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f(x) and g(x) are two differentiable functions on [0,2] such that f^{\prime \prime}(x)-g^{\prime \prime}(x)=0, f^{\prime}(1)=2 g^{\prime}(1)=4, f(2)=3g(2)=9 then f(x)-g(x) at x=\frac3 2 is:

Option: 1

0


Option: 2

2


Option: 3

10


Option: 4

5


Answers (1)

best_answer

Given f^{\prime \prime}(x)-g^{\prime \prime}(x)=0

f^{\prime \prime}(x)=g^{\prime \prime}(x)

Integrate both side

f^{ \prime}(x)+C_1=g^{ \prime}(x)+C_2

f^{ \prime}(x)-g^{ \prime}(x)=C_2-C_1\qquad\qquad{\color{DarkRed} \ldots(1)}

But {\color{DarkGreen} f^{\prime}(1)=2 g^{\prime}(1)=4}

Or

{\color{DarkGreen} f^{\prime}(1)=4\text{ and } g^{\prime}(1)=2}

Therefore at x = 1

C_2-C_1=f^{ \prime}(1)-g^{ \prime}(1)

C_2-C_1=2f^{\prime}(x)=g^{\prime}(x)+2

Equation 1 becomes

f^{ \prime}(x)-g^{ \prime}(x)=2

f^{ \prime}(x)=2+g^{ \prime}(x)

Again integrate both sides

f(x)+K_1=g(x)+K_2+2x

f(x)-g(x)=K_2-K_1+2x\qquad\qquad{\color{DarkRed} \ldots(2)}

Since f(2)=3g(2)=9

Therefore at x = 2

f\left ( 2 \right )-g\left ( 2 \right )=2\times 2+\left ( k_{2}-k_{1} \right )

9-3=4+\left ( k_{2}-k_{1} \right )

2=\left ( k_{2}-k_{1} \right )

Using Equation (2)

f\left ( x \right )-g\left ( x \right )= 3+2=5

Posted by

Riya

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