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Limits Continuity and Differentiability question

2021 · 1 Sep · Shift 2 · Q39

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Limits Continuity and Differentiability
2021 · 1 Sep · Shift 2 · Q39

Limits Continuity and Differentiability question

2021 · 1 Sep · Shift 2 · Q39

JEE MainMathematicsLimits Continuity and DifferentiabilityNumerical+4 / −1

Let

f(x) = {x^6} + 2{x^4} + {x^3} + 2x + 3

, x

\in

R. Then the natural number n for which

\mathop {\lim }\limits_{x \to 1} {{{x^n}f(1) - f(x)} \over {x - 1}} = 44

is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 7

We need to evaluate

\\lim_{x\to 1} \frac{x^n f(1)-f(x)}{x-1}=44.

Given

f(x)=x^6+2x^4+x^3+2x+3.

First compute $f(1)$ :

f(1)=1^6+2\cdot 1^4+1^3+2\cdot 1+3=1+2+1+2+3=9.

So the limit becomes

\lim_{x\to 1} \frac{9x^n-f(x)}{x-1}=44.

For this limit to be finite, the numerator must vanish at $x=1$ :

9\cdot 1^n-f(1)=9-9=0,

which is true.

Hence we can use the fact that if

g(x)=9x^n-f(x),

then

\lim_{x\to 1}\frac{g(x)-g(1)}{x-1}=g'(1).

Since $g(1)=0$ , the given limit is simply $g'(1)$ .

Differentiate:

g'(x)=9n x^{n-1}-f'(x).

Now

f'(x)=6x^5+8x^3+3x^2+2.

So

f'(1)=6+8+3+2=19.

Therefore,

g'(1)=9n-19.

Given that the limit equals $44$ ,

9n-19=44.

So,

9n=63

n=7.

Therefore, the required natural number is

\boxed{7}.

Comparison with stored answer: stored correct answer is $7$ , which matches our result.

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