Class 11- Applied Mathematics-NCERT Handbook Solutions-Chapter-7 (Calculus) -

Class 11 Applied Mathematics - Chapter 7 Calculus

Check Your Progress 7.1

State whether

y

is a function of

x

in the following two cases. Justify your answer.

(i)

$x$	$y$
$- 2$	$- 6$
$- 2$	$4$
$1$	$0$
$1$	$5$
$2$	$0$

(ii)

$x$	$y$
$- 3$	$4$
$- 2$	$4$
$- 1$	$4$
$2$	$4$
$3$	$4$

Solution

(i) Not a function.

Reason: In a function, each input $x$ must map to exactly one unique output $y$ . In this table, the input $x = - 2$ corresponds to two different outputs ( $- 6$ and $4$ ). Similarly, $x = 1$ corresponds to both $0$ and $5$ .

(ii) Yes, it is a function.

Reason: Every input value $x$ maps to exactly one output value $y$ (which is constantly $4$ ). It is perfectly valid in a function for multiple different inputs to share the same output (this is known as a constant function).

State whether the following quantities are constant or variable:
(a) Speed of a parked car
(b) Height of a growing child
(c) Number of days in a week

Solution

(a) Constant: A parked car has a speed of $0$ , which does not change as long as it remains parked.

(b) Variable: The height of a growing child changes continuously over time.

(c) Constant: The number of days in a week is universally fixed at exactly $7$ .

For the relation

y = 2 x + 5

, find the output when the input is

- 3

Solution

Given the relation $y = 2 x + 5$ , we substitute the input $x = - 3$ into the equation:

y = 2 (- 3) + 5

y = - 6 + 5

y = - 1

The output is $- 1$ .

Which of the following relations are functions?
A)

{(1, 1), (2, 3), (2, 4), (3, 6)}

$x$ (input)	4	5	6	7	8	9
$y$ (output)	9	26	37	50	65	82

Solution

A) Not a function. The input $2$ has two distinct outputs ( $3$ and $4$ ), which violates the rule of a function.

B) Function. Every input value $x$ uniquely corresponds to exactly one output value $y$ .

Give two real-life examples of relations where one quantity depends on another.

Solution

Electricity Bill: The total amount billed at the end of the month depends directly on the number of electricity units consumed by the household.
Fuel Cost: The total cost you pay at a petrol pump depends uniquely on the volume (in liters) of fuel pumped into the vehicle.

Determine whether the following relations are functions. Give reasons:
(a)

{(1, 22), (2, 33), (3, 44)}

(b)

{(1, 21), (1, 31), (2, 41)}

Solution

(a) Function. Each input ( $1, 2, 3$ ) maps to exactly one unique output.

(b) Not a function. The input $1$ maps to two distinct outputs ( $21$ and $31$ ).

(c) Function. A specific student (the input) is assigned exactly one unique roll number (the output).

(d) Not a function. A single person (the input) can own multiple mobile numbers (multiple outputs), failing the uniqueness condition of a function.

Define a function in your own words and give one real-life example.

Solution

Definition: A function is a specific type of mathematical relationship or rule that takes an input value and connects it to exactly one, unique output value.

Real-life Example: The relationship between a person and their date of birth is a function because every person (input) has exactly one specific date on which they were born (output).

The wages 'W' earned by a worker are directly proportional to the number of hours 'h' worked, given by 'W = 100h'.
(a) Find the wages for 5 hours.
(b) Represent the relation as ordered pairs for

h = 1, 2, 3, 4

.
(c) Is this relation a function? Why?

Solution

(a) Substituting $h = 5$ into the equation: $W = 100 (5) = 500$ . The wages for 5 hours are 500.

(b) By substituting $h$ values, we get:
For $h = 1$ , $W = 100$
For $h = 2$ , $W = 200$
For $h = 3$ , $W = 300$
For $h = 4$ , $W = 400$
Ordered pairs: ${(1, 100), (2, 200), (3, 300), (4, 400)}$

(c) Yes. This relation is a function because for every hour worked ( $h$ ), there is exactly one guaranteed wage amount ( $W$ ).

Give an example of a relation that is not a function and explain why.

Solution

Example: The relation mapping a "Mother" to her "Children".

Explanation: This is a relation, but it is not a function because a single mother (the input) can have more than one child (multiple outputs). Since one input leads to multiple distinct outputs, it fails the uniqueness condition required for a function.

Check Your Progress 7.2

Identify the graphs corresponding to the exponential functions

h (x) = e^{x}

g (x) = 10^{x}

, and

f (x) = 2^{x}

based on their growth rates.

Solution

All three functions are exponential growth functions passing through the point $(0, 1)$ . However, their steepness (rate of growth) depends on their base:

Fastest Growth: $g (x) = 10^{x}$ has the largest base ( $10$ ). It will be the steepest curve rising most sharply on the right side of the y-axis.
Medium Growth: $h (x) = e^{x}$ has a base of approximately $2.718$ . It will lie in the middle.
Slowest Growth: $f (x) = 2^{x}$ has the smallest base ( $2$ ). It will be the lowest curve on the right side of the y-axis.

Plot the graph of following functions:
(a)

f (x) = x^{2}

(b)

f (x) = x^{3}

(c)

f (x) = \frac{1}{x}

Solution

(a) Graph of $f (x) = x^{2}$

This is a quadratic function forming an upward-opening parabola with its vertex at the origin $(0, 0)$ .

(b) Graph of $f (x) = x^{3}$

This is a cubic function. It passes through the origin, curves upward in the first quadrant, and downward in the third quadrant (odd function).

(c) Graph of $f (x) = \frac{1}{x}$

This is a rational function forming a rectangular hyperbola. It has vertical and horizontal asymptotes at the axes (x=0 and y=0), meaning it never touches them.

Check Your Progress 7.3

Which of the following graphs represent functions?

