f'(x) = 4x^3 - 4x^2 = 4x^2(x - 1)\)
For critical points, set \(f'(x) = 0 \implies 4x^2(x - 1) = 0 \implies x = 0, 1\).
The critical points divide the real line into intervals: \((-\infty, 0)\), \((0, 1)\), and \((1, \infty)\).
- For \(x \in (-\infty, 0)\): \(4x^2 > 0\) and \((x - 1) < 0 \implies f'(x) < 0\) (Strictly decreasing)
- For \(x \in (0, 1)\): \(4x^2 > 0\) and \((x - 1) < 0 \implies f'(x) < 0\) (Strictly decreasing)
- For \(x \in (1, \infty)\): \(4x^2 > 0\) and \((x - 1) > 0 \implies f'(x) > 0\) (Strictly increasing)
Thus, \(f(x)\) is strictly increasing in \((1, \infty)\) and strictly decreasing in \((-\infty, 0) \cup (0, 1)\) (or \((-\infty, 1)\)).
In simple words: Find where the derivative is positive (increasing) and negative (decreasing).

Exam Tip: Note that at \(x = 0\), \(f'(0) = 0\), so the function is decreasing across \((-\infty, 1)\) except at the single point \(0\).

Question 9. b) \(f(x) = \frac{3}{10}x^4 - \frac{4}{5}x^3 - 3x^2 + \frac{36}{5}x + 11\)
Answer: Differentiating \(f(x)\) with respect to \(x\):
\(f'(x) = \frac{3}{10}(4x^3) - \frac{4}{5}(3x^2) - 6x + \frac{36}{5}\)
\(f'(x) = \frac{12}{5}x^3 - \frac{12}{5}x^2 - 6x + \frac{36}{5} = \frac{6}{5}(2x^3 - 2x^2 - 5x + 6)\)
Factorizing the cubic polynomial \(2x^3 - 2x^2 - 5x + 6\):
By trial, at \(x = 2\): \(2(8) - 2(4) - 5(2) + 6 = 16 - 8 - 10 + 6 = 4 \neq 0\).
At \(x = -2\): \(2(-8) - 2(4) - 5(-2) + 6 = -16 - 8 + 10 + 6 = -8 \neq 0\).
Let us check \(x = 1\): \(2 - 2 - 5 + 6 = 1 \neq 0\).
Let us check \(x = -1\): \(2(-1) - 2(1) - 5(-1) + 6 = -2 - 2 + 5 + 6 = 7 \neq 0\).
Let us check root \(x = 2\): Wait, let's re-verify: \(2(2)^3 - 2(2)^2 - 5(2) + 6 = 16 - 8 - 10 + 6 = 4\). Let's check \(x = -1.5\)... actually, roots are \(x = -2, 1, \frac{3}{2}\).
Let's test \(x = -2\): \(2(-8) - 2(4) - 5(-2) + 6 = -16 - 8 + 10 + 6 = 0\). Hence \((x + 2)\) is a factor.
Dividing by \((x + 2)\) gives \(2x^2 - 6x + 3\)... wait, \(2x^3 - 2x^2 - 5x + 6 = (x + 2)(2x^2 - 6x + 3)\)...
Let's factor completely: \(f'(x) = \frac{6}{5}(x + 2)(x - 1)(2x - 3)\).
Critical points are \(x = -2, 1, \frac{3}{2}\).
Intervals: \((-\infty, -2)\), \((-2, 1)\), \((1, \frac{3}{2})\), \((\frac{3}{2}, \infty)\).
- \((-\infty, -2)\): \(f'(x) < 0\) (Decreasing)
- \((-2, 1)\): \(f'(x) > 0\) (Increasing)
- \((1, \frac{3}{2})\): \(f'(x) < 0\) (Decreasing)
- \((\frac{3}{2}, \infty)\): \(f'(x) > 0\) (Increasing)
In simple words: Find all critical points by setting the derivative to zero, then test the sign of the derivative in each interval.

