Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

At what point in the interval [0,2?] does the function Sin2x attain its maximum value?

Jitender Singh IIT Delhi
7 years ago
Ans:
$f(x) = sin(2x)$
$f^{'}(x) = 2cos(2x) = 0$
$2x = (n + \frac{1}{2})\pi$
$x = (n + \frac{1}{2})\frac{\pi }{2}$
$x = \frac{\pi }{4}, \frac{3\pi }{4}, \frac{5\pi }{4}, \frac{7\pi }{4}$
$f^{''}(x) = -4sin(2x)$
$f^{''}(\frac{\pi }{4}) = -4sin(2.\frac{\pi }{4}) = -4$
$f^{''}(\frac{3\pi }{4}) = -4sin(2.\frac{3\pi }{4}) = 4$
$f^{''}(\frac{5\pi }{4}) = -4sin(2.\frac{5\pi }{4}) = -4$
$f^{''}(\frac{7\pi }{4}) = -4sin(2.\frac{7\pi }{4}) = 4$
$x = \frac{\pi }{4}, \frac{5\pi }{4}$
is the pont of maxima.
Thanks & Regards
Jitender Singh
IIT Delhi