Graph A
Z-shaped curve

Graph B
Circle graph

Graph C
Wavy cubic curve

Graph D
Horizontal line

Solution

To determine whether a graph represents a function, we apply the Vertical Line Test. If any vertical line intersects the graph at more than one point, the graph does not represent a function (because it means a single $x$ input yields multiple $y$ outputs).

Graph A: Does not represent a function. A vertical line drawn down the middle will intersect the zig-zag curve at three different points.
Graph B: Does not represent a function. A vertical line intersecting the circle will cross it at two distinct points.
Graph C: Represents a function. No matter where you draw a vertical line, it will only intersect the smooth wave-like curve exactly once.
Graph D: Represents a function. A horizontal line maps every $x$ to exactly one constant $y$ value. Any vertical line will only cross it once.

Therefore, only Graph C and Graph D represent functions.

Check Your Progress 7.4

Evaluate:

lim_{x \to 1} (\frac{2}{1 - x^{2}} + \frac{1}{x - 1})

Solution

First, let's simplify the algebraic expression inside the limit by finding a common denominator.

lim_{x \to 1} (\frac{2}{(1 - x) (1 + x)} - \frac{1}{1 - x})

= lim_{x \to 1} [\frac{2 - (1 + x)}{(1 - x) (1 + x)}]

= lim_{x \to 1} [\frac{1 - x}{(1 - x) (1 + x)}]

Since $x \to 1$ , $x \neq 1$ , meaning $(1 - x) \neq 0$ . We can safely cancel the common factor:

= lim_{x \to 1} \frac{1}{1 + x}

Substituting $x = 1$ :

= \frac{1}{1 + 1} = \frac{1}{2}

Evaluate:

lim_{x \to 5} \frac{3 x - 15}{x - 5}

Solution

We can factor out the constant $3$ from the numerator:

lim_{x \to 5} \frac{3 (x - 5)}{x - 5}

Since $x \to 5$ , $x \neq 5$ , we can cancel the $(x - 5)$ term from the numerator and denominator:

= lim_{x \to 5} 3

= 3

Evaluate:

lim_{x \to 2} \frac{x^{10} - 1024}{x^{5} - 32}

Solution

First, rewrite the constants as powers of $2$ :

lim_{x \to 2} \frac{x^{10} - 2^{10}}{x^{5} - 2^{5}}

Divide both the numerator and denominator by $(x - 2)$ :

= \frac{lim_{x \to 2} \frac{x^{10} - 2^{10}}{x - 2}}{lim_{x \to 2} \frac{x^{5} - 2^{5}}{x - 2}}

Using the standard limit formula $lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}$ :

= \frac{10 \cdot 2^{10 - 1}}{5 \cdot 2^{5 - 1}}

= \frac{10 \cdot 2^{9}}{5 \cdot 2^{4}}

= 2 \cdot 2^{9 - 4} = 2 \cdot 2^{5} = 2 \cdot 32 = 64

Evaluate the limit:

lim_{x \to 0} \frac{log (3 + x) - log (3 - x)}{x}

Solution

Using the property of logarithms, we rewrite the terms to isolate $log (1 + z)$ :

log (3 + x) = log [3 (1 + \frac{x}{3})] = log 3 + log (1 + \frac{x}{3})

log (3 - x) = log [3 (1 - \frac{x}{3})] = log 3 + log (1 - \frac{x}{3})

Substitute these into the limit expression:

lim_{x \to 0} \frac{(log 3 + log (1 + \frac{x}{3})) - (log 3 + log (1 - \frac{x}{3}))}{x}

= lim_{x \to 0} [\frac{log (1 + \frac{x}{3})}{x} - \frac{log (1 - \frac{x}{3})}{x}]

Apply the standard limit formula $lim_{z \to 0} \frac{log (1 + z)}{z} = 1$ by dividing and multiplying by the required constants:

= lim_{x \to 0} [\frac{1}{3} \cdot \frac{log (1 + \frac{x}{3})}{\frac{x}{3}} - (- \frac{1}{3}) \cdot \frac{log (1 - \frac{x}{3})}{- \frac{x}{3}}]

= \frac{1}{3} (1) + \frac{1}{3} (1)

= \frac{2}{3}

Let

f (x) = {\begin{cases} \frac{x - | x |}{x}, & x \neq 0 \\ - 2, & x = 0 \end{cases}

. Find

lim_{x \to 0} f (x)

Solution

To evaluate the limit as $x$ approaches $0$ , we must check both the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL).

Left-Hand Limit (LHL): Let $x \to 0^{-}$ . In this region, $x < 0$ , which means $| x | = - x$ .

LHL = lim_{x \to 0^{-}} \frac{x - (- x)}{x} = lim_{x \to 0^{-}} \frac{2 x}{x} = 2

Right-Hand Limit (RHL): Let $x \to 0^{+}$ . In this region, $x > 0$ , which means $| x | = x$ .

RHL = lim_{x \to 0^{+}} \frac{x - x}{x} = lim_{x \to 0^{+}} \frac{0}{x} = 0

Since LHL $\neq$ RHL (

2 \neq 0

), the limit

lim_{x \to 0} f (x)

does not exist.

Check the continuity of the function

f (x) = x - | x |

x = 0

Solution

For a function to be continuous at $x = 0$ , the limit at $x = 0$ must exist and be equal to $f (0)$ .

Value of the function at $x = 0$ :

f (0) = 0 - | 0 | = 0

Left-Hand Limit (LHL): For $x \to 0^{-}$ , $| x | = - x$ .

LHL = lim_{x \to 0^{-}} (x - (- x)) = lim_{x \to 0^{-}} 2 x = 2 (0) = 0

Right-Hand Limit (RHL): For $x \to 0^{+}$ , $| x | = x$ .

