Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

If $\mathrm{f(x)=x(\sqrt{x}-\sqrt{x+1})}$ then

Option: 1
f is continuous but not differentiable at x=0

Option: 2
f is differentiable at x=0

Option: 3
f is differentiable but not continuous at x=0

Option: 4
f is not differentiable at x=0

Answers (1)

best_answer

We have $\mathrm{f(x)=x(\sqrt{x}-\sqrt{x+1})}$
$\mathrm{\Rightarrow f^{\prime}(x)=(\sqrt{x}-\sqrt{x+1})+\frac{x(\sqrt{x+1}-\sqrt{x})}{2 \sqrt{x} \sqrt{x+1}}~ for ~x \neq 0,-1}$ Also
$\mathrm{f^{\prime}\left(0^{+}\right)=-1}$ So, f(x) is differentiable and hence continuous at x=0

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

5 g of Na₂SO₄ was dissolved in x g of H₂O. The change in freezing point was found to be 3.82⁰C. If Na₂SO₄ is 81.5% ionised, the value of x (K

A capacitor is made of two square plates each of side 'a' making a very small angle $\alpha$

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO₃ to give fraction A. The leftover organic phase was extracted with d

Trending Articles/News

anam-khan

JEE Main 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks

anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule

anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question