If f - R - R be a continuos function such that f x -1 x t f t dt- then correct statement isA- -xx fx d x-2 xB- -x- x f x dx - xC- -33 fx d x-xD- -33 f x dx -12

The correct option is C ∫^3_ 3 f(x) dx = xf'(x) = f(x), nbsp; f(1) = 0 f'(x)/f(x) = 1 ⇒ ln (f(x)) = x + c f(x) = k.e^x 0 = k × e ⇒ k = 0 So f(x) = 0 ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

If f : R → R ...

Question

If $f : R \to R$ be a continuos function such that $f (x) = \int_{1}^{x} t f (t) d t,$ then correct statement is

\int_{- x}^{x} f (x) d x = 2 x

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

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\int_{- 3}^{3} f (x) d x = x

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\int_{- 3}^{3} f (x) d x = 12

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Solution

The correct option is C $\int_{- 3}^{3} f (x) d x = x$
$f^{'} (x) = f (x), f (1) = 0$
$\frac{f^{'} (x)}{f (x)} = 1 \Rightarrow l n (f (x)) = x + c$
$f (x) = k . e^{x}$
$0 = k \times e \Rightarrow k = 0$
So $f (x) = 0$

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Similar questions

Q. Let

f (x)

be a continuous function such that

\int_{n}^{n + 1} f (x) d x = n^{3}, n ϵ Z .

Then the value of

\int_{- 3}^{3} f (x) d x

Q. If

f : R \to R

be given by

f (x) = (3 - x^{3})^{\frac{1}{3}},

then

f \circ f (x)

Q. If the function

f : R \to R

be such that

f (x) = x - [x]

, where [x] denotes the greatest integer less than or equal to x, then

f^{- 1} (x)

is

(a)

\frac{1}{x - [x]}

(b) [x] − x

(c) not defined

(d) none of these

Q. Let

f : R \to R

be defined by

f (x) = 2 x + | x |

. Then

f (2 x) + f (- x) - f (x)

L e t g (x) = m i n i m u m o f {(x + | x |), (x - | x |)}

then

\int_{- 3}^{3} f (x) d x

equals

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If f:R→R be a continuos function such that f(x)=∫x1t f(t)dt, then correct statement is

The correct option is C ∫3−3f(x)dx=xf′(x)=f(x), f(1)=0 f′(x)f(x)=1⇒ln(f(x))=x+c f(x)=k.ex 0=k×e⇒k=0 So f(x)=0

If $f : R \to R$ be a continuos function such that $f (x) = \int_{1}^{x} t f (t) d t,$ then correct statement is

The correct option is C $\int_{- 3}^{3} f (x) d x = x$
$f^{'} (x) = f (x), f (1) = 0$
$\frac{f^{'} (x)}{f (x)} = 1 \Rightarrow l n (f (x)) = x + c$
$f (x) = k . e^{x}$
$0 = k \times e \Rightarrow k = 0$
So $f (x) = 0$