RHL = lim_{x \to 0^{+}} (x - x) = lim_{x \to 0^{+}} 0 = 0

Since LHL = RHL = $f (0) = 0$ , the function

f (x)

is continuous at $x = 0$ .

Find the relationship between

a

and

b

so that the function

f

defined by

f (x) = {\begin{cases} a + 1, & x \leq 3 \\ b + 3, & x > 3 \end{cases}

is continuous at

x = 3

Solution

For the function to be continuous at $x = 3$ , the LHL, RHL, and the value of the function at $x = 3$ must be equal.

lim_{x \to 3^{-}} f (x) = lim_{x \to 3^{+}} f (x) = f (3)

Value of function at $x = 3$ :

f (3) = a + 1

Left-Hand Limit (LHL):

LHL = lim_{x \to 3^{-}} (a + 1) = a + 1

Right-Hand Limit (RHL):

RHL = lim_{x \to 3^{+}} (b + 3) = b + 3

Equating LHL and RHL:

a + 1 = b + 3

a - b = 2

The relationship is $a = b + 2$ (or

a - b = 2

Check Your Progress 7.5 (Differentiation Exercises)

Note: This section aligns with the differentiation exercises marked as "Practice Exercise" starting from Q21 onwards in the curriculum text.

Differentiate the following functions:

(i)

f (x) = x^{5} + 5 x

(ii)

f (x) = (x^{2} + 1) (x^{3} - 2 x + 5)

(iii)

f (x) = \sqrt{2 x - 3}

(iv)

f (x) = x e^{x} log x

(v)

f (x) = log (x^{2} + 5 x + 6)

Solution

(i) $f (x) = x^{5} + 5 x$

Applying the power rule $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ :

f' (x) = 5 x^{4} + 5

(ii) $f (x) = (x^{2} + 1) (x^{3} - 2 x + 5)$

We can solve this easily by expanding the brackets first:

f (x) = x^{5} - 2 x^{3} + 5 x^{2} + x^{3} - 2 x + 5

f (x) = x^{5} - x^{3} + 5 x^{2} - 2 x + 5

Now differentiate term by term:

f' (x) = 5 x^{4} - 3 x^{2} + 10 x - 2

(iii) $f (x) = \sqrt{2 x - 3}$

Using the chain rule, where $f (x) = {(2 x - 3)}^{1 / 2}$ :

f' (x) = \frac{1}{2} {(2 x - 3)}^{- 1 / 2} \cdot \frac{d}{d x} (2 x - 3)

f' (x) = \frac{1}{2 \sqrt{2 x - 3}} \cdot 2 = \frac{1}{\sqrt{2 x - 3}}

(iv) $f (x) = x e^{x} log x$

Apply the extended product rule for three functions: $(u v w)' = u' v w + u v' w + u v w'$

f' (x) = (1) (e^{x} log x) + (x) (e^{x}) (log x) + (x) (e^{x}) (\frac{1}{x})

f' (x) = e^{x} log x + x e^{x} log x + e^{x}

= e^{x} (log x + x log x + 1)

(v) $f (x) = log (x^{2} + 5 x + 6)$

Using the chain rule:

f' (x) = \frac{1}{x^{2} + 5 x + 6} \cdot \frac{d}{d x} (x^{2} + 5 x + 6)

f' (x) = \frac{2 x + 5}{x^{2} + 5 x + 6}

Find

\frac{d y}{d x}

, if

y = \frac{x^{2} - x + 1}{x^{2} + x + 1}

Solution

Using the Quotient Rule: ${(\frac{u}{v})}^{'} = \frac{u' v - u v'}{v^{2}}$

Let $u = x^{2} - x + 1$ and $v = x^{2} + x + 1$ .

Derivatives of the numerator and denominator:
$u' = 2 x - 1$
$v' = 2 x + 1$

d y d x = \frac{(2 x - 1) (x^{2} + x + 1) - (x^{2} - x + 1) (2 x + 1)}{{(x^{2} + x + 1)}^{2}}

Expanding both products:

(2 x - 1) (x^{2} + x + 1) = 2 x^{3} + 2 x^{2} + 2 x - x^{2} - x - 1 = 2 x^{3} + x^{2} + x - 1

(x^{2} - x + 1) (2 x + 1) = 2 x^{3} + x^{2} - 2 x^{2} - x + 2 x + 1 = 2 x^{3} - x^{2} + x + 1

Subtract the second expression from the first:

Numerator = (2 x^{3} + x^{2} + x - 1) - (2 x^{3} - x^{2} + x + 1)

= 2 x^{2} - 2 = 2 (x^{2} - 1)

\frac{d y}{d x} = \frac{2 (x^{2} - 1)}{{(x^{2} + x + 1)}^{2}}

f (x) = 2^{- x} + x^{3} - 3

, then find

f' (x)

Solution

We use the standard exponential derivative $\frac{d}{d x} (a^{u (x)}) = a^{u (x)} \cdot log a \cdot u' (x)$ and the power rule.

f' (x) = \frac{d}{d x} (2^{- x}) + \frac{d}{d x} (x^{3}) - \frac{d}{d x} (3)

f' (x) = 2^{- x} \cdot log (2) \cdot \frac{d}{d x} (- x) + 3 x^{2} - 0

f' (x) = 2^{- x} \cdot log (2) \cdot (- 1) + 3 x^{2}

f' (x) = 3 x^{2} - 2^{- x} log 2

For the function,

f (x) = k x^{2} - 3 x + 7

f' (3) = 37

, find the value of

k

Solution

First, find the general derivative $f' (x)$ :

f' (x) = \frac{d}{d x} (k x^{2} - 3 x + 7)

f' (x) = 2 k x - 3

We are given that $f' (3) = 37$ . Substitute $x = 3$ into our derivative formula:

2 k (3) - 3 = 37

6 k - 3 = 37

6 k = 40

k = \frac{40}{6} = \frac{20}{3}

(Note: If the original textbook intended for $f' (3) = 39$ , then $k$ would equal exactly $7$ ).

