JEE Advanced 1981

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Let $a, b, c$ be non-zero real numbers such that $\int_{0}^{1} (1 + cos^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + cos^{8} x) (a x^{2} + b x + c) d x$ . Then the quadratic equation $a x^{2} + b x + c = 0$ has

Select Answer:

Visualized Solution

The Given Integral Equation

We are given the equality:
$\int_{0}^{1} (1 + cos^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + cos^{8} x) (a x^{2} + b x + c) d x$
Our goal is to analyze this to find the nature of the roots of $a x^{2} + b x + c = 0$ .

Rearranging the Terms

Let's bring both integrals to one side by subtracting:
$\int_{0}^{2} (1 + cos^{8} x) (a x^{2} + b x + c) d x - \int_{0}^{1} (1 + cos^{8} x) (a x^{2} + b x + c) d x = 0$

Interval Subtraction Property

Using the definite integral property:
$\int_{a}^{c} h (x) d x - \int_{a}^{b} h (x) d x = \int_{b}^{c} h (x) d x$
We get:
$\int_{1}^{2} (1 + cos^{8} x) (a x^{2} + b x + c) d x = 0$

Separating the Integrand

Let $g (x) = 1 + cos^{8} x$
Let $f (x) = a x^{2} + b x + c$
The equation becomes:
$\int_{1}^{2} g (x) f (x) d x = 0$

Bounding $g (x)$

We know that for any real $x$ , $cos^{2} x \geq 0$ .
Therefore, $cos^{8} x \geq 0$ .
Adding $1$ to both sides:
$1 + cos^{8} x \geq 1$
So, $g (x) \geq 1 > 0$ for all $x \in [1, 2]$ .

The Zero Integral Condition

We have $\int_{1}^{2} g (x) f (x) d x = 0$ with $g (x) > 0$ .
If $f (x)$ were always positive in $(1, 2)$ , the integral would be strictly positive.
If $f (x)$ were always negative in $(1, 2)$ , the integral would be strictly negative.

Sign Change of $f (x)$

For the integral to be exactly zero, the product $g (x) f (x)$ must have both positive and negative areas that cancel out.
Since $g (x)$ is always positive, $f (x)$ must change sign in the interval $(1, 2)$ .

Applying Intermediate Value Theorem

$f (x) = a x^{2} + b x + c$ is a continuous polynomial function.
Since $f (x)$ changes sign in $(1, 2)$ , by the Intermediate Value Theorem, it must cross the x-axis.
Therefore, there exists at least one root $x_{0} \in (1, 2)$ such that $f (x_{0}) = 0$ .

Locating the Root

We found that $a x^{2} + b x + c = 0$ has at least one root in the interval $(1, 2)$ .
Notice that the interval $(1, 2)$ is completely contained within the interval $(0, 2)$ .
Thus, the equation has at least one root in $(0, 2)$ .

00:00 / 00:00

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

We are presented with the following integral equation:

\int_{0}^{1} (1 + cos^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + cos^{8} x) (a x^{2} + b x + c) d x

At first glance, this appears to be a daunting calculation. However, by rearranging the terms, we can simplify the expression significantly.

By subtracting the first integral from the second, we obtain:

\int_{1}^{2} (1 + cos^{8} x) (a x^{2} + b x + c) d x = 0

The Master Equation

Let us define two functions to simplify our analysis:

g (x) = 1 + cos^{8} x

f (x) = a x^{2} + b x + c

We observe that for any real

x

cos^{2} x \geq 0

, which implies that

cos^{8} x \geq 0

. Consequently,

g (x) = 1 + cos^{8} x \geq 1

This is a crucial realization: our function

g (x)

is strictly positive and never touches the

x

-axis.

Applying the Intermediate Value Theorem

We are now analyzing the integral:

\int_{1}^{2} g (x) f (x) d x = 0

Since

g (x) > 0

throughout the interval

[1, 2]

, the sign of the integrand

g (x) f (x)

is determined entirely by the sign of

f (x)

f (x)

were strictly positive, the integral would be positive. If

f (x)

were strictly negative, the integral would be negative.

