Maths

Let S be the region bonded by the curves y = x³ and y² = x. The curve y = 2|x| divides S into two regions of areas R₁, and R₂ _________.

If max {R₁, R₂} = R₂, then $\frac{R_{2}}{R_{1}}$ is equal to _________.

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Contributor-Level 10

$R_{2} = \int_{0}^{1} (x - x^{3}) d x = {(\frac{x^{2}}{2} - \frac{x^{4}}{4})}_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

R_{1} = \int_{0}^{\frac{1}{4}} (\sqrt[]{x} - 2 x) d x = {(\frac{2 x^{3 / 2}}{3} - x^{2})}_{0}^{\frac{1}{4}} = \frac{1}{12} - \frac{1}{16}

= \frac{4 - 3}{48} = \frac{1}{48}

R_{1} + R_{2} = {\int_{0}^{1} (\sqrt[]{x} - x^{3}) d x = (\frac{2 x^{3 / 2}}{3} - \frac{x^{4}}{4})}_{0}^{1} = \frac{2}{3} - \frac{1}{4} = \frac{5}{12}

\frac{R_{1} + R_{2}}{R_{1}} = 1 + \frac{R_{2}}{R_{1}} = \frac{\frac{5}{12}}{\frac{1}{48}} = 20

∴ \frac{R_{2}}{R_{1}} = 19

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New answer posted

a year ago

If two tangents drawn from a point (a, b) lying on the ellipse 25x² + 4y² = 1 to the parabola y² = 4x are such that the slope of one tangent is four times the other, then the value of ${(10 α + 5)}^{2} + {(16 β^{2} + 50)}^{2}$ equals __________.

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Contributor-Level 10

$\frac{x^{2}}{\frac{1}{25}} + \frac{y^{2}}{\frac{1}{4}} = 1$

$25 α^{2} + 4 β^{2} = 1 . . . . . . . . . . (i)$

Equation of tangent to parabola y = mx + $\frac{1}{m}$ passes

though (a, b)

$α m^{2} - β m + 1 = 0$

$m_{1} + m_{2} = \frac{β}{α}, 4 m_{1}^{2} = \frac{1}{α}$

$∴ f r o m (i) & (i i)$

25(a² + a) = 1 …………(iii)

$∴ {(10 α + 5)}^{2} + {(16 β^{2} + 50)}^{2} = 2929$

$∴ a = 1 + 0 + b {(s i n - \frac{π}{2})}^{\frac{π}{2}} + 2 b + 1$

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New answer posted

a year ago

Let $\underset{0 \leq x \leq 2}{M a x} {\frac{9 - x^{2}}{5 - x}} = α a n d \underset{0 \leq x \leq 2}{M i n} = {\frac{9 - x^{2}}{5 - x}} = β$

If $\int_{β - \frac{8}{3}}^{2 α - 1} M a x {\frac{9 - x^{2}}{5 - x}, x} d x = α_{1} + α_{2} l o g_{e} (\frac{8}{15})$ then $α_{1} + α_{2}$ is equal to __________.

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Contributor-Level 10

Let f(x) = $\frac{x^{2} - 9}{x - 5}$

$f' (x) = \frac{2 x (x - 5) - (x^{2} - 9)}{{(x - 5)}^{2}}$

$= \frac{x^{2} - 10 x + 9}{{(x - 5)}^{2}} = \frac{(x - 1) (x - 9)}{{(x - 5)}^{2}}$

$α = f (1) = 2, β = {f (0), f (2)} = \frac{5}{3}$

$∴ \int_{- 1}^{3} m a x {\frac{x^{2} - 9}{x - 5}, x} d x + {[\frac{x^{2}}{2}]}_{9 / 5}^{3}$

a₁ = 18, a₂ = 16

a₁ + a₂ = 34

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New answer posted

a year ago

Let f (θ) = $s i n θ + \int_{- \frac{π}{2}}^{\frac{π}{2}} (s i n θ + t c o s θ) f (t) d t .$ Then the value of $| \int_{0}^{\frac{π}{2}} f (θ) d θ |$ is

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Contributor-Level 10

$f (θ) = s i n θ + \int_{- \frac{π}{2}}^{\frac{π}{2}} (s i n θ + t c o s θ) f (t) d t$

$= s i n θ + s i n θ \int_{- \frac{π}{2}}^{\frac{π}{2}} f (t) d t + c o s θ \int_{- \frac{π}{2}}^{\frac{π}{2}} t f (t) d t$

$= (1 + \int_{- \frac{π}{2}}^{\frac{π}{2}} f (t) d t) s i n θ + (\int_{- \frac{π}{2}}^{\frac{π}{2}} t f (t) d t) c o s θ$

f(q) = a sin q + b cos q

$∴ a = 1 + \int_{- \frac{π}{2}}^{\frac{π}{2}} f (t) d t = 1 + \int_{- \frac{π}{2}}^{\frac{π}{2}} (a s i n t + b c o s t) d t$

$∴ a = 2 b + 1 . . . . . . . . . . (i)$

$b = \int_{- \frac{π}{2}}^{\frac{π}{2}} t f (t) d t = \int_{- \frac{π}{2}}^{\frac{π}{2}} (a s i n t + b c o s t) d t$

Solving (i) & (ii) we get a = $- \frac{1}{3}, b = - \frac{2}{3}$

$∴ | \int_{0}^{\frac{π}{2}} f (θ) d θ | = 1$

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New answer posted

a year ago

The number of points where the function f(x) = ${\begin{cases} | 2 x^{2} - 3 x - 7 | i f x \leq - 1 \\ [4 x^{2} - 1] i f - 1 < x < 1 \\ | x + 1 | + | x - 2 | i f x \geq 1 \end{cases}$ [t] denotes the greatest integer $\leq$ t is discontinuous is __________-.

