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

Total Price: Rs.

There are no items in this cart.
Continue Shopping
piyush narang Grade: 12

The eq. (cosp - 1)x2 +cospx +sinp = 0 where x is a variable, has real roots. then the interval of p is.........................

7 years ago

Answers : (1)

suryakanth AskiitiansExpert-IITB
105 Points

Dear piyush,

The given equation is (cosp - 1)x2 + cos px + (1 - sin p) = 0,

Discriminant D = b2 - 4ac ≥ 0.  given x has real roots

=> cos2 p - 4(1 - sin p)(cos p - 1) ≥ 0
cos2 p is always positive and (1 - sin p) is also positive. because (sin p <1)

(cos p - 1) is always negative.(cos p <1)

Hence the given expression is positive for all the values of p.And p can be in any of the quadrants.

 =>   p ∈ (- ∞ to ∞)

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..


Askiitians Expert

Suryakanth –IITB

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details