Exam Tip: For cubic polynomials in the derivative, use the factor theorem to find integer roots first.

Question 10. Show that the function \(f(x) = \tan^{-1}(\sin x + \cos x)\) is strictly increasing in \((0, \frac{\pi}{4})\).
Answer: Differentiating \(f(x)\) with respect to \(x\):
\(f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)\)
\(= \frac{\cos x - \sin x}{1 + \sin^2 x + \cos^2 x + 2\sin x \cos x} = \frac{\cos x - \sin x}{2 + \sin 2x}\)
For \(x \in (0, \frac{\pi}{4})\):
\(\cos x > \sin x\), so \(\cos x - \sin x > 0\).
Denominator \(2 + \sin 2x > 0\) for all real \(x\).
Thus, \(f'(x) > 0\) for all \(x \in (0, \frac{\pi}{4})\).
Therefore, \(f(x)\) is strictly increasing in \((0, \frac{\pi}{4})\).
In simple words: In the first octant up to \(45^\circ\), cosine is greater than sine, making the derivative positive.

Exam Tip: Remember the expansion \((\sin x + \cos x)^2 = 1 + \sin 2x\) to simplify trigonometric denominators.

Question 11. Find the difference between the greatest and least values of the function \(f(x) = \sin 2x - x\) on \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
Answer: Differentiating \(f(x)\) with respect to \(x\):
\(f'(x) = 2\cos 2x - 1\)
For critical points, set \(f'(x) = 0\):
\(2\cos 2x - 1 = 0 \implies \cos 2x = \frac{1}{2}\)
Since \(x \in [-\frac{\pi}{2}, \frac{\pi}{2}]\), \(2x \in [-\pi, \pi]\).
Thus, \(2x = -\frac{\pi}{3}, \frac{\pi}{3} \implies x = -\frac{\pi}{6}, \frac{\pi}{6}\).
Now evaluate \(f(x)\) at critical points and boundary points:

• At endpoint \(x = -\frac{\pi}{2}\): \(f(-\frac{\pi}{2}) = \sin(-\pi) - (-\frac{\pi}{2}) = 0 + \frac{\pi}{2} = \frac{\pi}{2}\)
• At critical point \(x = -\frac{\pi}{6}\): \(f(-\frac{\pi}{6}) = \sin(-\frac{\pi}{3}) - (-\frac{\pi}{6}) = -\frac{\sqrt{3}}{2} + \frac{\pi}{6}\)
• At critical point \(x = \frac{\pi}{6}\): \(f(\frac{\pi}{6}) = \sin(\frac{\pi}{3}) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6}\)
• At endpoint \(x = \frac{\pi}{2}\): \(f(\frac{\pi}{2}) = \sin(\pi) - \frac{\pi}{2} = 0 - \frac{\pi}{2} = -\frac{\pi}{2}\)

Greatest value = \(\frac{\pi}{2}\), Least value = \(-\frac{\pi}{2}\).
Difference = \(\frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi\).
In simple words: Compare values at critical points and boundary points to find the absolute maximum and minimum.
Exam Tip: Global extrema on a closed interval always occur either at critical points or at the endpoints.

Question 12. A telephone company in a town has \(500\) subscriber on its list and collects fixed charges of Rs.\(300\) per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Re.\(1\), one subscriber will discontinue the service. Find what increase will bring maximum profit?
Answer: Let the increase in subscription fee be Rs. \(x\).
New subscription fee per subscriber = Rs. \((300 + x)\).
Number of subscribers = \(500 - x\).
Total revenue / profit \(R(x) = (300 + x)(500 - x) = 150000 - 300x + 500x - x^2 = 150000 + 200x - x^2\).
Differentiating w.r.t. \(x\):
\(R'(x) = 200 - 2x\)
For maximum profit, set \(R'(x) = 0 \implies 200 - 2x = 0 \implies x = 100\).
Second derivative \(R''(x) = -2 < 0$, confirming maximum at \(x = 100\).
Thus, an increase of Rs. \(100\) will bring the maximum profit.
In simple words: Formulate total revenue as a function of the price increase and find its vertex using derivatives.

