Application of DerivativeshardComprehensionFree

Application of Derivatives: Let Thrice Differentiable Function Interval Define Auxiliary

JEE Maths reading comprehension with full step-by-step solutions.

Passage

Let

f(x)

be a thrice differentiable function on the interval

[a, e]

. Define the auxiliary functions:

G(x) = [f'(x)]^2 + f(x)f''(x), \qquad H(x) = e^{-x}(f'(x) - f(x))

It is given that

f(a) = f(e) = 0

. For constants

b, c, d

satisfying

a < b < c < d < e

, the function takes the values

f(b) = 2

f(c) = -1

, and

f(d) = 2

Question 1 · Single correct

The minimum number of zeros of the function

G(x)

in the open interval

(a, e)

is:

4

5

6

correct

3

Solution

Step 1: Identify

G(x)

as an exact derivative Observe that:

\frac{d}{dx}\bigl[f(x)f'(x)\bigr] = [f'(x)]^2 + f(x)f''(x) = G(x)

Define

\phi(x) = f(x)f'(x)

, so that

G(x) = \phi'(x)

. Zeros of

G

are found by applying Rolle's theorem to

\phi

. Step 2: Locate the zeros of

f(x)

The given boundary conditions and intermediate values are:

f(a) = 0

and

f(e) = 0

. Since

f(b) = 2 > 0

and

f(c) = -1 < 0

, the Intermediate Value Theorem guarantees a zero

x_1 \in (b, c)

. Since

f(c) = -1 < 0

and

f(d) = 2 > 0

, the Intermediate Value Theorem guarantees a zero

x_2 \in (c, d)

. Therefore

f

has at least four zeros on

[a, e]

with the ordering

a < x_1 < x_2 < e

. Step 3: Locate the zeros of

f'(x)

Applying Rolle's theorem to

f

on each consecutive pair of its zeros

f(x)

\exists\; y_1 \in (a,\, x_1) \text{ with } f'(y_1) = 0

\exists\; y_2 \in (x_1,\, x_2) \text{ with } f'(y_2) = 0

\exists\; y_3 \in (x_2,\, e) \text{ with } f'(y_3) = 0

Step 4: Count the zeros of

\phi(x) = f(x)f'(x)

Since

\phi(x) = 0

whenever

f(x) = 0

f'(x) = 0

, the function

\phi

vanishes at the seven points:

a < y_1 < x_1 < y_2 < x_2 < y_3 < e

These are all distinct (each

y_i

lies strictly inside a sub-interval between consecutive zeros of

f

), so

\phi

has exactly seven distinct zeros on

[a, e]

. Step 5: Apply Rolle's theorem to

\phi

Since

\phi

has seven distinct zeros, Rolle's theorem applies to each of the six consecutive sub-intervals:

[a, y_1],\quad [y_1, x_1],\quad [x_1, y_2],\quad [y_2, x_2],\quad [x_2, y_3],\quad [y_3, e]

Each sub-interval contains at least one zero of

\phi'(x) = G(x)

in its interior, and all these interiors are subsets of

(a, e)

. Therefore

G(x)

has at least

\mathbf{6}

zeros in

(a, e)

. Answer: (3)

Question 2 · Single correct

The derivative

H'(x)

can be expressed as:

e^{-x}f''(x)

e^{-x}\bigl(f''(x) - 2f'(x) + f(x)\bigr)

correct

f''(x) - f(x)

e^{-x}\bigl(f''(x) + f(x)\bigr)

Solution

Step 1: Apply the product rule

H'(x) = \frac{d}{dx}\!\left(e^{-x}\right)\cdot\bigl(f'(x)-f(x)\bigr) + e^{-x}\cdot\frac{d}{dx}\!\left(f'(x)-f(x)\right)

= -e^{-x}\bigl(f'(x)-f(x)\bigr) + e^{-x}\bigl(f''(x)-f'(x)\bigr)

Step 2: Expand and collect terms

H'(x) = e^{-x}\bigl[-f'(x)+f(x)+f''(x)-f'(x)\bigr]

= e^{-x}\bigl(f''(x) - 2f'(x) + f(x)\bigr)

Answer: (2)

Question 3 · Single correct

Let

I = \displaystyle\int_{a}^{e} f(x)\,f'(x)\,dx

. The value of

I

is:

1

0

correct

e - a

\dfrac{1}{2}

Solution

Step 1: Recognize the integrand as an exact derivative

f(x)f'(x) = \frac{1}{2}\frac{d}{dx}\bigl[f(x)^2\bigr]

Step 2: Evaluate using the Fundamental Theorem of Calculus

I = \frac{1}{2}\Bigl[f(x)^2\Bigr]_{a}^{e} = \frac{1}{2}\bigl(f(e)^2 - f(a)^2\bigr) = \frac{1}{2}(0^2 - 0^2) = 0

Note: The intermediate values

f(b) = 2

f(c) = -1

f(d) = 2

do not affect this integral. Only the boundary values

f(a) = f(e) = 0

are relevant. Answer: (2)

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