Let $f(x)=2^{x}-x^{2}, x \in R$. If $m$ and $n$ are respectively the number of points at which the curves $y=f(x)$ and $y=f^{\prime}(x)$ intersects the $x$-axis, then the value of $m+n$ is

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Application of Derivatives

Let

f (x) = 2^{x} - x^{2}, x \in R

. If

m

and

n

are respectively the number of points at which the curves

y = f (x)

and

y = f^{'} (x)

intersects the

x

-axis, then the value of

m + n

Answer: 5

f (x) = 2^{x} - x^{2}

f (x) = 0

: numerically,

f (- 1) = - 0.5

f (- 0.5) \approx 0.46

(one root between

- 1

and

- 0.5

);

f (2) = 4 - 4 = 0

(exact root);

f (3) = 8 - 9 = - 1

;

f (4) = 16 - 16 = 0

(exact root);

f (x) > 0

for all

x > 4

. So

f

has exactly

m = 3

real roots: one negative, and

x = 2, 4

f^{'} (x) = 2^{x} ln 2 - 2 x

. Checking signs:

f^{'} (0) = ln 2 > 0

f^{'} (1) = 2 ln 2 - 2 < 0

(root in

(0, 1)

);

f^{'} (2) < 0

f^{'} (3) < 0

f^{'} (4) > 0

(root in

(3, 4)

); for

x < 0

f^{'} (x)

stays positive throughout. So

f^{'}

has exactly

n = 2

real roots.

m + n = 3 + 2 = 5

Key confirmed: 5.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Integer answer