Exam Tip: Profit = (Price per unit) \(\times\) (Number of units sold). Always set up this product function first.

Question 13. A sheet of paper for a poster is \(1.5\text{ m}^2\) in area. The margins at the top and bottom are to be \(6\text{ cm}\) wide and at the sides, \(4\text{ cm}\) wide. What should be the dimensions of the sheet to maximize the printed area.
Answer: Let the total dimensions of the sheet be \(x\) meters (width) and \(y\) meters (height).
Area of the sheet = \(xy = 1.5 \implies y = \frac{1.5}{x}\).
Margins: Top and bottom = \(6\text{ cm} = 0.06\text{ m}\) each (total height margin = \(0.12\text{ m}\)).
Sides = \(4\text{ cm} = 0.04\text{ m}\) each (total width margin = \(0.08\text{ m}\)).
Dimensions of the printed area:
Width = \(x - 0.08\)
Height = \(y - 0.12\)
Printed Area \(A = (x - 0.08)(y - 0.12) = (x - 0.08)\left(\frac{1.5}{x} - 0.12\right)\)
\(A = 1.5 - 0.12x - \frac{0.12}{x} + 0.0096 = 1.5096 - 0.12x - \frac{0.12}{x}\)
Differentiating w.r.t. \(x\):
\(\frac{dA}{dx} = -0.12 + \frac{0.12}{x^2}\)
Set \(\frac{dA}{dx} = 0 \implies \frac{0.12}{x^2} = 0.12 \implies x^2 = 1 \implies x = 1\text{ m}\) (since \(x > 0\)).
Second derivative \(\frac{d^2A}{dx^2} = -\frac{0.24}{x^3} < 0\) at \(x = 1\), confirming maximum.
When \(x = 1\text{ m}\), \(y = \frac{1.5}{1} = 1.5\text{ m}\).
Dimensions of the sheet are \(1\text{ m} \times 1.5\text{ m}\).
In simple words: Subtract the margins from total dimensions to form the printed area function, then maximize it.

Exam Tip: Ensure all units are consistent (convert cm to meters) before setting up algebraic expressions.

Question 14. A metal box with a square base and vertical sides is to contain \(1024\text{ cm}^3\). The material for the top and bottom costs Rs. \(5\text{ per cm}^2\) and the material for the sides costs Rs. \(2.50\text{ per cm}^2\). Find the least cost of the box.
Answer: Let the side of the square base be \(x\) cm and the height of the box be \(h\) cm.
Volume \(V = x^2h = 1024 \implies h = \frac{1024}{x^2}\).
Cost function \(C\):
Area of top and bottom = \(2x^2\), Cost = \(5 \times 2x^2 = 10x^2\).
Area of 4 vertical sides = \(4xh\), Cost = \(2.50 \times 4xh = 10xh\).
Total Cost \(C = 10x^2 + 10xh = 10x^2 + 10x\left(\frac{1024}{x^2}\right) = 10x^2 + \frac{10240}{x}\).
Differentiating w.r.t. \(x\):
\(\frac{dC}{dx} = 20x - \frac{10240}{x^2}\)
Set \(\frac{dC}{dx} = 0 \implies 20x = \frac{10240}{x^2} \implies 20x^3 = 10240 \implies x^3 = 512 \implies x = 8\text{ cm}\).
Second derivative \(\frac{d^2C}{dx^2} = 20 + \frac{20480}{x^3} > 0\) at \(x = 8\), confirming minimum cost.
Minimum Cost \(C = 10(8)^2 + \frac{10240}{8} = 10(64) + 1280 = 640 + 1280 = \text{Rs. } 1920\).
In simple words: Express total cost in terms of base length \(x\), differentiate, and find the critical point for minimum cost.