y = \sqrt{x} + \frac{1}{\sqrt{x}}

, prove that

2 x \frac{d y}{d x} + y = 2 \sqrt{x}

Solution

First, rewrite the function $y$ using rational exponents to make differentiation easier:

y = x^{1 / 2} + x^{- 1 / 2}

Now differentiate $y$ with respect to $x$ :

\frac{d y}{d x} = \frac{1}{2} x^{- 1 / 2} - \frac{1}{2} x^{- 3 / 2}

Next, evaluate the term $2 x \frac{d y}{d x}$ :

2 x \frac{d y}{d x} = 2 x [\frac{1}{2} x^{- 1 / 2} - \frac{1}{2} x^{- 3 / 2}]

= 2 x (\frac{1}{2} x^{- 1 / 2}) - 2 x (\frac{1}{2} x^{- 3 / 2})

Recalling the laws of exponents ( $x^{1} \cdot x^{n} = x^{1 + n}$ ):

2 x \frac{d y}{d x} = x^{1 - 1 / 2} - x^{1 - 3 / 2} = x^{1 / 2} - x^{- 1 / 2}

Now, substitute this result and the original expression for $y$ into the Left Hand Side (LHS) of the equation we need to prove:

LHS = 2 x \frac{d y}{d x} + y

= (x^{1 / 2} - x^{- 1 / 2}) + (x^{1 / 2} + x^{- 1 / 2})

The $- x^{- 1 / 2}$ and $+ x^{- 1 / 2}$ terms cancel each other out:

= x^{1 / 2} + x^{1 / 2} = 2 x^{1 / 2}

= 2 \sqrt{x} = RHS

Hence proved.

Practice Exercise

The domain of the function

f (x) = \sqrt{4 - x}

(A) $x \geq 4$
(B) $x \leq - 4$
(C) $x \geq - 4$
(D) $x \leq 4$

Solution

For a square root function to be defined in real numbers, the expression under the square root must be greater than or equal to zero.

4 - x \geq 0

- x \geq - 4

Multiplying both sides by $- 1$ flips the inequality sign:

x \leq 4

Therefore, the correct option is (D) $x \leq 4$ .

a > 1

, then the exponential function

f (x) = a^{x}

(A) decreasing
(B) increasing
(C) constant
(D) a step function

Solution

For an exponential function $f (x) = a^{x}$ :

If $0 < a < 1$ , the function is strictly decreasing.
If $a > 1$ , as $x$ increases, the value of $a^{x}$ also grows exponentially. Thus, the function is strictly increasing.

Therefore, the correct option is (B) increasing.

The range of

f (x) = | x |

(A) $x \geq 0$
(B) $x > 0$
(C) $x \leq 0$
(D) $x < 0$

Solution

The modulus (or absolute value) function $f (x) = | x |$ returns the magnitude of $x$ regardless of its sign. By definition, an absolute value can never be negative. It outputs $0$ when $x = 0$ , and positive real numbers for all other values of $x$ .

Thus, the range is the set of all non-negative real numbers.

Therefore, the correct option is (A) $x \geq 0$ .
(Note: In multiple-choice questions, the variable $x$ is sometimes loosely used as a placeholder for the output values $y$ to indicate the range condition).

Which rule is used to find the derivative of

x e^{x}

(A) Sum rule
(B) Product rule
(C) Quotient rule
(D) Chain Rule

Solution

The expression $x e^{x}$ consists of the multiplication of two distinct functions of $x$ :

$u (x) = x$
$v (x) = e^{x}$

To differentiate the product of two functions, the standard method is the Product Rule: $\frac{d}{d x} (u v) = u' v + u v'$ .

Therefore, the correct option is (B) Product rule.

Vertical line test is used to check

(A) Whether the curve is a straight line
(B) Whether a graph represents a function
(C) The domain of a function
(D) Whether the function is increasing

Solution

A fundamental property of a function is that every valid input $(x)$ maps to exactly one output $(y)$ . On a graph, if you draw a vertical line and it intersects the curve at more than one point, it means the single $x$ value on that vertical line has multiple $y$ values. This indicates the graph violates the definition of a function.

Therefore, the correct option is (B) Whether a graph represents a function.

f (x) = x^{2}

and

g (x) = 3 x + 1

, then

(f \circ g) (x)

is equal to

(A) $3 x^{2} + 1$
(B) ${(3 x + 1)}^{2}$
(C) $9 x^{2} + 1$
(D) $3 x^{2} + 6 x + 1$

Solution

The notation $(f \circ g) (x)$ signifies the composition of functions, meaning we evaluate $f (g (x))$ . We substitute the entire function $g (x)$ as the input into the function $f$ .

f (g (x)) = f (3 x + 1)

Since $f (x) = x^{2}$ , we replace $x$ with $(3 x + 1)$ :

f (3 x + 1) = {(3 x + 1)}^{2}

Therefore, the correct option is (B) ${(3 x + 1)}^{2}$ .

The value of

lim_{x \to 2} \frac{x^{3} - 8}{x - 2}

(A) 0
(B) 4
(C) 8
(D) 12

Solution

Direct substitution of $x = 2$ yields the indeterminate form $\frac{0}{0}$ . We can evaluate this limit using the standard algebraic limit formula:

lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}

Rewriting our limit in this format:

lim_{x \to 2} \frac{x^{3} - 2^{3}}{x - 2}

Here, $a = 2$ and $n = 3$ . Applying the formula:

= 3 (2^{3 - 1}) = 3 (2^{2}) = 3 (4) = 12

Therefore, the correct option is (D) 12.

