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Differentiation question

2023 · 1 Feb · Shift 1 · Q43
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  5. /2023 · 1 Feb · Shift 1 · Q43

Differentiation question

2023 · 1 Feb · Shift 1 · Q43

JEE MainMathematicsDifferentiationNumerical+4 / −1
If f(x)=x2+g′(1)x+g′′(2)f(x)=x^{2}+g^{\prime}(1) x+g^{\prime \prime}(2)f(x)=x2+g′(1)x+g′′(2) and g(x)=f(1)x2+xf′(x)+f′′(x)g(x)=f(1) x^{2}+x f^{\prime}(x)+f^{\prime \prime}(x)g(x)=f(1)x2+xf′(x)+f′′(x), then the value of f(4)−g(4)f(4)-g(4)f(4)−g(4) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 14

Let f(x)=x2+g′(1)x+g′′(2)f(x)=x^2+g'(1)x+g''(2)f(x)=x2+g′(1)x+g′′(2) and g(x)=f(1)x2+xf′(x)+f′′(x).g(x)=f(1)x^2+x f'(x)+f''(x).g(x)=f(1)x2+xf′(x)+f′′(x).

We need to find f(4)−g(4)f(4)-g(4)f(4)−g(4).

1. Write f(x)f(x)f(x) in a simpler form

Let a=g′(1),b=g′′(2).a=g'(1), \qquad b=g''(2).a=g′(1),b=g′′(2). Then f(x)=x2+ax+b.f(x)=x^2+ax+b.f(x)=x2+ax+b.

So its derivatives are f′(x)=2x+a,f'(x)=2x+a,f′(x)=2x+a, f′′(x)=2.f''(x)=2.f′′(x)=2.

Also, f(1)=1+a+b.f(1)=1+a+b.f(1)=1+a+b.

2. Express g(x)g(x)g(x) using a,ba,ba,b

Given g(x)=f(1)x2+xf′(x)+f′′(x),g(x)=f(1)x^2+x f'(x)+f''(x),g(x)=f(1)x2+xf′(x)+f′′(x), substitute the values found above: g(x)=(1+a+b)x2+x(2x+a)+2.g(x)=(1+a+b)x^2+x(2x+a)+2.g(x)=(1+a+b)x2+x(2x+a)+2.

Now simplify: g(x)=(1+a+b)x2+2x2+ax+2,g(x)=(1+a+b)x^2+2x^2+ax+2,g(x)=(1+a+b)x2+2x2+ax+2, g(x)=(3+a+b)x2+ax+2.g(x)=(3+a+b)x^2+ax+2.g(x)=(3+a+b)x2+ax+2.

3. Find g′(x)g'(x)g′(x) and use g′(1)=ag'(1)=ag′(1)=a

Differentiate g(x)g(x)g(x): g′(x)=2(3+a+b)x+a.g'(x)=2(3+a+b)x+a.g′(x)=2(3+a+b)x+a.

Hence, g′(1)=2(3+a+b)+a=6+3a+2b.g'(1)=2(3+a+b)+a=6+3a+2b.g′(1)=2(3+a+b)+a=6+3a+2b.

But by definition, a=g′(1).a=g'(1).a=g′(1). Therefore, a=6+3a+2b.a=6+3a+2b.a=6+3a+2b. So, 2a+2b+6=0,2a+2b+6=0,2a+2b+6=0,

4. Find g′′(x)g''(x)g′′(x) and use g′′(2)=bg''(2)=bg′′(2)=b

Differentiate again: g′′(x)=2(3+a+b).g''(x)=2(3+a+b).g′′(x)=2(3+a+b).

Thus, g′′(2)=2(3+a+b).g''(2)=2(3+a+b).g′′(2)=2(3+a+b). But by definition, b=g′′(2).b=g''(2).b=g′′(2). So, b=2(3+a+b)=6+2a+2b.b=2(3+a+b)=6+2a+2b.b=2(3+a+b)=6+2a+2b. Hence, 2a+b+6=0.2a+b+6=0.2a+b+6=0.

Using a+b=−3a+b=-3a+b=−3, subtract: (2a+b+6)−(a+b+3)=0(2a+b+6)-(a+b+3)=0(2a+b+6)−(a+b+3)=0 which gives a+3=0,a+3=0,a+3=0, so a=−3.a=-3.a=−3. Then from a+b=−3a+b=-3a+b=−3, −3+b=−3⇒b=0.-3+b=-3 \Rightarrow b=0.−3+b=−3⇒b=0.

5. Find f(x)f(x)f(x) and g(x)g(x)g(x)

Now substitute a=−3a=-3a=−3, b=0b=0b=0.

For f(x)f(x)f(x):

f(x)=x2−3x.f(x)=x^2-3x.f(x)=x2−3x.

For g(x)g(x)g(x):

Since g(x)=(3+a+b)x2+ax+2,g(x)=(3+a+b)x^2+ax+2,g(x)=(3+a+b)x2+ax+2, we get g(x)=(3−3+0)x2−3x+2=−3x+2.g(x)=(3-3+0)x^2-3x+2=-3x+2.g(x)=(3−3+0)x2−3x+2=−3x+2.

6. Compute f(4)−g(4)f(4)-g(4)f(4)−g(4)

First, f(4)=42−3⋅4=16−12=4.f(4)=4^2-3\cdot 4=16-12=4.f(4)=42−3⋅4=16−12=4.

Next, g(4)=−3⋅4+2=−12+2=−10.g(4)=-3\cdot 4+2=-12+2=-10.g(4)=−3⋅4+2=−12+2=−10.

Therefore, f(4)−g(4)=4−(−10)=14.f(4)-g(4)=4-(-10)=14.f(4)−g(4)=4−(−10)=14.

7. Compare with stored answer

Derived answer = 141414.

Stored correct answer = 141414.

They agree.

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