JEE Main 2021 (16 March Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ and $g : R \to R$ be defined as $f (x) = {x + a, ∣ x - 1∣, x < 0 x \geq 0$ and $g (x) = {x + 1, (x - 1)^{2} + b, x < 0 x \geq 0$ where $a, b$ are non-negative real numbers. If $(g \circ f) (x)$ is continuous for all $x \in R$ , then $a + b$ is equal to ____

Q: Wait, why do we check continuity at $x = -a$ and $x = 0$? Why not other points?

Great question! The function $f(x)$ is continuous everywhere except potentially at $x=0$, and $g(x)$ is continuous everywhere except potentially at $x=0$. The composition $g(f(x))$ is continuous everywhere except potentially where $f(x)$ hits the discontinuity point of $g$ (which is $0$) or where $f$ itself is discontinuous. Thus, we only need to check the points where the 'inner' function $f(x)$ crosses $0$ (which happens at $x=-a$) and where $f(x)$ itself changes definition (at $x=0$).

Q: Why did we assume $f(x) < 0$ implies $g(f(x)) = f(x) + 1$?

Because the definition of $g(u)$ is $u+1$ when $u < 0$. If $f(x) < 0$, then $g(f(x))$ must use the first branch of $g$. It is all about tracking the input to the outer function $g$!

Q: Can we just differentiate to check continuity?

No, absolutely not! Differentiability implies continuity, but continuity does not imply differentiability. We need the limit definition, not the derivative. Stick to the limit definition $\lim_{x \to c^-} h(x) = \lim_{x \to c^+} h(x) = h(c)$ to be safe.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. To ensure the composite function

g (f (x))

is continuous everywhere, you must look beyond the individual functions

f (x)

and

g (x)

. The continuity of a composition

g (f (x))

is guaranteed if

f

is continuous and

g

is continuous, but here, the pieces are defined piecewise. You must identify the 'handover' points where

f (x)

crosses the boundary of

g (x)

's definition (which is

0

) and where

f (x)

itself changes its definition (which is

0

Hint 2

Hint 2: The Arsenal. Recall the definition of continuity at a point

c

lim_{x \to c^{-}} h (x) = lim_{x \to c^{+}} h (x) = h (c)

. You need to apply this at the critical points

x = - a

(where

f (x)

changes sign) and

x = 0

(where

f (x)

changes its definition).

Hint 3

Hint 3: The Execution. Start by defining the intervals for

g (f (x))

. For

x < 0

f (x) = x + a

. The function

g (f (x))

changes its rule when

f (x) = 0

, which occurs at

x = - a

. Set up the limit equations:

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

and

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

to solve for

a

and

b

Visualized Solution

Introduction to $g (f (x))$ Continuity

Given: $f (x) = {x + a, ∣ x - 1∣, x < 0 x \geq 0$
Given: $g (x) = {x + 1, (x - 1)^{2} + b, x < 0 x \geq 0$
Condition: $a, b \geq 0$ and $g (f (x))$ is continuous for all $x \in R$ .
Goal: Find the value of $a + b$ .

Structure of $g (f (x))$

$g (f (x)) = {f (x) + 1, (f (x) - 1)^{2} + b, f (x) < 0 f (x) \geq 0$
The definition of $g (f (x))$ depends on the sign of $f (x)$ .
We must analyze $f (x)$ to find where it is negative and where it is non-negative.

Interval 1: $x < - a$

For $x < 0$ , $f (x) = x + a$ .
When is $f (x) < 0$ ?
$x + a < 0 \Rightarrow x < - a$ .
In this region, $g (f (x)) = f (x) + 1 = (x + a) + 1$ .

Interval 2: $- a \leq x < 0$

Still for $x < 0$ , when is $f (x) \geq 0$ ?
$x + a \geq 0 \Rightarrow x \geq - a$ .
So, for $- a \leq x < 0$ , $f (x) \geq 0$ .
Here, $g (f (x)) = (f (x) - 1)^{2} + b = (x + a - 1)^{2} + b$ .

