JEE Main 2020 (8 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $f : R \to R$ be such that for all $x \in R$ , $(2^{1 + x} + 2^{1 - x})$ , $f (x)$ and $(3^{x} + 3^{- x})$ are in A.P., then the minimum value of $f (x)$ is :

Q: Wait, why can't we just differentiate $f(x)$ to find the minimum?

You absolutely can! Differentiating $f(x)$ and setting $f'(x) = 0$ is a perfectly valid path. However, using the AM-GM inequality is often faster and more elegant for functions involving sums of reciprocals. It saves you from the messy chain rule calculations involving natural logarithms.

Q: Is it possible for $f(x)$ to be less than 3?

No. Because $g(x) \ge 1$ and $h(x) \ge 2$ for all real $x$, their sum $f(x)$ is strictly bounded below by $1+2=3$. Since we found a specific value ($x=0$) where $f(x)=3$, we have mathematically proven that 3 is the absolute minimum.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. When three terms

a, b, c

are in an Arithmetic Progression, they are bound by the rigid geometric constraint that the middle term is the average of the two outer terms. Visualize this as a balance scale where

f (x)

sits perfectly in the center of two exponential growth functions.

Hint 2

Hint 2: The Arsenal. Once you express

f (x)

as the sum of two separate expressions, notice that each expression consists of a variable term and its reciprocal. This is the classic signature of the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Recall that for any positive

a, b

\frac{a + b}{2} \geq ab

Hint 3

Hint 3: The Execution. Start by writing

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

. Now, apply the AM-GM inequality separately to the terms involving base 2 and base 3. Remember, the minimum of the sum is the sum of the minimums only if the equality condition (where the terms are equal) occurs at the same value of

x

Visualized Solution

$The Arithmetic Progression Setup$

Given terms in A.P.: $(2^{1 + x} + 2^{1 - x})$ , $f (x)$ , and $(3^{x} + 3^{- x})$

$Property of an A.P.$

If $a, b, c$ are in A.P., then the middle term is the arithmetic mean:
$b = \frac{a + c}{2}$

$Expressing f (x)$

Substituting the given terms into the A.P. property:
$f (x) = \frac{( 2 ^{1 + x} + 2 ^{1 - x} ) + ( 3 ^{x} + 3 ^{- x} )}{2}$

$Decomposing the Function$

Splitting $f (x)$ into two parts for easier analysis:
$f (x) = \frac{3 ^{x} + 3 ^{- x}}{2} + \frac{2 ^{1 + x} + 2 ^{1 - x}}{2}$
Let $g (x) = \frac{3 ^{x} + 3 ^{- x}}{2}$ and $h (x) = \frac{2 ^{1 + x} + 2 ^{1 - x}}{2}$

$The AM-GM Inequality$

For positive real numbers, Arithmetic Mean $\geq$ Geometric Mean:
$A.M. \geq G.M.$
$\frac{a + b}{2} \geq ab$

$Applying AM-GM to g (x)$

Applying AM-GM to the terms of $g (x)$ :
$\frac{3 ^{x} + 3 ^{- x}}{2} \geq 3^{x} \cdot 3^{- x}$

$Minimum of g (x)$

Since $3^{x} \cdot 3^{- x} = 3^{0} = 1$ :
$g (x) \geq 1$
$g (x) \geq 1$

$Applying AM-GM to h (x)$

Applying AM-GM to the terms of $h (x)$ :
$\frac{2 ^{1 + x} + 2 ^{1 - x}}{2} \geq 2^{1 + x} \cdot 2^{1 - x}$

$Minimum of h (x)$

Simplifying the product: $2^{1 + x} \cdot 2^{1 - x} = 2^{(1 + x) + (1 - x)} = 2^{2} = 4$
$h (x) \geq 4$
$h (x) \geq 2$

$Combining the Inequalities$

Since $f (x) = g (x) + h (x)$ :
$f (x) \geq 1 + 2$
$f (x) \geq 3$

$Verifying the Minimum Point$

Equality in AM-GM holds when the terms are equal.
For $g (x)$ : $3^{x} = 3^{- x} ⟹ x = 0$
For $h (x)$ : $2^{1 + x} = 2^{1 - x} ⟹ x = 0$
Since both reach their minimum at $x = 0$ , the minimum value of $f (x)$ is $3$ .

