SATHEE: Definite Integration Question 64

Definite Integration Question 64

Question 64

Match List I with List II and select the correct answer using codes given below the lists.

(2014)

List I			List II
P.	The number of polynomials $f (x)$ with non-negative integer coefficients of degree $\leq 2$ , satisfying $f (0) = 0$ and $\int_{0}^{1} f (x) d x = 1$ , is	(i)	8
Q.	The number of points in the interval $[- \sqrt{13}, \sqrt{13}]$ at which $f (x) = \sin (x^{2}) + \cos (x^{2})$ attains its maximum value, is	(ii)	2
R.	$\int_{- 2}^{2} \frac{3 x^{2}}{1 + e^{x}} d x$ equals	(iii)	4
S.	$\frac{\int_{- 1 / 2}^{1 / 2} \cos 2 x \log \frac{1 + x}{1 - x} d x}{\int_{0}^{1 / 2} \cos 2 x \log \frac{1 + x}{1 - x} d x}$ equals	(iv)	0

Codes $\begin{array}{llll} P & Q & R & S \end{array}$ $\begin{array}{llll} P & Q & R & S \end{array}$ (a) (iii) (ii) (iv) (i) (b) (ii) (iii) (iv) (i) (c) (iii) (ii) (i) (iv) (d) (ii) (iii) (i) (iv)

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Solution:

(P) PLAN (i) A polynomial satisfying the given conditions is taken.

(ii) The other conditions are also applied and the number of polynomial is taken out.

Let $f (x) = a x^{2} + b x + c$

$f (0) = 0 \Rightarrow c = 0$

Now, $\int_{0}^{1} f (x) d x = 1$

$\begin{aligned} \Rightarrow \frac{a x^{3}}{3} + {\frac{b x^{2}}{2}}_{0}^{1} = 1 \Rightarrow \frac{α}{3} + \frac{β}{2} = 1 & \Rightarrow 2 a + 3 b = 6 \end{aligned}$

As $a, b$ are non-negative integers.

So, $a = 0, b = 2$ or $a = 3, b = 0$

$∴ f (x) = 2 x or f (x) = 3 x^{2}$

(Q) PLAN Such type of questions are converted into only sine or cosine expression and then the number of points of maxima in given interval are obtained.

$\begin{aligned} f (x) & = \sin (x^{2}) + \cos (x^{2}) & = \sqrt{2} \frac{1}{\sqrt{2}} \cos (x^{2}) + \frac{1}{\sqrt{2}} \sin (x^{2}) & = \sqrt{2} \cos x^{2} \cos \frac{π}{4} + \sin \frac{π}{4} \sin (x^{2}) & = \sqrt{2} \cos x^{2} - \frac{π}{4} \end{aligned}$

For maximum value, $x^{2} - \frac{π}{4} = 2 n π \Rightarrow x^{2} = 2 n π + \frac{π}{4}$

$\Rightarrow x = \pm \sqrt{\frac{π}{4}}$ , for $n = 0 \Rightarrow x = \pm \sqrt{\frac{9 π}{4}}$ , for $n = 1$

So, $f (x)$ attains maximum at 4 points in $[- \sqrt{13}, \sqrt{13}]$ .

(R) PLAN

(i) $\int_{- a}^{a} f (x) d x = \int_{- a}^{a} f (- x) d x$

(ii) $\int_{- a}^{a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ , if $f (- x) = f (x)$ , i.e. $f$ is an even function.

$I = \int_{- 2}^{2} \frac{3 x^{2}}{1 + e^{x}} d x$

and $I = \int_{- 2}^{2} \frac{3 x^{2}}{1 + e^{- x}} d x$

$\Rightarrow 2 I = \int_{- 2}^{2} \frac{3 x^{2}}{1 + e^{x}} + \frac{3 x^{2} (e^{x})}{e^{x} + 1} d x$

$2 I = \int_{- 2}^{2} 3 x^{2} d x \Rightarrow 2 I = 2 \int_{0}^{2} 3 x^{2} d x$

$I = {[x^{3}]}_{0}^{2} = 8$

(S) PLAN

$f (x) d x = 0$

If $f (- x) = - f (x)$ , i.e. $f (x)$ is an odd function.

Let

$f (x) = \cos 2 x \log \frac{1 + x}{1 - x}$

$f (- x) = \cos 2 x \log \frac{1 - x}{1 + x} = - f (x)$

Hence, $f (x)$ is an odd function.

So, $\int_{- 1 / 2}^{1 / 2} f (x) d x = 0$

$(P) \to$ (ii); (Q) $\to$ (iii); (R) $\to$ (i); (S) $\to$ (iv)

Definite Integration Question 64

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Definite Integration Question 64

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