Interval 3: $x \geq 0$

For $x \geq 0$ , $f (x) = ∣ x - 1∣$ .
The absolute value is always non-negative: $∣ x - 1∣ \geq 0$ .
Thus, $f (x) \geq 0$ for all $x \geq 0$ .
Here, $g (f (x)) = (∣ x - 1∣ - 1)^{2} + b$ .

Critical Points $x = - a$ and $x = 0$

The piecewise function $g (f (x))$ changes its definition at two points:
$x = - a$ (where $f (x)$ changes sign).
$x = 0$ (where $f (x)$ changes its definition).
For $g (f (x))$ to be continuous everywhere, it must be continuous at these critical points.

Limits at $x = - a$

Left Hand Limit (LHL) as $x \to - a^{-}$ :
$lim_{x \to - a^{-}} (x + a + 1) = - a + a + 1 = 1$
Right Hand Limit (RHL) as $x \to - a^{+}$ :
$lim_{x \to - a^{+}} ((x + a - 1)^{2} + b) = (- a + a - 1)^{2} + b = 1 + b$

Solving for $b$

For continuity at $x = - a$ , LHL must equal RHL.
$1 = 1 + b$
$b = 0$

Limits at $x = 0$

Left Hand Limit (LHL) as $x \to 0^{-}$ :
$lim_{x \to 0^{-}} ((x + a - 1)^{2} + b) = (0 + a - 1)^{2} + 0 = (a - 1)^{2}$ (since $b = 0$ )
Right Hand Limit (RHL) as $x \to 0^{+}$ :
$lim_{x \to 0^{+}} ((∣ x - 1∣ - 1)^{2} + b) = (∣0 - 1∣ - 1)^{2} + 0 = (1 - 1)^{2} = 0$

Solving for $a$

For continuity at $x = 0$ , LHL must equal RHL.
$(a - 1)^{2} = 0$
Taking the square root: $a - 1 = 0$
$a = 1$

Finding $a + b$

We found $a = 1$ and $b = 0$ .
The question asks for the value of $a + b$ .
$a + b = 1 + 0 = 1$ .
Final Answer: $1$

00:00 / 00:00

The Sigma Insight: Continuity at a Point and in an Interval

Solution Diagram

Analyzing the Setup

We are given two piecewise functions:

f (x) = {x + a, ∣ x - 1∣, x < 0 x \geq 0

g (x) = {x + 1, (x - 1)^{2} + b, x < 0 x \geq 0

Our objective is to determine the values of

a

and

b

such that the composite function

g (f (x))

is continuous for all

x \in R

. The continuity of

g (f (x))

depends on the behavior of

f (x)

relative to the transition point of

g (x)

, which occurs at

0

Mapping the Terrain

We must track the sign of

f (x)

to determine which branch of

g

to apply. For

x < 0

f (x) = x + a

x < - a

, then

f (x) < 0

, and we use the first branch of

g

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

- a \leq x < 0

, then

f (x) \geq 0

, and we use the second branch of

g

g (f (x)) = (f (x) - 1)^{2} + b = (x + a - 1)^{2} + b

For

x \geq 0

f (x) = ∣ x - 1∣

. Since

∣ x - 1∣ \geq 0

, we always use the second branch of

g

g (f (x)) = (∣ x - 1∣ - 1)^{2} + b

The Junctions of Continuity

The function

g (f (x))

potentially changes its definition at

x = - a

and

x = 0

. We enforce continuity at these points by equating the left-hand limit (LHL) and the right-hand limit (RHL).

x = - a

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

RHL = x \to - a^{+} lim ((x + a - 1)^{2} + b) = (- a + a - 1)^{2} + b = 1 + b

Setting

LHL = RHL

gives

1 = 1 + b

, which implies $b = 0$ .

x = 0

LHL = x \to 0^{-} lim ((x + a - 1)^{2} + b) = (a - 1)^{2} + b

RHL = x \to 0^{+} lim ((∣ x - 1∣ - 1)^{2} + b) = (∣0 - 1∣ - 1)^{2} + b = 0 + b

Substituting

b = 0

into the continuity condition

(a - 1)^{2} + 0 = 0

, we find

(a - 1)^{2} = 0

, which implies $a = 1$ .

