Maths

Let a function f : R ® R be defined as:

$f (x) = {\begin{matrix} \int_{0}^{x} (5 - | t - 3 |) d t, & x > 4 \\ x^{2} + b x, & x \leq 4 \end{matrix}$

where $b \in R .$ If f is continuous at x = 4, then which of the following statements is NOT true?

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$f (x) {\begin{cases} \int_{0}^{3} (2 - t) d t + \int_{3}^{x} (8 - t) d t; x > 4 \\ x^{2} + b x; x \leq 4 \end{cases}$

f(x) is continuous at x = 4

$16 + 4 b = \int_{0}^{3} (2 - t) d t + \int_{3}^{4} (8 - t) d t$

16 + 4b = 15

$\Rightarrow b = \frac{- 1}{4}$

$f o r x \leq 4,$ $\begin{array}{l} f (x) = x^{2} - \frac{x}{4} \\ f' (x) = 2 x - \frac{1}{4} \end{array}$

f(x) is increasing in $(\frac{1}{8}, \infty)$

rate change at $x = \frac{1}{8},$ from -ve to +ve. So minima occurs.

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New answer posted

a year ago

If the circle x² + y² – 2gx + 6y – 19c = 0, g, c $\in$ R passes through the point (6, 1) and its centre lies on the line x – 2cy = 8, then the length of intercept made by the circle on x-axis is

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Circle passes through (6, 1)

12 g – 19 c = 43 …. (i)

Centre lies on x – 2xy = 8

->g + 6c = 8 …. (ii)

From (i) & (ii), c = 1, 9 = 2

Length of x – intercept - $2 \sqrt[]{g^{2} - C}$

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New answer posted

a year ago

Let A(1, 1) B(-4, 3), C(-2, -5) be vertices of a triangle ABC, P be a point on side BC, and $Δ_{1} a n d Δ_{2}$ be the area of triangles APB and ABC, respectively. If $Δ_{1} : Δ_{2} = 4 : 7,$ then the area enclosed by the lines AP, AC and the x-axis is

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$\frac{Δ_{1}}{Δ_{2}} = \frac{| \begin{matrix} 1 & 1 & 1 \\ x & - 4 x - x & 1 \\ - 4 & 3 & 1 \end{matrix} |}{| \begin{matrix} 1 & 1 & 1 \\ - 4 & 3 & 1 \\ - 2 & - 5 & 1 \end{matrix} |} = \frac{4}{7}$

->14 x – 35 y = -95 …. (ii)

Solve (i) & (ii), x = $\frac{- 20}{7}, y = \frac{- 11}{7}$

$a r Δ A Q R$

$= \frac{1}{2} * 1 * 1 = \frac{1}{2}$

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New answer posted

a year ago

If the plane P passes through the intersection of two mutually perpendicular planes 2x + ky – 5z = 1 and 3kx – ky + 5, k < 3 and intercepts a unit length on positive x-axis, then the intercept made by the plane P on the y-axis is

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$π_{1} \equiv 2 x + k y - 5 z = 1$

$π_{2} \equiv 3 k x - k y + z = 5$

Given $π_{1} ⊥ π_{2} \Rightarrow 6 k - k^{2} - 5 = 0$

k = 1, 5

Now required plane is p₁ + $λ π_{2} = 0$

y intercept = $\frac{1 + 5 λ}{1 - λ}$

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New answer posted

a year ago

$(p \land r) \Leftrightarrow (p \land (\sim q))$ is equivalent to $(\sim p)$ when r is

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Draw the truthtable

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

T F T F

...more

Draw the truthtable

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

T F T F

T F T F F

F T F T

F F T

So, r = q

less

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New answer posted

a year ago

Let a vertical tower AB of height 2h stands on a horizontal ground. Let from a point P on the ground a man can see upto height h of the tower with an angle of elevation 2a. When from P, he moves a distance d in the direction of $\vec{A P}$ , he can see the top B of the tower with an angle of elevation a. If d = $\sqrt[]{7} h,$ then tan a is equal to

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Solve tan 2a = $\frac{h}{b}$

$t a n α = \frac{2 h}{\sqrt[]{7} h + b}$

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New answer posted

a year ago

Let S be the sample space of all five digit numbers. It p is the probability that a randomly selected number form S, is a multiple of 7 but not divisible by 5, then 9p is equal to

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Find total (5 digit) number of divisors of 7. Find total (5 digit) number of divisors of 35

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New answer posted

a year ago

Maths Determinants

Let $\vec{a} = 2 \hat{i} - \hat{j} + 5 \hat{k} a n d \vec{b} = α \hat{i} + β \hat{j} + 2 \hat{k} .$ If $((\vec{a} * \vec{b}) * \hat{i}) . \hat{k} = \frac{23}{2},$ then $| \vec{b} * 2 \hat{j} |$ is equal to

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$(\vec{a} * \vec{b}) * \hat{i} = (\vec{a} . \hat{i}) \vec{b} - (\vec{b} . \hat{i}) \vec{a}$

$((\vec{a} * \vec{b}) * \hat{i}) . \hat{k} = (\vec{a} . \hat{i}) (\vec{b} . \hat{k}) - (\vec{b} . \hat{i}) (\vec{a} . \hat{k})$

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New answer posted

a year ago

Let $\vec{a} = α \hat{i} + \hat{j} + β \hat{k} a n d \vec{b} = 3 \hat{i} - 5 \hat{j} + 4 \hat{k}$ be two vectors, such that $\vec{a} * \vec{b} = - \hat{i} + 9 \hat{i} + 12 \hat{k} .$ Then the projection of $\vec{b} - 2 \vec{a} o n \vec{b} + \vec{a}$ is equal to

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$\vec{a} * \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ α & 1 & β \\ 3 & - 5 & 4 \end{matrix} | = (4 + 5 β) \hat{i} + (3 β - 4 α) \hat{j} + (- 5 α - 3) \hat{k}$

Compare with $\vec{a} * \vec{b} = - \hat{i} + 9 \hat{j} + 12 \hat{k}$

b = -1, a = -3

Projection of $\vec{b} - 2 \vec{a} o n \vec{b} + \vec{a}$

$= \frac{(\vec{b} - 2 \vec{a}) . (\vec{b} + \vec{a})}{| \vec{b} + \vec{a} |}$

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New answer posted

a year ago

Let P(a, b) be a point on the parabola y² = 8x such that the tangent at P passes through the centre of the circle x² + y² – 10x – 14y + 65 = 0. Let A be the product of all possible values of a and B be the product of all possible values of b. Then the value of A + B is equal to

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