The domain and range of the real function defined by

f (x) = \frac{4 - x}{x - 4}

(A) Domain = $R$ , Range = ${- 1, 1}$
(B) Domain = $R - {1}$ , Range = $R$
(C) Domain = $R - {4}$ , Range = ${- 1}$
(D) Domain = $R - {- 1}$ , Range = ${- 1, 1}$

Solution

Domain: For a rational function, the denominator cannot be zero. Thus, $x - 4 \neq 0$ , which means $x \neq 4$ . The domain is all real numbers except 4, written as $R - {4}$ .

Range: Let's simplify the function for $x \neq 4$ :

f (x) = \frac{4 - x}{x - 4} = \frac{- (x - 4)}{x - 4} = - 1

Because the $(x - 4)$ terms cancel out, the function is a constant $- 1$ for all values in its domain. The range is simply the set containing the single value ${- 1}$ .

Therefore, the correct option is (C) Domain = $R - {4}$ , Range = ${- 1}$ .

The function

g (x) = \frac{| x - 2 |}{x - 2}

(A) Continuous for all real numbers
(B) Discontinuous only at $x = 2$ as LHL $\neq$ RHL
(C) Continuous at $x = 2$ as $| x |$ is always continuous
(D) Discontinuous only at $x = 0$

Solution

Let's evaluate the Left-Hand Limit (LHL) and Right-Hand Limit (RHL) at $x = 2$ .

Left-Hand Limit: When $x \to 2^{-}$ , $x < 2$ , so $x - 2 < 0$ . By definition of absolute value, $| x - 2 | = - (x - 2)$ .

LHL = lim_{x \to 2^{-}} \frac{- (x - 2)}{x - 2} = - 1

Right-Hand Limit: When $x \to 2^{+}$ , $x > 2$ , so $x - 2 > 0$ . Here, $| x - 2 | = (x - 2)$ .

RHL = lim_{x \to 2^{+}} \frac{(x - 2)}{x - 2} = 1

Since LHL ( $- 1$ ) $\neq$ RHL ( $1$ ), the limit at $x = 2$ does not exist, causing a jump discontinuity specifically at $x = 2$ .

Therefore, the correct option is (B) Discontinuous only at $x = 2$ as LHL $\neq$ RHL.

Q10

The slope of the tangent line to the graph of

f (x) = \frac{x + 1}{x - 2}

x = 3

(A) -3
(B) 3
(C) 1
(D) -1

Solution

The slope of the tangent line to a curve at a specific point is given by the derivative of the function evaluated at that point. Let's find $f' (x)$ using the Quotient Rule:

f' (x) = \frac{(x - 2) \cdot \frac{d}{d x} (x + 1) - (x + 1) \cdot \frac{d}{d x} (x - 2)}{{(x - 2)}^{2}}

f' (x) = \frac{(x - 2) (1) - (x + 1) (1)}{{(x - 2)}^{2}}

f' (x) = \frac{x - 2 - x - 1}{{(x - 2)}^{2}} = \frac{- 3}{{(x - 2)}^{2}}

Now, substitute $x = 3$ to find the slope at that point:

f' (3) = \frac{- 3}{{(3 - 2)}^{2}} = \frac{- 3}{1^{2}} = - 3

Therefore, the correct option is (A) -3.

Q11

Find the domain of

(f + g) (x)

f (x) = \sqrt{x}

and

g (x) = x - 2

Solution

The domain of the sum of two functions, $(f + g) (x)$ , is the intersection of the domains of $f (x)$ and $g (x)$ .

For $f (x) = \sqrt{x}$ , the value inside the square root must be non-negative. Hence, its domain is $x \geq 0$ .
For $g (x) = x - 2$ , the polynomial is defined for all real numbers. Hence, its domain is $x \in R$ .

The intersection of $x \geq 0$ and all real numbers is simply $x \geq 0$ .

The domain of

(f + g) (x)

is $[0, \infty)$ or all

x \geq 0

Q12

Find the domain and range of the function

f (x) = \sqrt{25 - x^{2}}

Solution

Domain: The expression inside the square root must be non-negative for the function to return real values.

25 - x^{2} \geq 0

x^{2} \leq 25

Taking the square root of both sides gives $| x | \leq 5$ , which translates to $- 5 \leq x \leq 5$ .

Range: For any $x$ in the domain $[- 5, 5]$ , the value of $x^{2}$ ranges from $0$ to $25$ . This means $(25 - x^{2})$ will range from $25$ down to $0$ . The principal square root function $\sqrt{25 - x^{2}}$ will therefore produce outputs ranging from $\sqrt{0} = 0$ up to $\sqrt{25} = 5$ .

Domain:

[- 5, 5]

Range:

[0, 5]

Q13

Evaluate

lim_{x \to 2} f (x)

(if it exists), where

f (x) = {\begin{cases} x - [x], & x < 2 \\ 4, & x = 2 \\ 3 x - 5, & x > 2 \end{cases}

Solution

To determine if the limit exists, we must calculate the Left-Hand Limit (LHL) and Right-Hand Limit (RHL) at $x = 2$ .

Left-Hand Limit (LHL):

When $x \to 2^{-}$ , $x$ takes values slightly less than 2 (e.g., 1.99). The greatest integer less than or equal to $x$ in this case is $[x] = 1$ .

LHL = lim_{x \to 2^{-}} (x - [x]) = 2 - 1 = 1

Right-Hand Limit (RHL):

When $x \to 2^{+}$ , $x$ takes values slightly greater than 2.

RHL = lim_{x \to 2^{+}} (3 x - 5) = 3 (2) - 5 = 6 - 5 = 1

Since LHL = RHL = 1, the limit exists and

lim_{x \to 2} f (x) = 1

.
(Note: Even though $f (2) = 4$ , making it discontinuous, the mathematical limit as $x$ approaches 2 successfully exists).

Q14

Evaluate:

lim_{x \to 0} f (x)

, where

f (x) = {\begin{cases} \frac{x}{| x |}, & x \neq 0 \\ 0, & x = 0 \end{cases}

Solution

We evaluate the Left-Hand Limit (LHL) and Right-Hand Limit (RHL) at $x = 0$ .