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Contributor-Level 10

$y = 2 | x^{2} - \frac{3}{2} x - \frac{7}{2} |$

$= 2 | {(x - \frac{3}{4})}^{2} - \frac{65}{16} |$

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New answer posted

a year ago

Let a line having direction ratios 1, -4, 2 intersect the lines $\frac{x - 7}{3} = \frac{y - 1}{- 1} = \frac{z + 2}{1}$ and $\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1}$ at the points A and B. Then (AB)² is equal to _________.

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Contributor-Level 10

$\frac{x - 7}{3} = \frac{y - 1}{- 1} = \frac{z + 2}{1} = r_{1}$

$A (3 r_{1} + 7, 1 - r_{1}, - 2 + r_{1})$

and $\frac{x}{2} = \frac{y - 7}{3} = \frac{z}{1} = r_{2}$

$B (2 r_{2}, 7 + 3 r_{2}, r_{2})$

$A / q, \frac{3 r_{1} - 2 r_{2} + 7}{1} = \frac{3 r_{2} + r_{1} + 6}{4} = \frac{r_{1} - r_{2} - 2}{2}$

$\Rightarrow r_{1} = - 5, r_{2} = - 3$

$∴ A (- 8, 6, - 7) a n d B (- 6, - 2, - 3)$

AB² = 84

$f (x) = {\begin{cases} | 2 x^{2} - 3 x - 7 |, x \leq - 1 \\ [4 x^{2} - 1], - 1 < x < 1 \\ | x + 1 | + | x - 2 |, x \geq 1 \end{cases}$

f(-1) = 1

f(1) = 3

Hence f(x) will be discontinuous at x = 1 and also 4x² – 1 = 0 , 1 , 2

$\Rightarrow x = \pm \frac{1}{2}, \pm \frac{1}{\sqrt[]{2}}, \pm \frac{\sqrt[]{3}}{2}$

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New answer posted

a year ago

Let $A (\frac{3}{\sqrt[]{a}}, \sqrt[]{a}), a > 0$ , be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If D (3 cos q, a sin q) is a point in the fourth quadrant such that the maximum area of $Δ A C D$ is 12 square units, then a is equal to _________.

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Contributor-Level 10

$B (- \frac{3}{\sqrt[]{a}}, \sqrt[]{a}) a n d c (- \frac{3}{\sqrt[]{a}}, - \sqrt[]{a})$

$∴ A r e a o f Δ A C D = \frac{1}{2} | \begin{matrix} \frac{3}{\sqrt[]{a}} & \sqrt[]{a} & 1 \\ \frac{- 3}{\sqrt[]{a}} & - \sqrt[]{a} & 1 \\ 3 c o s θ & a s i n θ & 1 \end{matrix} |$ = 12

$\Rightarrow Δ = 3 \sqrt[]{a} | c o s θ + s i n θ | = 12$

$∴ Δ m a x = 3 \sqrt[]{a} . \sqrt[]{2} = 12 \Rightarrow a = 8$

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New answer posted

a year ago

In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, -2 marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is __________

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Contributor-Level 10

Let x = correct answer, y = incorrect answer

$∴ 3 x - 2 y = 5, x + y \leq 5, x, y \in w$

only possible (x, y) is (3, 2)

$∴$ Required number of ways =

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New answer posted

a year ago

Consider the following two propositions:

$P 1 : \sim (p \to \sim q)$

$P 2 : (p \land \sim q) \land ((\sim p) \lor q)$

If the proposition p $((\sim p) \lor q)$ is evaluated as FALSE, then :

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Vishal Baghel

Contributor-Level 10

$_{}$ $p_{1} : \sim (p \leftrightarrow \sim q)$ $p_{2} : (p \land \sim q) \land ((\sim p) \lor q)$

$= \sim ((\sim p) \lor (\sim q))$

$= \sim (\sim (p \land q))$

$p \land q$

$(\sim p) \lor ((\sim p) \lor q)$ is false (given)

p q p₁ p₂ $(\sim p) \lor q$

T F T

T F F

...more

$_{}$ $p_{1} : \sim (p \leftrightarrow \sim q)$ $p_{2} : (p \land \sim q) \land ((\sim p) \lor q)$

$= \sim ((\sim p) \lor (\sim q))$

$= \sim (\sim (p \land q))$

$p \land q$

$(\sim p) \lor ((\sim p) \lor q)$ is false (given)

p q p₁ p₂ $(\sim p) \lor q$

T F T

T F F

F T F T

F T

less

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New answer posted

a year ago

The number of one-one functions f:{a, b, c, d} ® {0, 1, 2, ……., 10} such that 2f (a) – f(b) + 3f(c) + f(d) = 0 is ___________.

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