Sign in
Please select an account to continue using cracku.in
↓ →
Let f(x)= $$\max(5x, 52-2x^2)$$, where x is any positive real number. Then the minimum possible value of f(x)
Correct Answer: 20
The minimum value of the function will occur when the expressions inside the function are equal.
So, 5$$x$$ = $$52 - 2x^2$$
or, $$2x^2 + 5x - 52$$ = 0
On solving, we get $$x$$ = 4 or $$-\dfrac{13}{2}$$
But, it is given that $$x$$ is a positive number.
So, $$x$$ = 4
And the minimum value = 5*4 = 20
Hence, 20 is the correct answer.
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
Book Free CAT Mentorship
Get personalized CAT strategy from a 99%iler
500+ students mentored
OTP Verification
Enter the 6-digit code sent to your phone
Booking Summary
Enter OTP
Didn't receive the OTP?
Start your IIM journey with the right preparation and crack CAT 2026.
Educational materials for CAT preparation