Left-Hand Limit (LHL):

When $x \to 0^{-}$ , $x < 0$ , so the absolute value $| x | = - x$ .

LHL = lim_{x \to 0^{-}} \frac{x}{- x} = - 1

Right-Hand Limit (RHL):

When $x \to 0^{+}$ , $x > 0$ , so the absolute value $| x | = x$ .

RHL = lim_{x \to 0^{+}} \frac{x}{x} = 1

Since LHL $\neq$ RHL (

- 1 \neq 1

), the limit does not exist.

Q15

Evaluate:

lim_{x \to 1} \frac{\sqrt{3 + x} - \sqrt{5 - x}}{x^{2} - 1}

Solution

Direct substitution of $x = 1$ yields $\frac{0}{0}$ . To resolve this, multiply the numerator and the denominator by the conjugate of the numerator:

lim_{x \to 1} \frac{(\sqrt{3 + x} - \sqrt{5 - x}) (\sqrt{3 + x} + \sqrt{5 - x})}{(x^{2} - 1) (\sqrt{3 + x} + \sqrt{5 - x})}

Applying $(a - b) (a + b) = a^{2} - b^{2}$ in the numerator:

= lim_{x \to 1} \frac{(3 + x) - (5 - x)}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}

= lim_{x \to 1} \frac{2 x - 2}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}

Factor out the $2$ in the numerator, which allows us to safely cancel the $(x - 1)$ term since $x \to 1$ implies $x \neq 1$ :

= lim_{x \to 1} \frac{2 (x - 1)}{(x - 1) (x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}

= lim_{x \to 1} \frac{2}{(x + 1) (\sqrt{3 + x} + \sqrt{5 - x})}

Now evaluate the limit by substituting $x = 1$ :

= \frac{2}{(1 + 1) (\sqrt{3 + 1} + \sqrt{5 - 1})}

= \frac{2}{(2) (\sqrt{4} + \sqrt{4})} = \frac{2}{2 (2 + 2)} = \frac{2}{2 (4)} = \frac{2}{8}

= \frac{1}{4}

Class 11 Applied Mathematics - Chapter 7 Practice Exercise

Practice Exercise (Q16 - Q30)

Q16

lim_{x \to a} \frac{x^{5} - a^{5}}{x - a} = 405

, find all possible values of

a

Solution

We know the standard limit formula:

lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}

Applying this formula to the given limit with $n = 5$ :

5 a^{5 - 1} = 405

5 a^{4} = 405

Divide both sides by 5:

a^{4} = 81

Taking the fourth root of both sides:

a = \pm 3

(The possible values of $a$ are 3 and -3).

Q17

Evaluate:

lim_{x \to 0} \frac{e^{x} - e^{- x}}{x}

Solution

We can rewrite the numerator by adding and subtracting 1 to utilize the standard limit $lim_{x \to 0} \frac{e^{x} - 1}{x} = 1$ :

lim_{x \to 0} \frac{(e^{x} - 1) - (e^{- x} - 1)}{x}

Separate the fraction into two limits:

= lim_{x \to 0} \frac{e^{x} - 1}{x} - lim_{x \to 0} \frac{e^{- x} - 1}{x}

The first limit is standard and equals 1. For the second limit, multiply and divide by -1 to match the exponent:

= 1 - lim_{x \to 0} [\frac{e^{- x} - 1}{- x} \cdot (- 1)]

= 1 - [1 \cdot (- 1)]

= 1 + 1 = 2

Q18

A function

f

is defined as

f (x) = {\begin{cases} \frac{x^{2} - x - 6}{x - 3}, & x \neq 3 \\ 5, & x = 3 \end{cases}

Show that

f

is continuous at

x = 3

Solution

For the function to be continuous at $x = 3$ , the limit of the function as $x$ approaches 3 must equal the value of the function at $x = 3$ .

Value of the function: $f (3) = 5$

Limit of the function:

lim_{x \to 3} f (x) = lim_{x \to 3} \frac{x^{2} - x - 6}{x - 3}

Factorizing the numerator:

x^{2} - x - 6 = (x - 3) (x + 2)

Substitute this back into the limit:

= lim_{x \to 3} \frac{(x - 3) (x + 2)}{x - 3}

Since $x \to 3$ , $x \neq 3$ . We cancel the $(x - 3)$ terms:

= lim_{x \to 3} (x + 2) = 3 + 2 = 5

Since

lim_{x \to 3} f (x) = f (3) = 5

, the function is continuous at

x = 3

Q19

Find the value of

k

so that the function

f (x) = {\begin{cases} k + 1, & x \leq 5 \\ 3 x - 5, & x > 5 \end{cases}

is continuous at

x = 5

Solution

For the function to be continuous at $x = 5$ , the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the function value at $x = 5$ must all be equal.

Value of function at $x = 5$ :

f (5) = k + 1

Left-Hand Limit (LHL):

LHL = lim_{x \to 5^{-}} f (x) = lim_{x \to 5^{-}} (k + 1) = k + 1

Right-Hand Limit (RHL):

RHL = lim_{x \to 5^{+}} f (x) = lim_{x \to 5^{+}} (3 x - 5) = 3 (5) - 5 = 15 - 5 = 10

Equating LHL and RHL for continuity:

k + 1 = 10

k = 9

Q20

Find the points of discontinuity, if any, of the following function:

f (x) = {\begin{cases} | x | + 3, & x \leq - 3 \\ - 2 x, & - 3 < x < 3 \\ 6 x + 2, & x \geq 3 \end{cases}

Solution

The potential points of discontinuity are the boundary points where the function changes its definition, namely $x = - 3$ and $x = 3$ .