For the integral to equal zero,

f (x)

must change sign within the interval

(1, 2)

. Because

f (x)

is a polynomial, it is continuous.

By the Intermediate Value Theorem, if a continuous function changes sign, it must cross the

x

-axis. Therefore, there exists at least one root of

f (x)

in the interval

(1, 2)

Final Conclusion

Since the interval

(1, 2)

is a subset of

(0, 2)

, we have successfully proven that there exists at least one root of the polynomial

f (x) = a x^{2} + b x + c

in the interval

(0, 2)

This elegant approach demonstrates that we do not need to solve for the specific constants

a

b

, and

c

to understand the fundamental behavior of the function.

Common Traps & FAQs

Wait, why do we reject the negative root here? Or why can't we just say the integral is zero so the function is zero?

That is a classic trap! If the integral of a function is zero, it does not mean the function itself is zero everywhere. It only means the net signed area is zero. In our case, the product

g (x) f (x)

must have positive and negative parts that cancel out. Since

g (x)

is always positive,

f (x)

must be the one changing sign.

Why is

1 + cos^{8} x

always positive?

Great question! Remember that any real number raised to an even power is non-negative. Since

cos^{2} x \geq 0

, it follows that

(cos^{2} x)^{4} = cos^{8} x \geq 0

. Adding

1

to this non-negative value ensures that

1 + cos^{8} x \geq 1

, making it strictly positive for all

x

How does changing sign imply a root?

This is the beauty of the Intermediate Value Theorem (IVT). Since

f (x) = a x^{2} + b x + c

is a polynomial, it is continuous everywhere. If a continuous function takes both positive and negative values in an interval, it must cross the x-axis at some point within that interval to get from one sign to the other. That crossing point is your root!

Animated Solution for Mathematics - Definite Integration: Let $a, b, c$ be non-zero real numbers such that $\int_{0}^{1} (1 + cos^{8} x) (a x^{2} + b x + c) d x = \int_{0}^{2} (1 + cos^{8} x) (a x^{2} + b x + c) d x$ . Then the quadratic equation $a x^{2} + b x + c = 0$ has

Select Answer:

Visualized Solution

The Given Integral Equation

Rearranging the Terms

Interval Subtraction Property

Separating the Integrand

Bounding $g (x)$

The Zero Integral Condition

Sign Change of $f (x)$

Applying Intermediate Value Theorem

Locating the Root

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Master Equation

Applying the Intermediate Value Theorem

Final Conclusion

Similar Questions

Let $α = \int_{α}^{l o g_{e} 4} \frac{d x}{e ^{x} - 1} = \frac{π}{6}$ . Then $e^{α}$ and $e^{- α}$ are the roots of the equation :

If $\int_{0}^{π /4} \frac{s i n ^{2} x}{1 + s i n x c o s x} d x = \frac{1}{a} lo g_{e} (\frac{a}{3}) + \frac{π}{b 3}$ , where $a, b \in N$ , then $a + b$ is equal to

If $\int_{- π /2}^{π /2} \frac{8 2 c o s x d x}{( 1 + e ^{s i n x} ) ( 1 + s i n ^{4} x )} = α π + β lo g_{e} (3 + 22)$ , where $α, β$ are integers, then $α^{2} + β^{2}$ equals

If $\int_{0}^{π /3} cos^{4} x d x = aπ + b 3$ , where $a$ and $b$ are rational numbers, then $9 a + 8 b$ is equal to :

If $\int_{0}^{1} \frac{1}{3 + x + 1 + x} d x = a + b 2 + c 3$ , where $a, b, c$ are rational numbers, then $2 a + 3 b - 4 c$ is equal to :

Let for $f (x) = 7 tan^{8} x + 7 tan^{6} x - 3 tan^{4} x - 3 tan^{2} x$ , $I_{1} = \int_{0}^{π /4} f (x) d x$ and $I_{2} = \int_{0}^{π /4} x f (x) d x$ . Then $7 I_{1} + 12 I_{2}$ is equal to:

$\int_{- 2}^{0} {x^{3} + 3 x^{2} + 3 x + 3 + (x + 1) cos (x + 1)} d x$ is equal to

