Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The longest interval in which f(x)=x\sqrt{4ax-x^{2}}(a<0) is decreasing is -

Option: 1

\left [ 4a,0 \right ]
 


Option: 2

\left [ 3a,0 \right ]

 


Option: 3

(-\infty,3]


Option: 4

none of these


Answers (1)

best_answer

 

Condition for Decreasing function -

The tangents drawn at any point on it make an obtuse angle with positive direction of  the x-axis.

M_{T}=tan\theta\leq 0

\therefore \:\:\frac{dy}{dx}\leq 0\:\:\:\:\:x\epsilon (a,b)

- wherein

y=f(x) is continuous and differentiable in (a,b)

 

 

f'(x)=\sqrt{4ax-x^{2}}+\frac{x(4a-2x)}{2\sqrt{4ax-x^{2}}}=\frac{6ax-2x^{2}}{\sqrt{4ax-x^{2}}}<0,\forall \:x\:\varepsilon (4a,3a)

so f(x) is decreasing in \left [ 4a,3a \right ]

Posted by

himanshu.meshram

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question