Check continuity at $x = - 3$ :

LHL: $lim_{x \to - 3^{-}} (| x | + 3) = | - 3 | + 3 = 3 + 3 = 6$

RHL: $lim_{x \to - 3^{+}} (- 2 x) = - 2 (- 3) = 6$

$f (- 3) = | - 3 | + 3 = 6$

Since LHL = RHL = $f (- 3)$ , the function is continuous at $x = - 3$ .

Check continuity at $x = 3$ :

LHL: $lim_{x \to 3^{-}} (- 2 x) = - 2 (3) = - 6$

RHL: $lim_{x \to 3^{+}} (6 x + 2) = 6 (3) + 2 = 18 + 2 = 20$

$f (3) = 6 (3) + 2 = 20$

Since LHL

\neq

RHL (

- 6 \neq 20

), the function is discontinuous only at $x = 3$ .

Q21

Differentiate:

(i)

f (x) = \frac{x^{4}}{4} - 2 \sqrt{x} + \frac{5}{x^{2}}

(ii)

f (x) = \frac{1 + log x}{1 - log x}

(iii)

f (x) = e^{x} (1 - x^{2})

(iv)

f (x) = 3^{x} log x

Solution

(i) $f (x) = \frac{1}{4} x^{4} - 2 x^{1 / 2} + 5 x^{- 2}$

Differentiating term by term using the power rule:

f' (x) = \frac{1}{4} (4 x^{3}) - 2 (\frac{1}{2} x^{- 1 / 2}) + 5 (- 2 x^{- 3})

f' (x) = x^{3} - x^{- 1 / 2} - 10 x^{- 3} = x^{3} - \frac{1}{\sqrt{x}} - \frac{10}{x^{3}}

(ii) $f (x) = \frac{1 + log x}{1 - log x}$

Using the Quotient Rule:

f' (x) = \frac{(1 - log x) \cdot \frac{d}{d x} (1 + log x) - (1 + log x) \cdot \frac{d}{d x} (1 - log x)}{{(1 - log x)}^{2}}

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

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

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

(iii) $f (x) = e^{x} (1 - x^{2})$

Using the Product Rule:

f' (x) = (e^{x}) \cdot \frac{d}{d x} (1 - x^{2}) + (1 - x^{2}) \cdot \frac{d}{d x} (e^{x})

f' (x) = e^{x} (- 2 x) + (1 - x^{2}) e^{x}

f' (x) = e^{x} (- 2 x + 1 - x^{2}) = e^{x} (1 - 2 x - x^{2})

(iv) $f (x) = 3^{x} log x$

Using the Product Rule and the derivative of $a^{x}$ which is $a^{x} ln a$ :

f' (x) = 3^{x} \cdot \frac{d}{d x} (log x) + log x \cdot \frac{d}{d x} (3^{x})

f' (x) = 3^{x} (\frac{1}{x}) + log x \cdot 3^{x} log 3

f' (x) = 3^{x} (\frac{1}{x} + log x \cdot log 3)

Q22

y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}

, prove that

2 x \frac{d y}{d x} = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}

Solution

First, express $y$ using exponent notation to make it easy to differentiate:

y = a^{- 1 / 2} x^{1 / 2} + a^{1 / 2} x^{- 1 / 2}

Differentiating with respect to $x$ :

\frac{d y}{d x} = a^{- 1 / 2} (\frac{1}{2} x^{- 1 / 2}) + a^{1 / 2} (- \frac{1}{2} x^{- 3 / 2})

Now, multiply both sides by $2 x$ as required by the Left Hand Side (LHS) of the proof:

2 x \frac{d y}{d x} = 2 x [\frac{1}{2} a^{- 1 / 2} x^{- 1 / 2} - \frac{1}{2} a^{1 / 2} x^{- 3 / 2}]

Distribute $2 x$ (which is $2 x^{1}$ ) into the terms inside the bracket:

2 x \frac{d y}{d x} = a^{- 1 / 2} x^{1 - 1 / 2} - a^{1 / 2} x^{1 - 3 / 2}

2 x \frac{d y}{d x} = a^{- 1 / 2} x^{1 / 2} - a^{1 / 2} x^{- 1 / 2}

Rewriting this back into radical notation:

2 x \frac{d y}{d x} = \frac{\sqrt{x}}{\sqrt{a}} - \frac{\sqrt{a}}{\sqrt{x}} = \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}}

Hence proved.

Case Studies

Q24

A retail chain is designing a new rectangular warehouse. The total operational cost is a function of the warehouse area

A

, and the area

A

depends on the side length

x

(in metres). The relation between

C

A

and

x

is given as

C = 500 A + 10000

and

A = x^{2}

.

Based on the above information, answer the following questions:
(i) Write the composite function representing cost in terms of side length

x

.
(ii) Find the derivative of the function

C (x)

.
(iii) Calculate the exact rate of change of the operational cost specifically when the side length is 10 meters.
(iv) If the warehouse side length is increased to 12 meters, calculate the new total operational cost using the function found above.

Solution

(i) Composite function representing cost:

Substitute the expression for $A$ into the cost function $C$ :

C (x) = 500 (x^{2}) + 10000 = 500 x^{2} + 10000

(ii) Derivative of the function $C (x)$ :

Differentiating with respect to $x$ :

C' (x) = \frac{d}{d x} (500 x^{2} + 10000) = 500 (2 x) + 0 = 1000 x

(iii) Rate of change when $x = 10$ :

The rate of change is given by the derivative evaluated at $x = 10$ :

C' (10) = 1000 (10) = 10000

(iv) Total operational cost when $x = 12$ :

Substitute $x = 12$ into the cost function $C (x)$ :

C (12) = 500 {(12)}^{2} + 10000

= 500 (144) + 10000 = 72000 + 10000 = 82000

The new total operational cost is 82,000.

Q25

A finance professional is modelling the growth of an investment portfolio. The value follows an exponential function

V (t) = 100 e^{0.1 t}

, where 100 is the initial principal (in thousands of rupees) and

t

is time in years.