Final Calculation

Having determined the constants, we calculate the required sum:

a + b = 1 + 0 = 1

The value of

a + b

that ensures the continuity of the composite function

g (f (x))

is $1$ .

Common Traps & FAQs

Wait, why do we check continuity at

x = - a

and

x = 0

? Why not other points?

Great question! The function

f (x)

is continuous everywhere except potentially at

x = 0

, and

g (x)

is continuous everywhere except potentially at

x = 0

. The composition

g (f (x))

is continuous everywhere except potentially where

f (x)

hits the discontinuity point of

g

(which is

0

) or where

f

itself is discontinuous. Thus, we only need to check the points where the 'inner' function

f (x)

crosses

0

(which happens at

x = - a

) and where

f (x)

itself changes definition (at

x = 0

Why did we assume

f (x) < 0

implies

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

Because the definition of

g (u)

u + 1

when

u < 0

. If

f (x) < 0

, then

g (f (x))

must use the first branch of

g

. It is all about tracking the input to the outer function

g

Can we just differentiate to check continuity?

No, absolutely not! Differentiability implies continuity, but continuity does not imply differentiability. We need the limit definition, not the derivative. Stick to the limit definition

lim_{x \to c^{-}} h (x) = lim_{x \to c^{+}} h (x) = h (c)

to be safe.

Enter Numerical Value:

Visualized Solution

Introduction to $g (f (x))$ Continuity

Structure of $g (f (x))$

Interval 1: $x < - a$

Interval 2: $- a \leq x < 0$

Interval 3: $x \geq 0$

Critical Points $x = - a$ and $x = 0$

Limits at $x = - a$

Solving for $b$

Limits at $x = 0$

Solving for $a$

Finding $a + b$

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

Mapping the Terrain

The Junctions of Continuity

Final Calculation

Similar Questions

If $f (x) = {x + a, ∣ x - 4∣, x \leq 0 x > 0$ and $g (x) = {x + 1, (x - 4)^{2} + b, x < 0 x \geq 0$ are continuous on $R$ , then $(g \circ f) (2) + (f \circ g) (- 2)$ is equal to :

Let $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ 2 sin (- \frac{π}{2}), a x^{2} + x + b, sin (π x), if x < - 1 if - 1 \leq x \leq 1 if x > 1$ If $f (x)$ is continuous on $R$ , then $a + b$ equals :

For every integer $n$ , let $a_{n}$ and $b_{n}$ be real numbers. Let function $f : R \to R$ be given by $f (x) = {a_{n} + sin π x, b_{n} + cos π x, for x \in [2 n, 2 n + 1] for x \in (2 n - 1, 2 n)$ for all integers $n$ . If $f$ is continuous, then which of the following hold(s) for all $n$ ?

Let a function $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ sin x - e^{x} a + [- x] 2 x - b if x \leq 0 if 0 < x < 1 if x \geq 1$ Where $[x]$ is the greatest integer less than or equal to $x$ . If $f$ is continuous on $R$ , then $(a + b)$ is equal to:

If $f (x) = {\frac{a ∣ x ∣ + x ^{2} - 2 ( s i n ∣ x ∣ ) ( c o s ∣ x ∣ )}{x}, b, x \neq = 0 x = 0$ is continuous at $x = 0$ , then $a + b$ is equal to

Let $f : (- \frac{π}{4}, \frac{π}{4}) \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ (1 + ∣ sin x ∣)^{\frac{3 a}{∣ s i n x ∣}}, b, e^{c o t 4 x / c o t 2 x}, - \frac{π}{4} < x < 0 x = 0 0 < x < \frac{π}{4}$ If $f$ is continuous at $x = 0$ , then the value of $6 a + b^{2}$ is equal to:

For $a, b > 0$ , let $f (x) = ⎩ ⎨ ⎧ \frac{t a n (( a + 1 ) x ) + b t a n x}{x}, 3, \frac{a x + b ^{2} x ^{2} - a x}{b a x x}, x < 0 x = 0 x > 0$ be a continuous function at $x = 0$ . Then $\frac{b}{a}$ is equal to :

If the function $f (x) = ⎩ ⎨ ⎧ \frac{1}{x} lo g_{e} (\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}), k, \frac{c o s ^{2} x - s i n ^{2} x - 1}{x ^{2} + 1 - 1}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to :

Let $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ [e^{x}], a e^{x} + [x - 1], b + [sin (π x)], [e^{- x}] - c, x < 0 0 \leq x < 1 1 \leq x < 2 x \geq 2$ where $a, b, c \in R$ and $[t]$ denotes greatest integer less than or equal to $t$ . Then, which of the following statements is true ?