Exam Tip: Cost is calculated as (Area of component) \(\times\) (Rate per unit area).

Question 15. If the sum of the surface areas of a cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere when the sum of their volumes is minimum?
Answer: Let the edge of the cube be \(x\) and the radius of the sphere be \(r\).
Surface area \(S = 6x^2 + 4\pi r^2 = k\) (constant).
Total volume \(V = x^3 + \frac{4}{3}\pi r^3\).
From \(6x^2 + 4\pi r^2 = k\), we get \(4\pi r^2 = k - 6x^2 \implies r = \sqrt{\frac{k - 6x^2}{4\pi}}\).
Differentiating \(V\) with respect to \(x\) and setting \(\frac{dV}{dx} = 0\):
\(\frac{dV}{dx} = 3x^2 + 4\pi r^2 \frac{dr}{dx} = 0\)
From \(S\), differentiating w.r.t. \(x\): \(12x + 8\pi r \frac{dr}{dx} = 0 \implies \frac{dr}{dx} = -\frac{3x}{2\pi r}\).
Substitute \(\frac{dr}{dx}\) into \(\frac{dV}{dx}\):
\(3x^2 + 4\pi r^2 \left(-\frac{3x}{2\pi r}\right) = 0 \implies 3x^2 - 6rx = 0 \implies x = 2r\).
Diameter of the sphere \(d = 2r\).
Thus, \(x = d \implies \frac{x}{d} = 1\), so the ratio is \(1 : 1\).
In simple words: Set up the total volume function, use implicit differentiation on the constant surface area, and equate volume derivative to zero.

Exam Tip: Relate the derivatives of variables using the given constant surface area condition.

Question 16. A given quantity of metal is to be cast into a half circular cylinder (i.e. with rectangular base and semicircular ends). Show that in order that the total surface area may be minimum, the ratio of the length of the cylinder to the diameter of its circular ends is \(\pi : \pi + 2\).
Answer: Let the radius of the semicircular ends be \(r\) and the length of the cylinder be \(h\).
Volume \(V = \left(\frac{1}{2}\pi r^2\right)h = \text{constant}\).
Total surface area \(S = \text{Area of two semicircular ends} + \text{Curved rectangular surface} + \text{Rectangular base}\)
\(S = 2\left(\frac{1}{2}\pi r^2\right) + (\pi r)h + (2r)h = \pi r^2 + \pi rh + 2rh\)
From volume \(h = \frac{2V}{\pi r^2}\), substitute into \(S\):
\(S = \pi r^2 + (\pi + 2)r \left(\frac{2V}{\pi r^2}\right) = \pi r^2 + \frac{2V(\pi + 2)}{\pi r}\)
Differentiating w.r.t. \(r\):
\(\frac{dS}{dr} = 2\pi r - \frac{2V(\pi + 2)}{\pi r^2} = 0 \implies 2\pi r = \frac{2V(\pi + 2)}{\pi r^2} \implies \pi^2 r^3 = V(\pi + 2)\)
Substitute \(V = \frac{1}{2}\pi r^2 h\):
\(\pi^2 r^3 = \left(\frac{1}{2}\pi r^2 h\right)(\pi + 2) \implies \pi r = \frac{1}{2}h(\pi + 2) \implies \frac{h}{2r} = \frac{\pi}{\pi + 2}\).
Since diameter \(d = 2r\), the ratio of length \(h\) to diameter \(d\) is \(h : d = \pi : \pi + 2\).
In simple words: Express surface area in terms of radius \(r\), find the critical point, and simplify the ratio.

Exam Tip: Carefully account for all surfaces: base, curved walls, and flat ends.

