SATHEE: Area Question 31

Area Question 31

Question 31

For which of the following values of $m$ , is the area of the region bounded by the curve $y = x - x^{2}$ and the line $y = m x$ equals $\frac{9}{2}$ ?

$(1999, 3 M)$ (a) -4 (b) -2 (c) 2 (d) 4

Analytical & Descriptive Questions

$f (x)$ is a quadratic function and its maximum value occurs at a point $V$ . $A$ is a point of intersection of $y = f (x)$ with $X$ -axis and point $B$ is such that chord $A B$ subtends a right angle at $V$ . Find the area enclosed by $f (x)$ and chord $A B$ .

$(2005, 5$ M)

Show Answer

Solution:

Case I When $m = 0$

In this case, $y = x - x^{2}$

and

$y = 0$

are two given curves, $y > 0$ is total region above $X$ -axis.

Therefore, area between $y = x - x^{2}$ and $y = 0$

is area between $y = x - x^{2}$ and above the $X$ -axis

Hence, no solution exists.

Case II When $m < 0$

In this case, area between $y = x - x^{2}$ and $y = m x$ is $O A B C O$ and points of intersection are $(0, 0)$ and $1 - m, m (1 - m)$ .

$∴$ Area of curve $O A B C O = \int_{0}^{1 - m} [x - x^{2} - m x] d x$

$\begin{aligned} = (1 - m) \frac{x^{2}}{2} - {\frac{x^{3}}{3}}^{1 - m} & = \frac{1}{2} (1 - m)^{3} - \frac{1}{3} (1 - m)^{3} = \frac{1}{6} (1 - m)^{3} ∴ & \frac{1}{6} (1 - m)^{3} & = \frac{9}{2} \Rightarrow (1 - m)^{3} & = 27 \Rightarrow 1 - m & = 3 \Rightarrow & m & = - 2 \end{aligned}$

326 Area

Case III When $m > 0$

In this case, $y = m x$ and $y = x - x^{2}$ intersect in $(0, 0)$ and $(1 - m), m (1 - m)$ as shown in figure

$∴$ Area of shaded region $= \int_{1 - m}^{0} (x - x^{2} - m x) d x$

$\begin{aligned} = (1 - m) \frac{x^{2}}{2} - {\frac{x^{3}}{3}}^{0} & = - \frac{1}{2} (1 - m) (1 - m)^{2} + \frac{1}{3} (1 - m)^{3} & = - \frac{1}{6} (1 - m)^{3} \Rightarrow \frac{9}{2} & = - \frac{1}{6} (1 - m)^{3} \Rightarrow (1 - m)^{3} & = - 27 \Rightarrow (1 - m) & = - 3 \Rightarrow m & = 3 + 1 = 4 \end{aligned}$

[given]

Therefore, (b) and (d) are the answers.

$\begin{array}{rrr} \Rightarrow 4 a^{2} f (- 1) + 4 a f (1) + f (2) & = 3 a^{2} + 3 a, & 4 b^{2} f (- 1) + 4 b f (1) + f (2) = 3 b^{2} + 3 b and 4 c^{2} f (- 1) + 4 c f (1) + f (2) & = 3 c^{2} + 3 c \end{array}$

where, $f (x)$ is quadratic expression given by,

$f (x) = a x^{2} + b x + c and Eqs. (i), (ii) and (iii).$

$\Rightarrow 4 x^{2} f (- 1) + 4 x f (1) + f (2) = 3 x^{2} + 3 x$

or $4 f (- 1) - 3 x^{2} + 4 f (1) - 3 x + f (2) = 0$

As above equation has 3 roots $a, b$ and $c$ .

So, above equation is identity in $x$ .

i.e. coefficients must be zero.

$\begin{array}{rrrl} \Rightarrow & f (- 1) & = 3 / 4, f (1) = 3 / 4, f (2) = 0 & ∵ & f (x) & = a x^{2} + b x + c ∴ & a = - 1 / 4, b & = 0 and c = 1, using Eq. (v) \end{array}$

Thus, $f (x) = \frac{4 - x^{2}}{4}$ shown as,

Let $A (- 2, 0), B = (2 t, - t^{2} + 1)$

Since, $A B$ subtends right angle at vertex $V (0, 1)$ .

$\Rightarrow \frac{1}{2} \cdot \frac{t^{2}}{2 t} = - 1$

$\begin{array}{lc} \Rightarrow & t = 4 ∴ & B (8, - 15) \end{array}$

So, equation of chord $A B$ is $y = \frac{- (3 x + 6)}{2}$ .

$∴$ Required area $= | \int_{- 2}^{8} \frac{4 - x^{2}}{4} + \frac{3 x + 6}{2} d x |$

$\begin{aligned} = | x - \frac{x^{3}}{12} + \frac{3 x^{2}}{4} + 3 x_{- 2}^{8} | & =∣ 8 - \frac{128}{3} + 48 + 24 - 2 + \frac{2}{3} + 3 - 6 & = \frac{125}{3} sq units \end{aligned}$

Area Question 31

Question 31

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Area Question 31

Question 31

Analytical & Descriptive Questions

326 Area

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

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Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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