Match the statement in Column-I with the values in Column-II

List-I

List-II

The value of $I = \int_{0}^{π /2} \frac{( s i n x + c o s x ) ^{2}}{1 + s i n 2 x} d x$ is

Let $f (x) = x + \frac{a}{π ^{2} - 4} sin x + \frac{b}{π ^{2} - 4} cos x, x \in R$ be a function which satisfies $f (x) = x + \int_{0}^{π /2} sin (x + y) f (y) d y$ . Then $(a + b)$ is equal to

Select Answer:

Visualized Solution

The Given Integral Equation

Rearranging the Terms

Interval Subtraction Property

Separating the Integrand

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

The Zero Integral Condition

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

Applying Intermediate Value Theorem

Locating the Root

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Master Equation

Applying the Intermediate Value Theorem

Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α=∫αloge​4​ex−1​dx​=6π​. Then eα and e−α are the roots of the equation :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫0π/4​1+sinxcosxsin2x​dx=a1​loge​(3a​)+b3​π​, where a,b∈N, then a+b is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫−π/2π/2​(1+esinx)(1+sin4x)82​cosxdx​=απ+βloge​(3+22​), where α,β are integers, then α2+β2 equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫0π/3​cos4xdx=aπ+b3​, where a and b are rational numbers, then 9a+8b is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∫01​3+x​+1+x​1​dx=a+b2​+c3​, where a,b,c are rational numbers, then 2a+3b−4c is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let for f(x)=7 tan8x+7 tan6x−3 tan4x−3 tan2x, I1​=∫0π/4​f(x)dx and I2​=∫0π/4​xf(x)dx. Then 7I1​+12I2​ is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫−20​{x3+3x2+3x+3+(x+1)cos(x+1)}dx is equal to

Match the statement in Column-I with the values in Column-II

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of I=∫0π/2​1+sin2x​(sinx+cosx)2​dx is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x+π2−4a​sinx+π2−4b​cosx,x∈R be a function which satisfies f(x)=x+∫0π/2​sin(x+y)f(y)dy. Then (a+b) is equal to

Bounding $g (x)$

Sign Change of $f (x)$

Let $α = \int_{α}^{l o g_{e} 4} \frac{d x}{e ^{x} - 1} = \frac{π}{6}$ . Then $e^{α}$ and $e^{- α}$ are the roots of the equation :

If $\int_{0}^{π /4} \frac{s i n ^{2} x}{1 + s i n x c o s x} d x = \frac{1}{a} lo g_{e} (\frac{a}{3}) + \frac{π}{b 3}$ , where $a, b \in N$ , then $a + b$ is equal to

If $\int_{- π /2}^{π /2} \frac{8 2 c o s x d x}{( 1 + e ^{s i n x} ) ( 1 + s i n ^{4} x )} = α π + β lo g_{e} (3 + 22)$ , where $α, β$ are integers, then $α^{2} + β^{2}$ equals

If $\int_{0}^{π /3} cos^{4} x d x = aπ + b 3$ , where $a$ and $b$ are rational numbers, then $9 a + 8 b$ is equal to :

If $\int_{0}^{1} \frac{1}{3 + x + 1 + x} d x = a + b 2 + c 3$ , where $a, b, c$ are rational numbers, then $2 a + 3 b - 4 c$ is equal to :

Let for $f (x) = 7 tan^{8} x + 7 tan^{6} x - 3 tan^{4} x - 3 tan^{2} x$ , $I_{1} = \int_{0}^{π /4} f (x) d x$ and $I_{2} = \int_{0}^{π /4} x f (x) d x$ . Then $7 I_{1} + 12 I_{2}$ is equal to:

$\int_{- 2}^{0} {x^{3} + 3 x^{2} + 3 x + 3 + (x + 1) cos (x + 1)} d x$ is equal to

The value of $I = \int_{0}^{π /2} \frac{( s i n x + c o s x ) ^{2}}{1 + s i n 2 x} d x$ is

Let $f (x) = x + \frac{a}{π ^{2} - 4} sin x + \frac{b}{π ^{2} - 4} cos x, x \in R$ be a function which satisfies $f (x) = x + \int_{0}^{π /2} sin (x + y) f (y) d y$ . Then $(a + b)$ is equal to