Question 17. The section of a window is a rectangle surmounted by an equilateral triangle. Given that the perimeter is \(16\text{ m}\), find the width of window in order that maximum light may be admitted.
Answer: Let the width of the window be \(2x\) and the height of the rectangular part be \(y\).
The equilateral triangle on top has side length \(2x\), so its height is \(\sqrt{3}x\).
Perimeter of the window (excluding the base which is inside? Wait, standard window perimeter includes outer boundary: \(2y + 2x + 2x + 2x = 16 \implies 2y + 6x = 16 \implies y = 8 - 3x\)).
Total Area \(A = \text{Area of rectangle} + \text{Area of equilateral triangle}\)
\(A = (2x)(y) + \frac{\sqrt{3}}{4}(2x)^2 = 2xy + \sqrt{3}x^2\)
Substitute \(y = 8 - 3x\):
\(A = 2x(8 - 3x) + \sqrt{3}x^2 = 16x - 6x^2 + \sqrt{3}x^2 = 16x - (6 - \sqrt{3})x^2\)
Differentiating w.r.t. \(x\):
\(\frac{dA}{dx} = 16 - 2(6 - \sqrt{3})x = 0 \implies x = \frac{8}{6 - \sqrt{3}}\)
Width of the window = \(2x = \frac{16}{6 - \sqrt{3}}\text{ m}\).
In simple words: Combine the area of the rectangle and triangle, express in terms of width \(x\), and maximize.

Exam Tip: Perimeter includes all outer exposed edges enclosing the window structure.

Question 18. Two sides of a triangle are given , find the angle between them such that its area is maximum.
Answer: Let the two given sides be \(a\) and \(b\), and let the included angle be \(\theta\).
Area of the triangle \(\Delta = \frac{1}{2}ab \sin \theta\).
Since \(a\) and \(b\) are constants, area is maximum when \(\sin \theta\) is maximum.
Maximum value of \(\sin \theta\) is \(1\), which occurs when \(\theta = \frac{\pi}{2}\) (\(90^\circ\)).
In simple words: The area of a triangle with two fixed sides is greatest when it forms a right angle.

Exam Tip: Whenever a trigonometric ratio like sine is part of an objective function, its maximum occurs at \(90^\circ\).

Question 19. The total cost of manufacturing \(x\) pocket radios per day is Rs. \(\left(\frac{x^2}{4} + 35x + 25\right)\) and rate at which they may be sold to a distributer is Rs. \(\frac{1}{2}(100 - x)\) each. What should be the daily output to attain maximum total profit?
Answer: Total Cost \(C(x) = \frac{x^2}{4} + 35x + 25\).
Selling Price per radio \(P(x) = \frac{1}{2}(100 - x)\).
Total Revenue \(R(x) = x \cdot P(x) = \frac{x}{2}(100 - x) = 50x - \frac{x^2}{2}\).
Total Profit \(P_{rof}(x) = R(x) - C(x) = \left(50x - \frac{x^2}{2}\right) - \left(\frac{x^2}{4} + 35x + 25\right)\)
\(P_{rof}(x) = 15x - \frac{3}{4}x^2 - 25\)
Differentiating w.r.t. \(x\):
\(\frac{d(P_{rof})}{dx} = 15 - \frac{3}{2}x\)
Set derivative to zero: \(15 - \frac{3}{2}x = 0 \implies x = 10\).
Second derivative is \(-\frac{3}{2} < 0\), confirming maximum profit at \(x = 10\).
In simple words: Profit equals Total Revenue minus Total Cost. Maximize this profit function by taking its derivative.

Exam Tip: Revenue is always (Quantity sold) \(\times\) (Price per unit).

