The intercepts on x-axis made by tangents to the curve, y = _ 0 ^ x | t | dt , x R …
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
The intercepts on x-axis made by tangents to the curve, \(\mathrm{y}=\int_{0}^{\mathrm{x}}|\mathrm{t}| \mathrm{dt}, \mathrm{x} \in \mathrm{R},\) which are parallel to the line y = 2x, are equal to
- ± 1
- ± 2
- ± 3
- ± 4
Answer
(A) ± 1
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- If f (x) = 2 tan –1 x + cos –1 ( 1-x^ 2 1+x^ 2 ), then
- The minimum value of x x is attained (where x is positive real number) when x is equal to :
- The points on the curve y = x 1- x ^ 2 –1 < x < 1 at which the tangent line is vertical are
- _ x 0 2^ x -1 (1+x)^ 1 2 -1 =
- The equation e x - x - 1 = 0 has, apart from x = 0
- _ x 0 e^ 1 x -1 e^ 1 x +1 =
- If f( a )=2_ 1 f^ ( a )=1_ 1 g( a )=-1_ 1 g^ ( a )=2 , then the value of _ x a g(x) f(a)-g(a) f(x) x-a is
- Evaluate, _ x / 4 x- x ( 4 -x )( x+ x) .
More Limit, Continuity and Differentiability questions · Browse all practice questions