let f 0infty r and f x 0 x f t dt if f x 2 x 2 1 x 493028585

Let $$f:\left( {0,\,\infty } \right) \to R$$ and $$F\left( x \right) = \int\limits_0^x {f\left( t \right)dt} .$$ If $$F\left( {{x^2}} \right) = {x^2}\left( {1 + x} \right),$$ then $$f\left( 4 \right)$$ equals :

A. $$\frac{5}{4}$$

B. $$7$$

C. $$4$$

D. $$2$$

Answer : $$4$$

Solution :
$$\eqalign{ & F'\left( x \right) = f\left( x \right) \cr & {\text{Also, }}F\left( t \right) = t\left( {1 + \sqrt t } \right) \cr & \Rightarrow F'\left( t \right) = 1 + \frac{3}{2}{t^{\frac{1}{2}}}\,; \cr & F'\left( 4 \right) = 1 + 3 = 4 \Rightarrow f\left( 4 \right) = 4 \cr} $$

Releted MCQ Question on
Calculus >> Definite Integration

Releted Question 1

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

A. $$ - 1$$

B. $$2$$

C. $$1 + {e^{ - 1}}$$

D. none of these

View Answer

Releted Question 2

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

A. no root in $$\left( {0,\,2} \right)$$

B. at least one root in $$\left( {0,\,2} \right)$$

C. a double root in $$\left( {0,\,2} \right)$$

D. two imaginary roots

View Answer

Releted Question 3

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

A. $$\frac{\pi }{4}$$

B. $$\frac{\pi }{2}$$

C. $$\pi $$

D. none of these

View Answer

Releted Question 4

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

A. $$\pi $$

B. $$1$$

C. $$0$$

D. none of these

View Answer

Let $$f:\left( {0,\,\infty } \right) \to R$$ and $$F\left( x \right) = \int\limits_0^x {f\left( t \right)dt} .$$ If $$F\left( {{x^2}} \right) = {x^2}\left( {1 + x} \right),$$ then $$f\left( 4 \right)$$ equals :

Releted MCQ Question on
Calculus >> Definite Integration

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

Practice More Releted MCQ Question on
Definite Integration

Practice More MCQ Question on Maths Section

Let $$f:\left( {0,\,\infty } \right) \to R$$ and $$F\left( x \right) = \int\limits_0^x {f\left( t \right)dt} .$$ If $$F\left( {{x^2}} \right) = {x^2}\left( {1 + x} \right),$$ then $$f\left( 4 \right)$$ equals :

Releted MCQ Question on Calculus >> Definite Integration

The value of the definite integral $$\int\limits_0^1 {\left( {1 + {e^{ - {x^2}}}} \right)} \,dx$$ is-

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$ Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-

For any integer $$n$$ the integral $$\int\limits_0^\pi {{e^{{{\cos }^2}x}}} {\cos ^3}\left( {2n + 1} \right)xdx$$ has the value-

Practice More Releted MCQ Question on Definite Integration

Practice More MCQ Question on Maths Section

Releted MCQ Question on
Calculus >> Definite Integration

Let $$a,\,b,\,c$$ be non-zero real numbers such that $$\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = } \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx.} $$
Then the quadratic equation $$a{x^2} + bx + c = 0$$ has-

Practice More Releted MCQ Question on
Definite Integration