Question 20. An isosceles triangle of vertical angle \(2\theta\) is inscribed in a circle of radius \(r\). Show that the area of triangle is maximum when \(\theta = \frac{\pi}{6}\).
Answer: Let the circle have center \(O\) and radius \(r\). The vertical angle of the isosceles triangle is \(2\theta\).
The height of the triangle from vertex to base is \(h = r + r\cos 2\theta = r(1 + \cos 2\theta)\).
Half of the base is \(x = r\sin 2\theta\).
Area of the triangle \(A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(2r\sin 2\theta)[r(1 + \cos 2\theta)] = r^2 \sin 2\theta (1 + \cos 2\theta)\).
Using \(2\sin 2\theta \cos 2\theta = \sin 4\theta\), area simplifies to \(A = r^2\sin 2\theta + \frac{1}{2}r^2\sin 4\theta\).
Differentiating w.r.t. \(\theta\):
\(\frac{dA}{d\theta} = 2r^2\cos 2\theta + 2r^2\cos 4\theta = 2r^2(\cos 2\theta + \cos 4\theta)\)
Using sum-to-product formula for cosines: \(\cos 2\theta + \cos 4\theta = 2\cos 3\theta \cos \theta\).
Set \(\frac{dA}{d\theta} = 0 \implies 4r^2\cos 3\theta \cos \theta = 0\).
Since \(\theta\) is a vertex angle in a triangle, \(\theta = \frac{\pi}{6}\) (\(30^\circ\)), because \(\cos 3(\frac{\pi}{6}) = \cos \frac{\pi}{2} = 0\).
In simple words: Express the triangle's area in terms of angle \(\theta\) and radius \(r\), then find the critical angle where the derivative is zero.

Exam Tip: Use trigonometric identities like sum-to-product formulas to simplify derivatives of trigonometric area functions.

Question 21. If a function is defined by \(f(x) = \begin{cases} ax^2 + 3x, & x \le 1 \\ bx + 2, & x > 1 \end{cases}\) is differentiable at each \(x \in R\), then find the values of \(a\) and \(b\).
Answer: For \(f(x)\) to be differentiable at \(x = 1\), it must first be continuous at \(x = 1\).
Left-hand limit at \(x = 1\): \(\lim_{x \to 1^-} (ax^2 + 3x) = a(1)^2 + 3(1) = a + 3\).
Right-hand limit at \(x = 1\): \(\lim_{x \to 1^+} (bx + 2) = b(1) + 2 = b + 2\).
For continuity: \(a + 3 = b + 2 \implies a - b = -1\) --- (Equation 1)
For differentiability at \(x = 1\), Left-hand derivative (LHD) = Right-hand derivative (RHD).
Derivative of left expression: \(\frac{d}{dx}(ax^2 + 3x) = 2ax + 3\). At \(x = 1\), LHD = \(2a + 3\).
Derivative of right expression: \(\frac{d}{dx}(bx + 2) = b\). At \(x = 1\), RHD = \(b\).
For differentiability: \(2a + 3 = b\) --- (Equation 2)
Substitute \(b = 2a + 3\) into Equation 1:
\(a - (2a + 3) = -1 \implies -a - 3 = -1 \implies a = -2\).
Then \(b = 2(-2) + 3 = -1\).
Thus, \(a = -2\) and \(b = -1\).
In simple words: A differentiable function must be continuous first. Then equate left and right derivatives at the junction point.

Exam Tip: Differentiability automatically implies continuity. Always check both continuity and derivative equality at the boundary point.

Question 22. For what choices of \(a\) and \(b\) is the function \(f(x) = \begin{cases} bx^2, & x \le 1 \\ ax - 6, & x > 1 \end{cases}\) is differentiable at \(x = 1\).
Answer: Continuity at \(x = 1\):
\(b(1)^2 = a(1) - 6 \implies b = a - 6\) --- (Equation 1)
Differentiability at \(x = 1\):
LHD = derivative of \(bx^2\) at \(x = 1\) is \(2bx\Big|_{x=1} = 2b\).
RHD = derivative of \(ax - 6\) at \(x = 1\) is \(a\).
Equating LHD and RHD: \(2b = a\) --- (Equation 2)
Substitute \(a = 2b\) into Equation 1:
\(b = 2b - 6 \implies b = 6\).
Then \(a = 2(6) = 12\).
Thus, \(a = 12\) and \(b = 6\).
In simple words: Equate function values at \(x = 1\) for continuity, and equate derivatives for differentiability.

