The slope of the tangent to the curve y=∫0x 11+t3dt Then, x1+y1=a at the point where x = 1, is

# The slope of the tangent to the curve $y={\int }_{0}^{x} \frac{1}{1+{t}^{3}}dt$ Then, $\sqrt{{x}_{1}}+\sqrt{{y}_{1}}=\sqrt{a}$ at the point where , is

1. A

$\frac{1}{2}$

2. B

1

3. C

$\frac{1}{4}$

4. D

non-existent

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

We have,

$y={\int }_{0}^{x} \frac{1}{1+{t}^{3}}dt⇒\frac{dy}{dx}=\frac{1}{1+{x}^{3}}⇒{\left(\frac{dy}{dx}\right)}_{x=1}=\frac{1}{2}$

## Related content

 How to Score 100 in Class 6 Maths using NCERT Solutions TS EAMCET Previous Year Question Papers CBSE Class 8 English Syllabus Academic Year 2023-2024 CBSE Class 7 English Syllabus Academic Year 2023-2024 CBSE Worksheets for Class 7 with Answers COMEDK UGET Mock Test 2024 (Available) – Free Mock Test Series Indian tribes Maurya Empire CBSE Class 10 Science Important Topics – You Should Not Miss in Board Exam 2024 CBSE Class 6 Social Science: Important Tips and Topics

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)