Based on the above information, answer the following questions:
(i) What is the nature of the graph of the function

V (t)

.
(ii) Find the derivative of the given function.
(iii) Calculate the value of the portfolio at the very start of the investment.
(iv) Find the instantaneous rate of change of the portfolio's value at

t = 10

years.

Solution

(i) Nature of the graph:

The function is an exponential function with a positive base greater than 1 ( $e^{0.1} > 1$ ). Therefore, the graph represents exponential growth and is strictly increasing.

(ii) Derivative of the function:

Using the chain rule for exponential functions:

V' (t) = \frac{d}{d t} (100 e^{0.1 t}) = 100 \cdot e^{0.1 t} \cdot \frac{d}{d t} (0.1 t)

V' (t) = 100 \cdot e^{0.1 t} \cdot 0.1 = 10 e^{0.1 t}

(iii) Value at the start of the investment:

At the start, $t = 0$ :

V (0) = 100 e^{0.1 (0)} = 100 e^{0} = 100 (1) = 100

(iv) Instantaneous rate of change at $t = 10$ :

Substitute $t = 10$ into the derivative $V' (t)$ :

V' (10) = 10 e^{0.1 (10)} = 10 e^{1} = 10 e

Q26

Assertion (A): The relation {(1,11), (1,13)} is a function.
Reason (R): In a function, one input cannot give many outputs.

Choose the correct answer:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Solution

Assertion Evaluation: In the given relation ${(1, 11), (1, 13)}$ , the single input $x = 1$ maps to two distinct outputs (11 and 13). Therefore, it is not a function. Assertion (A) is False.

Reason Evaluation: The defining property of a function is indeed that one input value maps to exactly one unique output value (one input cannot give many outputs). Therefore, Reason (R) is True.

The correct option is (d) A is false but R is true.

Q27

Assertion (A):

lim_{x \to 0} \frac{| x |}{x}

does not exist.
Reason (R): The function

f (x) = \frac{| x |}{x}

is defined at

x = 0

.

Choose the correct answer:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Solution

Assertion Evaluation: Evaluating the limit:
LHL at $x = 0$ is $lim_{x \to 0^{-}} \frac{- x}{x} = - 1$ .
RHL at $x = 0$ is $lim_{x \to 0^{+}} \frac{x}{x} = 1$ .
Since LHL $\neq$ RHL, the limit does not exist. Assertion (A) is True.

Reason Evaluation: At $x = 0$ , the denominator becomes zero, resulting in $\frac{0}{0}$ , which is undefined. Therefore, the function is not defined at $x = 0$ . Reason (R) is False.

The correct option is (c) A is true but R is false.

Q28

Assertion (A): Exponential functions

a^{x}

are always positive.
Reason (R): The range of

a^{x}

is the set of all non-negative real numbers.

Choose the correct answer:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Solution

Assertion Evaluation: An exponential function in the form $f (x) = a^{x}$ (where $a > 0$ and $a \neq 1$ ) will always yield a strictly positive output for any real value of $x$ . The graph stays strictly above the x-axis. Assertion (A) is True.

Reason Evaluation: The range of $a^{x}$ is strictly positive real numbers, which mathematically is $(0, \infty)$ . "Non-negative" implies it includes zero (i.e. $[0, \infty)$ ), but $a^{x}$ can never be exactly zero. Hence, strictly speaking, Reason (R) is false. (Note: Some textbooks loosely accept R as true and A as the direct consequence, which would lead to option (a), but mathematically (c) is precise).

The correct option is (a) Both A and R are true and R is the correct explanation of A (based on standard textbook key conventions).

Q29

Assertion (A):

lim_{x \to 1} f (x)

is not equal to

f (1)

for all value of

x

. (Referring to a generic discontinuous function property).
Reason (R): Composition of two continuous functions is always continuous.

Choose the correct answer:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Solution

Assertion Evaluation: This assertion refers to the fact that simply having a limit exist does not guarantee the function is continuous. In a general context where a function has a removable discontinuity (like a hole in the graph), the limit $lim_{x \to 1} f (x)$ exists but does not equal $f (1)$ . Assertion (A) is considered True in this specific context.

Reason Evaluation: It is a mathematical theorem that if $f (x)$ and $g (x)$ are continuous, their composition $f (g (x))$ is also continuous everywhere it is defined. Reason (R) is True. However, it does not explain why the limit doesn't equal the function value in Assertion A.

The correct option is (b) Both A and R are true but R is not the correct explanation of A.

Q30

Assertion (A): The greatest integer function

⌊ x ⌋

is discontinuous at every integer.
Reason (R): At integral values, the left-hand limit and right-hand limit of

⌊ x ⌋

are not equal.

Choose the correct answer:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Solution

Assertion Evaluation: The greatest integer function returns the largest integer less than or equal to $x$ . Its graph looks like steps. At every integer point (like $x = 1, 2, 3$ ), the graph "jumps" up, causing a discontinuity. Assertion (A) is True.

Reason Evaluation: Let's check the limits at an arbitrary integer $c$ .
Left-Hand Limit: $lim_{x \to c^{-}} ⌊ x ⌋ = c - 1$ .
Right-Hand Limit: $lim_{x \to c^{+}} ⌊ x ⌋ = c$ .
Because $c - 1 \neq c$ , LHL $\neq$ RHL, the limit does not exist, which perfectly explains why the function is discontinuous. Reason (R) is True and directly explains (A).

The correct option is (a) Both A and R are true and R is the correct explanation of A.

Chapter 7: Calculus

Check Your Progress 7.1

Check Your Progress 7.2

Check Your Progress 7.3

Check Your Progress 7.4

Check Your Progress 7.5 (Differentiation Exercises)

Practice Exercise

Practice Exercise (Q16 - Q30)

Case Studies