if a function f is such that f 0 2f 1 3f x 2 2f x f x 1

If a function $$F$$ is such that $$F\left( 0 \right) = 2,\,F\left( 1 \right) = 3,\,F\left( {x + 2} \right) = 2F\left( x \right) - F\left( {x + 1} \right)$$ for $$x \geqslant 0,$$ then $$F\left( 5 \right)$$ is equal to :

A. $$ - 7$$

B. $$ - 3$$

C. $$17$$

D. $$13$$

Answer : $$13$$

Solution :
$$\eqalign{ & {\text{We have, }}F\left( {x + 2} \right) = 2F\left( x \right) - F\left( {x + 1} \right).....\left( {\text{i}} \right) \cr & x = 0{\text{ in }}\left( 1 \right) \Rightarrow F\left( 2 \right) = 2F\left( 0 \right) - F\left( 1 \right) = 2\left( 2 \right) - 3 = 1 \cr & x = 1{\text{ in }}\left( 1 \right) \Rightarrow F\left( 3 \right) = 2F\left( 1 \right) - F\left( 2 \right) = 2\left( 3 \right) - 1 = 5 \cr & x = 2{\text{ in }}\left( 1 \right) \Rightarrow F\left( 4 \right) = 2F\left( 2 \right) - F\left( 3 \right) = 2\left( 1 \right) - 5 = - 3 \cr & x = 3{\text{ in }}\left( 1 \right) \Rightarrow F\left( 5 \right) = 2F\left( 3 \right) - F\left( 4 \right) = 2\left( 5 \right) - \left( { - 3} \right) = 13 \cr} $$
$$\therefore $$ The correct answer is (D).

Releted MCQ Question on
Calculus >> Function

Releted Question 1

Let $$R$$ be the set of real numbers. If $$f:R \to R$$ is a function defined by $$f\left( x \right) = {x^2},$$ then $$f$$ is:

A. Injective but not surjective

B. Surjective but not injective

C. Bijective

D. None of these.

View Answer

Releted Question 2

The entire graphs of the equation $$y = {x^2} + kx - x + 9$$ is strictly above the $$x$$-axis if and only if

A. $$k < 7$$

B. $$ - 5 < k < 7$$

C. $$k > - 5$$

D. None of these.

View Answer

Releted Question 3

Let $$f\left( x \right) = \left| {x - 1} \right|.$$ Then

A. $$f\left( {{x^2}} \right) = {\left( {f\left( x \right)} \right)^2}$$

B. $$f\left( {x + y} \right) = f\left( x \right) + f\left( y \right)$$

C. $$f\left( {\left| x \right|} \right) = \left| {f\left( x \right)} \right|$$

D. None of these

View Answer

Releted Question 4

If $$x$$ satisfies $$\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6,$$ then

A. $$0 \leqslant x \leqslant 4$$

B. $$x \leqslant - 2\,{\text{or}}\,x \geqslant 4$$

C. $$x \leqslant 0\,{\text{or}}\,x \geqslant 4$$

D. None of these

View Answer

If a function $$F$$ is such that $$F\left( 0 \right) = 2,\,F\left( 1 \right) = 3,\,F\left( {x + 2} \right) = 2F\left( x \right) - F\left( {x + 1} \right)$$ for $$x \geqslant 0,$$ then $$F\left( 5 \right)$$ is equal to :

Releted MCQ Question on
Calculus >> Function

Let $$R$$ be the set of real numbers. If $$f:R \to R$$ is a function defined by $$f\left( x \right) = {x^2},$$ then $$f$$ is:

The entire graphs of the equation $$y = {x^2} + kx - x + 9$$ is strictly above the $$x$$-axis if and only if

Let $$f\left( x \right) = \left| {x - 1} \right|.$$ Then

If $$x$$ satisfies $$\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6,$$ then

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If a function $$F$$ is such that $$F\left( 0 \right) = 2,\,F\left( 1 \right) = 3,\,F\left( {x + 2} \right) = 2F\left( x \right) - F\left( {x + 1} \right)$$ for $$x \geqslant 0,$$ then $$F\left( 5 \right)$$ is equal to :

Releted MCQ Question on Calculus >> Function

Let $$R$$ be the set of real numbers. If $$f:R \to R$$ is a function defined by $$f\left( x \right) = {x^2},$$ then $$f$$ is:

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Let $$f\left( x \right) = \left| {x - 1} \right|.$$ Then

If $$x$$ satisfies $$\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right| \geqslant 6,$$ then

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