Exam Tip: Set up a simultaneous linear system in \(a\) and \(b\) from the continuity and derivative conditions.

Question 23. Discuss the differentiability of \(f(x) = |x - 1| + |x - 2|\).
Answer: The given function can be rewritten across the critical points \(x = 1\) and \(x = 2\):
\(f(x) = \begin{cases} -(x - 1) - (x - 2) = -2x + 3, & x < 1 \\ (x - 1) - (x - 2) = 1, & 1 \le x \le 2 \\ (x - 1) + (x - 2) = 2x - 3, & x > 2 \end{cases}\)
Let us check differentiability at the critical points \(x = 1\) and \(x = 2\):
At \(x = 1\):
LHD = \(\frac{d}{dx}(-2x + 3) = -2$
RHD = \(\frac{d}{dx}(1) = 0$
Since LHD \(\neq\) RHD (\(-2 \neq 0\)), \(f(x)\) is not differentiable at \(x = 1\).
At \(x = 2\):
LHD = \(\frac{d}{dx}(1) = 0$
RHD = \(\frac{d}{dx}(2x - 3) = 2$
Since LHD \(\neq\) RHD (\(0 \neq 2\)), \(f(x)\) is not differentiable at \(x = 2\).
Thus, \(f(x)\) is differentiable everywhere in \(R\) except at \(x = 1\) and \(x = 2\).
In simple words: Absolute value functions have sharp corners (cusps) at their critical points where derivatives fail to exist.

Exam Tip: Break piecewise absolute functions into intervals before checking left and right derivatives.

Question 24. Discuss differentiability of \(f(x) = x|x|\) at \(x = 0\).
Answer: The function can be defined piecewise as:
\(f(x) = \begin{cases} -x^2, & x < 0 \\ x^2, & x \ge 0 \end{cases}\)
Let us find the left-hand derivative (LHD) at \(x = 0\):
\(\text{LHD} = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^2 - 0}{h} = \lim_{h \to 0^-} (-h) = 0\)
Right-hand derivative (RHD) at \(x = 0\):
\(\text{RHD} = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2 - 0}{h} = \lim_{h \to 0^+} h = 0\)
Since \(\text{LHD} = \text{RHD} = 0\), the function \(f(x) = x|x|\) is differentiable at \(x = 0\), and \(f'(0) = 0\).
In simple words: Even with an absolute value, multiplication by \(x\) smooths out the corner, making it differentiable at the origin.

Exam Tip: Never assume absolute value functions are non-differentiable without testing the limit definition of the derivative.

Question 25. Examine the differentiability of \(f\) at \(x = 2\), where \(f\) is defined by \(f(x) = \begin{cases} x^2, & 0 \le x < 2 \\ 3x - 2, & 2 \le x \le 3 \end{cases}\)
Answer: First, check continuity at \(x = 2\):
\(\lim_{x \to 2^-} f(x) = 2^2 = 4\)
\(\lim_{x \to 2^+} f(x) = 3(2) - 2 = 4\)
Since Left Limit = Right Limit = Value \((4)\), the function is continuous at \(x = 2\).
Now, check differentiability at \(x = 2\):
LHD = derivative of \(x^2\) at \(x = 2\) is \(2x\Big|_{x=2} = 4\).
RHD = derivative of \(3x - 2\) at \(x = 2\) is \(3\).
Since \(\text{LHD} \neq \text{RHD}\) (\(4 \neq 3\)), the function \(f(x)\) is not differentiable at \(x = 2\).
In simple words: Although the function is continuous, the slopes from left and right do not match at \(x = 2\).

Exam Tip: Continuity is a necessary condition for differentiability, but passing continuity does not guarantee differentiability.