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The Sigma Insight: Relation Between A.M., G.M., and H.M.

Solution Diagram

Analyzing the Setup

We are given three terms:

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

f (x)

, and

(3^{x} + 3^{- x})

which form an Arithmetic Progression (A.P.).

In any A.P. where

a, b, c

are consecutive terms, the middle term is the arithmetic mean of the outer terms, expressed as

b = \frac{a + c}{2}

Applying this property to our sequence, we define the function

f (x)

as:

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

Decomposing the Function

To find the minimum value, we split the expression into two distinct components:

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

and

g (x) = \frac{3 ^{x} + 3 ^{- x}}{2}

The function can then be written as

f (x) = h (x) + g (x)

Applying the AM-GM Inequality

The Arithmetic Mean-Geometric Mean (AM-GM) inequality states that for positive real numbers

a

and

b

\frac{a + b}{2} \geq ab

We apply this to

g (x)

as follows:

g (x) = \frac{3 ^{x} + 3 ^{- x}}{2} \geq 3^{x} \cdot 3^{- x} = 3^{0} = 1

Next, we apply the same logic to

h (x)

h (x) = \frac{2 ^{1 + x} + 2 ^{1 - x}}{2} \geq 2^{1 + x} \cdot 2^{1 - x} = 2^{(1 + x) + (1 - x)} = 2^{2} = 2

The Synthesis and Final Result

Since

f (x) = h (x) + g (x)

, the minimum value of the sum is the sum of the individual minimums, provided they occur at the same value of

x

For

g (x)

, the minimum occurs when

3^{x} = 3^{- x}

, which simplifies to

x = 0

. For

h (x)

, the minimum occurs when

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

, which also simplifies to

x = 0

Because both components reach their minimum at

x = 0

, the minimum value of the function is:

f (0) = 1 + 2 = 3

The minimum value of the function

f (x)

is 3.

Common Traps & FAQs

Wait, why can't we just differentiate

f (x)

to find the minimum?

You absolutely can! Differentiating

f (x)

and setting

f^{'} (x) = 0

is a perfectly valid path. However, using the AM-GM inequality is often faster and more elegant for functions involving sums of reciprocals. It saves you from the messy chain rule calculations involving natural logarithms.

Does the equality condition for AM-GM always happen at the same

x

Not always! That is a brilliant observation. If the equality conditions occurred at different values of

x

, you would have to use calculus to find the global minimum. Here, we are lucky because both

3^{x} = 3^{- x}

and

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

are satisfied at

x = 0

Why did we split

f (x)

into

g (x)

and

h (x)

Splitting the function allows us to isolate the exponential components. By treating them as independent entities, we can apply the AM-GM inequality to each part individually, which simplifies the algebra significantly compared to trying to find the minimum of the entire expression at once.

Is it possible for

f (x)

to be less than 3?

No. Because

g (x) \geq 1

and

h (x) \geq 2

for all real

x

, their sum

f (x)

is strictly bounded below by

1 + 2 = 3

. Since we found a specific value (

x = 0

) where

f (x) = 3

, we have mathematically proven that 3 is the absolute minimum.

What if the base was negative? Would AM-GM still work?

Great question! The AM-GM inequality requires the terms to be positive. Since exponential functions like

a^{x}

are always positive for any real

x

(when

a > 0

), we are safe to use AM-GM here. If the terms could be negative, AM-GM would not be applicable.

Animated Solution for Mathematics - Sequence and Series: Let $f : R \to R$ be such that for all $x \in R$ , $(2^{1 + x} + 2^{1 - x})$ , $f (x)$ and $(3^{x} + 3^{- x})$ are in A.P., then the minimum value of $f (x)$ is :

Select Answer:

