Assertion: The equation of the tangent of the curve y = ∫ x 2 x 3 d t 1 + t 2 at x = 1 is y + 1 = x 2 . Reason: Tangent for curve y = f x is straight line passing through point x 1 , y 1 and slope is d y / d x at point x 1 , y 1 .

If both A and R are true, and R is the correct explanation of A .

If both A and R are true but R is not the correct explanation of A .

If f x = ∫ 0 x t sin x - sin t d t , then

f ''' x + f ' x = cos x - 2 x sin x

f ''' x - f '' x = cos x - 2 x sin x

f ''' x + f '' x - f ' x = cos x

Let f : R → R be a continuous function satisfying f x + ∫ 0 x t f t d t + x 2 = 0 . For all x ∈ R . Then-

f x has more than one point in common with the x -axis

A function f is continuous for all x (and not everywhere zero) such that f 2 x = ∫ 0 x f t cost 2 + sin t d t then f x is:

1 2 log e 2 + sin x 2 ; x ≠ n π , n ∈ I

1 2 log e x + cos x 2 ; x ≠ 0

1 2 log e 3 2 + cos x ; x ≠ 0

cos x + sin x 2 + sin x ; x ≠ n π + 3 π 4 , n ∈ I

HARD

Earn 100

Assertion: The equation of the tangent of the curve $y = \int_{x^{2}}^{x^{3}} \frac{d t}{\sqrt{1 + t^{2}}}$ at $x = 1$ is $y + 1 = x \sqrt{2} .$

Reason: Tangent for curve $y = f (x)$ is straight line passing through point $(x_{1}, y_{1})$ and slope is $(d y / d x)$ at point $(x_{1}, y_{1})$ .

50% studentsanswered this correctly

Important Questions on Application of Derivatives

EASY

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

Let

f : R \to R

be a continuous and differentiable function such that

f (2) = 6

and

f^{'} (2) = \frac{1}{48}

. If

\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x),

then

\underset{x \to 2}{l i m} g (x)

is equal to

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

For

x \in R, x \neq 0,

y (x)

is a differentiable function such that

x \int_{1}^{x} y (t) d t = (x + 1) \int_{1}^{x} t y (t) d t,

then

y (x)

equals (where

C

is a constant)

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

f : R \to R

is a differentiable function and

f (2) = 6,

then

\underset{x \to 2}{l i m} \int_{6}^{f (x)} \frac{2 t d t}{(x - 2)}

is:

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

\lim_{x \to 1} (\frac{\int_{0}^{(x - 1)^{2}} t \cos t^{2} d t}{(x - 1) \sin (x - 1)})

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

The limit

\underset{x \to \infty}{l i m} x^{2} \int_{0}^{x} e^{t^{3} - x^{3}} d t

equals

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

Let a function $f : [0, 5] \to R$ be continuous, $f (1) = 3$ and $F$ be defined as: $F (x) = \int_{1}^{x} t^{2} g (t) d t,$ where $g (t) = \int_{1}^{t} f (u) d u .$ Then for the function $F (x),$ the point $x = 1$ is:

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

Let

f : [0, 2] \to R

be a function which is continuous on

[0, 2]

and is differentiable on

(0, 2) f (0) = 1

. Let

F (x) = \int_{0}^{x^{2}} f (\sqrt{t}) d t

for

x \in [0, 2] .

F^{'} (x) = f^{'} (x)

for all

x \in (0, 2),

then

F (2)

equals

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

\lim_{x \to 0} \frac{\int_{0}^{x^{2}} (\sin \sqrt{t}) d t}{x^{3}}

is equal to:

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

f (x) = \int_{0}^{x} t (\sin x - \sin t) d t

, then

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

Let

f : R \to R

be a continuous function satisfying

f (x) + \int_{0}^{x} t f (t) d t + x^{2} = 0 .

For all

x \in R .

Then-

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

The least value of the function $f (x) = \int_{0}^{x} (3 \sin x + 4 \cos x) d x$ in the interval $[\frac{5 π}{4}, \frac{4 π}{3}]$ is

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

The intercepts on the

x

-axis made by tangents to the curve,

y = \int_{0}^{x} |t| d t, x \in R,

which are parallel to the line

y = 2 x

, are equal to

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

The value of

\lim_{x \to 0} \frac{1}{x} [\int_{y}^{a} e^{\sin^{2} t} d t - \int_{x + y}^{a} e^{\sin^{2} t} d t]

is equal to

EASY

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

\underset{x \to 0}{l i m} \frac{\int_{0}^{x} t s i n (10 t) d t}{x}

, is equal to

MEDIUM

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

Suppose a continuous function

f : (0, \infty) \to R

satisfies

f (x) = 2 \int_{0}^{x} t f (t) d t + 1, \forall x \geq 0

. Then,

f (1)

equals

EASY

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

A function

f

is continuous for all

x

(and not everywhere zero) such that

f^{2} (x) = \int_{0}^{x} f (t) \frac{cost}{2 + \sin t} d t

then

f (x)

is:

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

Let

f : R \to R

be defined as

f (x) = e^{- x} \sin x .

F : [0, 1] \to R

is a differentiable function such that

F (x) = \int_{0}^{x} f (t) d t,

then the value of

\int_{0}^{1} (F^{'} (x) + f (x)) e^{x} d x

lies in the interval

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

A continuous function

f : R \to R

satisfies equation

f (x) = x + \int_{0}^{x} f (t) d t .

Which of the following options is true?

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

Let

f : (- 1, 1) \to R

be a continuous function. If

\int_{0}^{\sin x} f (t) d t = \frac{\sqrt{3}}{2} x,

then

f (\frac{\sqrt{3}}{2})

is equal to:

HARD

Mathematics>Differential Calculus>Application of Derivatives>Tangents and Normals

\int_{0}^{x} f (t) d t = x^{2} + \int_{x}^{1} t^{2} f (t) d t,

then

f^{'} (\frac{1}{2})

Assertion: The equation of the tangent of the curve y=∫x2x3dt1+t2 at x=1 is y+1=x2. Reason: Tangent for curvey=fx is straight line passing through point x1, y1 and slope is dy/dx at point x1, y1 .

Important Questions on Application of Derivatives

Learn and Prepare for any exam you want !

Assertion: The equation of the tangent of the curve $y = \int_{x^{2}}^{x^{3}} \frac{d t}{\sqrt{1 + t^{2}}}$ at $x = 1$ is $y + 1 = x \sqrt{2} .$

Reason: Tangent for curve $y = f (x)$ is straight line passing through point $(x_{1}, y_{1})$ and slope is $(d y / d x)$ at point $(x_{1}, y_{1})$ .