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CBSE 2012 Mathematics Definite Integration Problem’s Solution

Evaluate:

\int\limits_{ - 1}^2 {\left| {{x^3} - x} \right|dx}

CBSE 2012

Solution

\int\limits_{ - 1}^2 {\left| {{x^3} - x} \right|dx}  = I = \int\limits_{ - 1}^1 {\left| {{x^3} - x} \right|dx}  + \int\limits_1^2 {\left| {{x^3} - x} \right|dx}

Since {\left| {{x^3} - x} \right|} is an even function,

\int\limits_{ - 1}^1 {\left| {{x^3} - x} \right|dx}  = 2\int\limits_0^1 {\left| {{x^3} - x} \right|dx}

I = 2\int\limits_0^1 {\left| {{x^3} - x} \right|dx}  + \int\limits_1^2 {\left| {{x^3} - x} \right|dx}

Since x > {x^3} when x lies between 0 to 1 and {x^3} > x when x lies between 1 and 2,

I = 2\int\limits_0^1 {\left( {x - {x^3}} \right)dx}  + \int\limits_1^2 {\left( {{x^3} - x} \right)dx}

 = 2\left. {\left( {\frac{{{x^2}}}{2} - \frac{{{x^4}}}{4}} \right)} \right|_0^1 + \left. {\left( {\frac{{{x^4}}}{4} - \frac{{{x^2}}}{2}} \right)} \right|_1^2

 = 2\left[ {\left( {\frac{1}{2} - \frac{1}{4}} \right) - 0} \right] + \left[ {\left( {\frac{{16}}{4} - \frac{4}{2}} \right) - \left( {\frac{1}{4} - \frac{1}{2}} \right)} \right]

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

 = \frac{3}{4} + 2 = \frac{{11}}{4}

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

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

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$