Visualized Solution

$The Arithmetic Progression Setup$

$Property of an A.P.$

$Expressing f (x)$

$Decomposing the Function$

$The AM-GM Inequality$

$Applying AM-GM to g (x)$

$Minimum of g (x)$

$Applying AM-GM to h (x)$

$Minimum of h (x)$

$Combining the Inequalities$

$Verifying the Minimum Point$

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

Decomposing the Function

Applying the AM-GM Inequality

The Synthesis and Final Result

Similar Questions

Let $f : R \to R$ be such that for all $x \in R (2^{1 + x} + 2^{1 - x}), f (x)$ and $(3^{x} + 3^{- x})$ are in A.P., then the minimum value of $f (x)$ is :

The minimum value of $f (x) = a^{a^{x}} + a^{1 - a^{x}}$ , where $a, x \in R$ and $a > 0$ , is equal to:

For $x \geq 0$ , the least value of $K$ , for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 1 6^{x} + 1 6^{- x}$ are three consecutive terms of an A.P., is equal to :

If $a_{1}, a_{2}, \dots, a_{n}$ are positive real numbers whose product is a fixed number $c$ , then the minimum value of $a_{1} + a_{2} + \dots + a_{n - 1} + 2 a_{n}$ is

Let $x, y > 0$ . If $x^{3} y^{2} = 2^{15}$ , then the least value of $3 x + 2 y$ is

The minimum value of the sum of real numbers $a^{- 5}, a^{- 4}, 3 a^{- 3}, 1, a^{8}$ and $a^{10}$ where $a > 0$ is .........

If $a, b, c$ are in A.P., $a^{2}, b^{2}, c^{2}$ are in H.P., then prove that either $a = b = c$ or $a, b, - c /2$ form a G.P.

The minimum value of $2^{s i n x} + 2^{c o s x}$ is:

If $ln (a + c), ln (a - c), ln (a - 2 b + c)$ are in A.P., then

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Arithmetic Progression Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Property of an A.P.

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Decomposing the Function

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The AM-GM Inequality

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying AM-GM to h(x)

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Combining the Inequalities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Verifying the Minimum Point

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

Decomposing the Function

Applying the AM-GM Inequality

The Synthesis and Final Result

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a1​,a2​,…,an​ are positive real numbers whose product is a fixed number c, then the minimum value of a1​+a2​+⋯+an−1​+2an​ is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The minimum value of the sum of real numbers a−5,a−4,3a−3,1,a8 and a10 where a>0 is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b,c are in A.P., a2,b2,c2 are in H.P., then prove that either a=b=c or a,b,−c/2 form a G.P.

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ln(a+c),ln(a−c),ln(a−2b+c) are in A.P., then

$The Arithmetic Progression Setup$

$Property of an A.P.$

$Expressing f (x)$

$Decomposing the Function$

$The AM-GM Inequality$

$Applying AM-GM to g (x)$

$Minimum of g (x)$

$Applying AM-GM to h (x)$

$Minimum of h (x)$

$Combining the Inequalities$

$Verifying the Minimum Point$

Let $f : R \to R$ be such that for all $x \in R (2^{1 + x} + 2^{1 - x}), f (x)$ and $(3^{x} + 3^{- x})$ are in A.P., then the minimum value of $f (x)$ is :

The minimum value of $f (x) = a^{a^{x}} + a^{1 - a^{x}}$ , where $a, x \in R$ and $a > 0$ , is equal to:

For $x \geq 0$ , the least value of $K$ , for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 1 6^{x} + 1 6^{- x}$ are three consecutive terms of an A.P., is equal to :

If $a_{1}, a_{2}, \dots, a_{n}$ are positive real numbers whose product is a fixed number $c$ , then the minimum value of $a_{1} + a_{2} + \dots + a_{n - 1} + 2 a_{n}$ is

Let $x, y > 0$ . If $x^{3} y^{2} = 2^{15}$ , then the least value of $3 x + 2 y$ is

The minimum value of the sum of real numbers $a^{- 5}, a^{- 4}, 3 a^{- 3}, 1, a^{8}$ and $a^{10}$ where $a > 0$ is .........

If $a, b, c$ are in A.P., $a^{2}, b^{2}, c^{2}$ are in H.P., then prove that either $a = b = c$ or $a, b, - c /2$ form a G.P.

The minimum value of $2^{s i n x} + 2^{c o s x}$ is:

If $ln (a + c), ln (a - c), ln (a - 2 b + c)$ are in A.P., then