Iff - R - R be a function satisfying the functional Rule f x - f y - f x - x - f x - y - - x - y - R thenbegin array c l c 1P - f 0 - A - 1 Q - - f 1- f 2- - B - 3R - - f 2- f -3- - C - 0S - - f 1- f -3- - D - 2lendarray A- P - B - Q - A - R - C - S - DB- P - A - Q - B - R - D - S - CC- P - C - Q - B - R - A - S - DD- P - C - Q - B - R - D - S - A

The correct option is C P → C, Q → B, R → A, S → DPut y = 0 in the given equation and let f(0) = k f(x+k) = 2f(x)+x Put x = k, k =2 f( k) k f( k) = k Put y = ...

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Byju's Answer

Standard XII

Mathematics

Composite Function

Iff : R → R b...

Question

$I f f : R \to R$ be a function satisfying the functional Rule $f (x + f (y)) = f (x) + x + f (x - y); \forall x, y \in R$ then
$\begin{matrix} Column I & Column II (P) & f (0) & (A) & 1 (Q) & | f (1) + f (2) | & (B) & 3 (R) & | f (2) + f (- 3) | & (C) & 0 (S) & | f (1) + f (- 3) | & (D) & 2 \end{matrix}$

P \to B, Q \to A, R \to C, S \to D

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P \to A, Q \to B, R \to D, S \to C

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P \to C, Q \to B, R \to A, S \to D

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P \to C, Q \to B, R \to D, S \to A

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Solution

The correct option is C $P \to C, Q \to B, R \to A, S \to D$
Put y = 0 in the given equation and let f(0) = k
f(x+k) = 2f(x)+x
Put x =-k, k =2 f(-k) -k
f(-k) = k
Put y = -k in functional definition, then
f(x+k) =f(x) + x+ f(x+k)
f(x) =-x

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Similar questions

I f f : R \to R

be a function satisfying the functional Rule

f (x + f (y)) = f (x) + x + f (x - y); \forall x, y \in R

then

\begin{matrix} Column I & Column II (P) & f (0) & (A) & 1 (Q) & | f (1) + f (2) | & (B) & 3 (R) & | f (2) + f (- 3) | & (C) & 0 (S) & | f (1) + f (- 3) | & (D) & 2 \end{matrix}

$\begin{matrix} Column - 1 & Column - II (a) & N_{2} O & (p) & Brown coloured gas (b) & N O_{2} & (q) & Blue coloured liquid at 243K (c) & N_{2} O_{3} & (r) & White crystalline solid (d) & N_{2} O_{5} & (s) & Colourless gas, produces hysterical laughter when inhaled \end{matrix}$

Q. Match the following by appropriately matching the lists based on the information given in

Column I

and

Column II

\begin{matrix} C o l u m n I & C o l u m n I I (T y p e o f △ A B C) a. cot \frac{A}{2} = \frac{b + c}{a} & p. always right angled b. a tan A + b tan B = (a + b) tan \frac{A + B}{2} & q. always isosceles c. a cos A = b cos B & r. may be right angled d. cos A = \frac{sin B}{2 sin C} & s. may be right angled isosceles \end{matrix}

$\begin{matrix} c o l u m n 1 & c o l u m n 2 a & p) x b & q) - x c & r) x^{2} d & s) x^{4} \end{matrix}$

Match the graphs with their corresponding functions.

Q. For an atom having atomic number

z

.
Match the following:

\begin{matrix} Column-I & Column-II (A) Radius of orbit & (p) is proportional to z (B) Current associated due to orbital motion of electron & (q) is inversely proportional to z (C) Magnetic field at the centre due to orbital motion of electron & {(r) is proportional to z}^{2} (D) Velocity of an electron & {(s) is proportional to z}^{3} \end{matrix}

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Iff:R→R be a function satisfying the functional Rule f(x+f(y))=f(x)+x+f(x−y);∀x,y∈R then Column IColumn II(P)f(0)(A)1(Q)|f(1)+f(2)|(B)3(R)|f(2)+f(−3)|(C)0(S)|f(1)+f(−3)|(D)2

The correct option is C P→C,Q→B,R→A,S→DPut y = 0 in the given equation and let f(0) = k f(x+k) = 2f(x)+x Put x =-k, k =2 f(-k) -k f(-k) = k Put y = -k in functional definition, then f(x+k) =f(x) + x+ f(x+k) f(x) =-x

The correct option is C $P \to C, Q \to B, R \to A, S \to D$
Put y = 0 in the given equation and let f(0) = k
f(x+k) = 2f(x)+x
Put x =-k, k =2 f(-k) -k
f(-k) = k
Put y = -k in functional definition, then
f(x+k) =f(x) + x+ f(x+k)
f(x) =-x