JEE Main 2003

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $f (x)$ be a polynomial function of second degree. If $f (1) = f (- 1)$ and $a, b, c$ are in A.P, then $f^{'} (a), f^{'} (b), f^{'} (c)$ are in

Q: Wait, why does $f(1) = f(-1)$ force $B=0$? Can't $B$ be something else?

Great question! Let's look at the algebra. $f(1) = A(1)^2 + B(1) + C = A + B + C$. Similarly, $f(-1) = A(-1)^2 + B(-1) + C = A - B + C$. Setting them equal, $A + B + C = A - B + C$. If you subtract $A$ and $C$ from both sides, you get $B = -B$, which means $2B = 0$, so $B$ must be $0$. Geometrically, this means the vertex of the parabola is on the y-axis, making it symmetric.

Q: What if $A=0$? The problem says it's a second-degree polynomial, so $A$ cannot be zero. Why is this important?

You hit the nail on the head! If $A$ were zero, the function would become $f(x) = C$, a constant function. Its derivative would be zero everywhere, and while a sequence of zeros is technically an A.P., the problem specifically defines it as a second-degree polynomial, which requires $A \neq 0$ to maintain the parabolic shape.

Select Answer:

Visualized Solution

$G e n er a l F or m o f f (x)$

Let the second-degree polynomial be $f (x) = A x^{2} + B x + C$
Here, $A, B, C$ are constants and $A \neq = 0$

$T h e S y mm e t r y C o n d i t i o n$

We are given the condition: $f (1) = f (- 1)$
This implies the function's values are equal at $x = 1$ and $x = - 1$

$S u b s t i t u t in g x = 1 an d x = - 1$

Substitute $x = 1$ : $f (1) = A (1)^{2} + B (1) + C = A + B + C$
Substitute $x = - 1$ : $f (- 1) = A (- 1)^{2} + B (- 1) + C = A - B + C$

$S o l v in g f or B$

Equating the two expressions: $A + B + C = A - B + C$
Canceling $A$ and $C$ from both sides: $2 B = 0$
Therefore, $B = 0$

$T h e S im pl i f i e d P o l y n o mia l$

Substituting $B = 0$ back into the general form
The simplified function is: $f (x) = A x^{2} + C$

$F in d in g t h eD er i v a t i v e f^{'} (x)$

The question asks about $f^{'} (a), f^{'} (b), f^{'} (c)$
Differentiating $f (x)$ with respect to $x$ : $f^{'} (x) = \frac{d}{d x} (A x^{2} + C)$
Using the power rule: $f^{'} (x) = 2 A x$

$T h e A r i t hm e t i c P r o g r ess i o n$

We are given that $a, b, c$ are in Arithmetic Progression (A.P.)
This means the difference between consecutive terms is constant.

$E v a l u a t in g f^{'} (x) a t a, b, c$

Substitute $x = a$ : $f^{'} (a) = 2 A a$
Substitute $x = b$ : $f^{'} (b) = 2 A b$
Substitute $x = c$ : $f^{'} (c) = 2 A c$

$A na l y z in g t h e S e q u e n ce$

The sequence of derivatives is: $2 A a, 2 A b, 2 A c$
Notice that each term is the original A.P. term multiplied by a constant $2 A$

$F ina l C o n c l u s i o n$

Property of A.P.: Multiplying each term of an A.P. by a non-zero constant results in another A.P.
Therefore, $2 A a, 2 A b, 2 A c$ are in A.P.
Conclusion: $f^{'} (a), f^{'} (b), f^{'} (c)$ are in A.P.

00:00 / 00:00

The Sigma Insight: Arithmetic Progression (A.P.)

Solution Diagram

Analyzing the Symmetry of the Parabola

Imagine you are standing before a perfectly symmetric parabola defined by the second-degree polynomial

f (x) = A x^{2} + B x + C

. We are given the condition

f (1) = f (- 1)

Geometrically, this means the height of the parabola at

x = 1

is identical to its height at

x = - 1

. This is the hallmark of symmetry.

Algebraically, substituting these values into the equation yields:

A (1)^{2} + B (1) + C = A (- 1)^{2} + B (- 1) + C

Simplifying this expression, we get

A + B + C = A - B + C

. This leads us directly to

2 B = 0

, which implies $B = 0$ .

Our polynomial simplifies to the elegant form

f (x) = A x^{2} + C

. The linear term vanishes, leaving us with a pure, symmetric quadratic.

The Derivative as a Linear Operator

Now, we consider the derivatives

f^{'} (a)

f^{'} (b)

, and

f^{'} (c)

. Applying the power rule to our simplified polynomial, we find:

f^{'} (x) = 2 A x

Notice how the derivative is a linear function passing through the origin. It is a straight line.

This is a powerful realization. We are no longer dealing with a curve; we are dealing with a linear transformation of the points

a

b

, and

c

The Beauty of Arithmetic Progressions

We are given that

a, b, c

are in an Arithmetic Progression (A.P.). This implies that the common difference is constant:

b - a = c - b

Now, let us evaluate the derivative values at these points:

f^{'} (a) = 2 A a, f^{'} (b) = 2 A b, f^{'} (c) = 2 A c

To check if these values form an A.P., we examine the differences between consecutive terms:

f^{'} (b) - f^{'} (a) = 2 A b - 2 A a = 2 A (b - a)

f^{'} (c) - f^{'} (b) = 2 A c - 2 A b = 2 A (c - b)

The Final Conclusion

Since

b - a = c - b

, it follows that

2 A (b - a) = 2 A (c - b)

. The common difference is preserved under the linear transformation

2 A x

We have taken a journey from the symmetry of a parabola to the linear nature of its derivative, and finally to the preservation of an A.P. under linear scaling.