Find the values of $a$ and $b$ so that the function $f (x) = ⎩ ⎨ ⎧ x + a 2 sin x, 2 x cot x + b, a cos 2 x - b sin x, 0 \leq x < π /4 π /4 \leq x \leq π /2 π /2 < x \leq π$ is continuous for $0 \leq x \leq π$ .

Enter Numerical Value:

Visualized Solution

Introduction to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(f(x)) Continuity

Structure of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(f(x))

Interval 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<−a

Interval 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −a≤x<0

Interval 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥0

Critical Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Limits at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−a

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Limits at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

Mapping the Terrain

The Junctions of Continuity

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={x+a,∣x−4∣,​x≤0x>0​ and g(x)={x+1,(x−4)2+b,​x<0x≥0​ are continuous on R, then (g∘f)(2)+(f∘g)(−2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined as f(x)=⎩⎨⎧​2sin(−2π​),ax2+x+b,sin(πx),​if x<−1if −1≤x≤1if x>1​ If f(x) is continuous on R, then a+b equals :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={xa∣x∣+x2−2(sin∣x∣)(cos∣x∣)​,b,​x=0x=0​ is continuous at x=0, then a+b is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a,b>0, let f(x)=⎩⎨⎧​xtan((a+1)x)+btanx​,3,bax​xax+b2x2​−ax​​,​x<0x=0x>0​ be a continuous function at x=0. Then ab​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the values of a and b so that the function f(x)=⎩⎨⎧​x+a2​sinx,2xcotx+b,acos2x−bsinx,​0≤x<π/4π/4≤x≤π/2π/2<x≤π​ is continuous for 0≤x≤π.

Introduction to $g (f (x))$ Continuity

Structure of $g (f (x))$

Interval 1: $x < - a$

Interval 2: $- a \leq x < 0$

Interval 3: $x \geq 0$

Critical Points $x = - a$ and $x = 0$

Limits at $x = - a$

Solving for $b$

Limits at $x = 0$

Solving for $a$

Finding $a + b$

If $f (x) = {x + a, ∣ x - 4∣, x \leq 0 x > 0$ and $g (x) = {x + 1, (x - 4)^{2} + b, x < 0 x \geq 0$ are continuous on $R$ , then $(g \circ f) (2) + (f \circ g) (- 2)$ is equal to :

Let $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ 2 sin (- \frac{π}{2}), a x^{2} + x + b, sin (π x), if x < - 1 if - 1 \leq x \leq 1 if x > 1$ If $f (x)$ is continuous on $R$ , then $a + b$ equals :

If $f (x) = {\frac{a ∣ x ∣ + x ^{2} - 2 ( s i n ∣ x ∣ ) ( c o s ∣ x ∣ )}{x}, b, x \neq = 0 x = 0$ is continuous at $x = 0$ , then $a + b$ is equal to

For $a, b > 0$ , let $f (x) = ⎩ ⎨ ⎧ \frac{t a n (( a + 1 ) x ) + b t a n x}{x}, 3, \frac{a x + b ^{2} x ^{2} - a x}{b a x x}, x < 0 x = 0 x > 0$ be a continuous function at $x = 0$ . Then $\frac{b}{a}$ is equal to :

Find the values of $a$ and $b$ so that the function $f (x) = ⎩ ⎨ ⎧ x + a 2 sin x, 2 x cot x + b, a cos 2 x - b sin x, 0 \leq x < π /4 π /4 \leq x \leq π /2 π /2 < x \leq π$ is continuous for $0 \leq x \leq π$ .