Let fx-A sin- x-2-B- f-1-2-2 and -01 f x dx -2 A - then A and B areA- -2- -2- 2- 3- 0-4 -D- 4- 0

F(x)=Asin(π x2)+B ⇒ f'(x)=Acos(π x2)·π2 ⇒ f'(12)=Acos(π4)·π2=√(2) ⇒ A= 4π Now, nbsp;∫_0^1f(x)dx=2Aπ ⇒∫_0^1[Asin(π x2)+B]dx=2Aπ ⇒ 2Aπcos[π x2]_0^1+B[x]_0^1=2A ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Standard Formulae - 2

Let fx=A sinπ...

Question

Let $f (x) = A sin (\frac{π x}{2}) + B, f^{'} (\frac{1}{2}) = \sqrt{2}$ and $1 \int 0 f (x) d x = \frac{2 A}{π}$ , then $A$ and $B$ are

\frac{π}{2}, \frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2}{π}, \frac{3}{π}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0, - 4 π

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4}{π}, 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{4}{π}, 0$
$f (x) = A sin (\frac{π x}{2}) + B$
$\Rightarrow f^{'} (x) = A cos (\frac{π x}{2}) \cdot \frac{π}{2}$
$\Rightarrow f^{'} (\frac{1}{2}) = A cos (\frac{π}{4}) \cdot \frac{π}{2} = \sqrt{2}$
$\Rightarrow A = \frac{4}{π}$
Now, $1 \int 0 f (x) d x = \frac{2 A}{π}$
$\Rightarrow 1 \int 0 [A sin (\frac{π x}{2}) + B] d x = \frac{2 A}{π}$
$\Rightarrow - \frac{2 A}{π} cos {[\frac{π x}{2}]}_{0}^{1} + B [x]_{0}^{1} = \frac{2 A}{π}$
$\Rightarrow B = 0$
Hence $A = \frac{4}{π}$ and $B = 0$

Suggest Corrections

Similar questions

Q. If

f (x) = A sin (\frac{π x}{2}) + B, f^{'} (\frac{1}{2}) = \sqrt{2}

and

\int_{0}^{1} f (x) d x = \frac{2 A}{π}

, then

A

and

B

are

Q. If

f (x) = A sin (\frac{π x}{2}) + B, f^{'} (\frac{1}{2}) = \sqrt{2}

and

\int_{0}^{1} f (x) d x = \frac{2 A}{π},

then the constant A and B are

Q. If

f (x) = A sin (\frac{π x}{2}) + B, f^{'} (\frac{1}{2}) = \sqrt{2}

and

\int_{0}^{1} f (x) d x = \frac{2 A}{π}

then the constants

A

and

B

are

Q. If

f (x) = A sin (\frac{π x}{2}) + B, f (\frac{1}{2}) = \sqrt{2}

and

\int_{0}^{1} f (x) d x = \frac{2 A}{π}

, then the constant

A

and

B

are

Q. Let

f : (- 1, 1) \to B

, be a function defined by

f (x) = t a n^{- 1} \frac{2 x}{1 - x^{2}}

, then f is both one - one and onto when B is the interval

Join BYJU'S Learning Program

Let f(x)=Asin(πx2)+B, f′(12)=√2 and 1∫0f(x)dx=2Aπ, then A and B are

The correct option is D 4π,0f(x)=Asin(πx2)+B ⇒f′(x)=Acos(πx2)⋅π2 ⇒f′(12)=Acos(π4)⋅π2=√2 ⇒A=4π Now, 1∫0f(x)dx=2Aπ ⇒1∫0[Asin(πx2)+B]dx=2Aπ ⇒−2Aπcos[πx2]10+B[x]10=2Aπ ⇒B=0 Hence A=4π and B=0

Let $f (x) = A sin (\frac{π x}{2}) + B, f^{'} (\frac{1}{2}) = \sqrt{2}$ and $1 \int 0 f (x) d x = \frac{2 A}{π}$ , then $A$ and $B$ are