Because the common difference remains constant, we can confidently conclude that $f^{'} (a), f^{'} (b), f^{'} (c)$ are in an A.P.

Common Traps & FAQs

Wait, why does

f (1) = f (- 1)

force

B = 0

? Can't

B

be something else?

Great question! Let's look at the algebra.

f (1) = A (1)^{2} + B (1) + C = A + B + C

. Similarly,

f (- 1) = A (- 1)^{2} + B (- 1) + C = A - B + C

. Setting them equal,

A + B + C = A - B + C

. If you subtract

A

and

C

from both sides, you get

B = - B

, which means

2 B = 0

, so

B

must be

0

. Geometrically, this means the vertex of the parabola is on the y-axis, making it symmetric.

Why are we differentiating? Does the derivative preserve the A.P. property?

We differentiate because the question asks about the sequence

f^{'} (a), f^{'} (b), f^{'} (c)

. The derivative

f^{'} (x) = 2 A x

is a linear function. A key property of an Arithmetic Progression is that if you multiply every term by a constant (in this case,

2 A

), the resulting sequence is still an A.P. Since

a, b, c

are in A.P.,

2 A a, 2 A b, 2 A c

will also be in A.P.

What if

A = 0

? The problem says it's a second-degree polynomial, so

A

cannot be zero. Why is this important?

You hit the nail on the head! If

A

were zero, the function would become

f (x) = C

, a constant function. Its derivative would be zero everywhere, and while a sequence of zeros is technically an A.P., the problem specifically defines it as a second-degree polynomial, which requires

A \neq = 0

to maintain the parabolic shape.

Animated Solution for Mathematics - Sequence and Series: Let $f (x)$ be a polynomial function of second degree. If $f (1) = f (- 1)$ and $a, b, c$ are in A.P, then $f^{'} (a), f^{'} (b), f^{'} (c)$ are in

Select Answer:

Visualized Solution

$G e n er a l F or m o f f (x)$

$T h e S y mm e t r y C o n d i t i o n$

$S u b s t i t u t in g x = 1 an d x = - 1$

$S o l v in g f or B$

$T h e S im pl i f i e d P o l y n o mia l$

$F in d in g t h eD er i v a t i v e f^{'} (x)$

$T h e A r i t hm e t i c P r o g r ess i o n$

$E v a l u a t in g f^{'} (x) a t a, b, c$

$A na l y z in g t h e S e q u e n ce$

$F ina l C o n c l u s i o n$

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Symmetry of the Parabola

The Derivative as a Linear Operator

The Beauty of Arithmetic Progressions

The Final Conclusion

Similar Questions

Let $T_{r}$ be the $r$ th term of an A.P., for $r = 1, 2, 3, \dots$ . If for some positive integers $m, n$ we have $T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $T_{mn}$ equals

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

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Comprehension Passage

The sum $V_{1} + V_{2} + \dots + V_{n}$ is

$T_{r}$ is always

Which one of the following is a correct statement ?

Let $a_{1} = 8, a_{2}, a_{3}, \dots, a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is

If $1, lo g_{9} (3^{1 - x} + 2), lo g_{3} (4 \cdot 3^{x} - 1)$ are in A.P. then $x$ equals

Select Answer:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitutingx=1andx=−1

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluatingf′(x)ata,b,c

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

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

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Symmetry of the Parabola

The Derivative as a Linear Operator

The Beauty of Arithmetic Progressions

The Final Conclusion

Similar Questions

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$G e n er a l F or m o f f (x)$

$T h e S y mm e t r y C o n d i t i o n$

$S u b s t i t u t in g x = 1 an d x = - 1$

$S o l v in g f or B$

$T h e S im pl i f i e d P o l y n o mia l$

$F in d in g t h eD er i v a t i v e f^{'} (x)$

$T h e A r i t hm e t i c P r o g r ess i o n$

$E v a l u a t in g f^{'} (x) a t a, b, c$

$A na l y z in g t h e S e q u e n ce$

$F ina l C o n c l u s i o n$

Let $T_{r}$ be the $r$ th term of an A.P., for $r = 1, 2, 3, \dots$ . If for some positive integers $m, n$ we have $T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $T_{mn}$ equals

Let $T_{r}$ be the $r$ th term of an A.P. whose first term is $a$ and common difference is $d$ . If for some positive integers $m, n, m \neq = n, T_{m} = 1/ n$ and $T_{n} = 1/ m$ , then $a - d$ equals

If $x = \sum_{n = 0}^{\infty} a^{n}, y = \sum_{n = 0}^{\infty} b^{n}, z = \sum_{n = 0}^{\infty} c^{n}$ , where $a, b, c$ are in A.P. and $∣ a ∣ < 1, ∣ b ∣ < 1, ∣ c ∣ < 1, ab c \neq = 0$ , then

If $lo g_{3} 2, lo g_{3} (2^{x} - 5), lo g_{3} (2^{x} - \frac{7}{2})$ are in an arithmetic progression, then the value of $x$ is equal to

The sum $V_{1} + V_{2} + \dots + V_{n}$ is

$T_{r}$ is always

Let $a_{1} = 8, a_{2}, a_{3}, \dots, a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170, then the product of its middle two terms is

If $1, lo g_{9} (3^{1 - x} + 2), lo g_{3} (4 \cdot 3^{x} - 1)$ are in A.P. then $x$ equals