Maths

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

Maths Class 11th

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

Maths Class 11th

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

Maths Class 11th

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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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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Contributor-Level 10

f(b) = 2f(a) + 3f(c) + f(d)

Value of f(c) Value of f(a) Number of functions

1 7

...more

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New answer posted

a year ago

The number of choices for $Δ \in {\land, \lor, \Rightarrow, \Leftrightarrow}$ , such that $(p Δ q) \Rightarrow ((p Δ ~ q) \lor (~ p) Δ q)$ is a tautology, is :

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Maths Inverse Trigonometric Functions

Contributor-Level 10

$(p Δ q) \Rightarrow (p Δ \sim q) \lor (\sim p Δ q)$

Case I

When $Δ$ is same as $\lor .$

Then $(p Δ \sim q) \lor (- p Δ q)$ becomes

$(p \lor \sim q) \lor (\sim p \lor q)$ which is always true, so x becomes tautology.

Case II

When $Δ$ is same as $\lor$

Then $(p \land q) \Rightarrow (p \land \sim q) \lor (\sim p \land q)$ becomes $p \land q$ is T, then $(p \land \sim q) \lor (\sim p \land q)$ is false, so x cannot be tautology.

Case III

When $Δ$ is same as $\Rightarrow$

Then $(p \Rightarrow \sim q) \lor (- p \Rightarrow q)$ is same as $(\sim p \lor \sim q) \lor (p \lor q)$ which is true, so x becomes tautology.

Case IV

When $Δ$ is same as $\Leftrightarrow$

Then $(p \Leftrightarrow q) \Rightarrow (p \Leftrightarrow \sim q) \lor (\sim p \Leftrightarrow q)$

$p \Leftrightarrow q$ is true when p and q have same truth values $p \Leftrightarrow \sim q a n d \sim p \Leftrightarrow q$ both are false. Hence x cannot be tautology.

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New answer posted

a year ago

Let f : R -> R be defined as f(x) = x³ + x – 5

If g(x) is a function such that f(g(x)) = x, $\forall$ 'x' $\in$ R, then g' (63) is equal to……….

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Contributor-Level 10

$f (g (x)) = x \forall x \in R$

$\Rightarrow g (x) = f^{- 1} (x)$

For y = g (x)

x = y³ + y – 5

$\frac{d x}{d y} = 3 y^{2} + 1 \Rightarrow g' (63) = \frac{1}{3 (16) + 1} = \frac{1}{49}$

$g' (63) = (\frac{d y}{d x})$

$(\begin{array}{l} x = 63 \\ y = 4 \end{array})$

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New answer posted

a year ago

Maths Matrices

Le A be 3 * 3 real matrix such that $A (\begin{matrix} 1 & 1 & 0 \end{matrix}) = (\begin{matrix} 1 & 1 & 0 \end{matrix}); A (\begin{matrix} 1 & 0 & 1 \end{matrix}) = (\begin{matrix} - 1 & 0 & 1 \end{matrix}) a n d A (\begin{matrix} 0 & 0 & 1 \end{matrix}) = (\begin{matrix} 1 & 1 & 2 \end{matrix})$ .

If x = (x₁, x₂, x₃)^T and I is an identity matrix of order 3, then the system (A – 2I)X = has:

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Contributor-Level 10

$A (\begin{array}{l} 1 \\ 1 \\ 0 \end{array}) = (\begin{array}{l} 1 \\ 1 \\ 0 \end{array})$

$A (\begin{array}{l} 1 \\ 0 \\ 1 \end{array}) = (\begin{array}{l} - 1 \\ 0 \\ 1 \end{array})$

$A (\begin{array}{l} 0 \\ 0 \\ 1 \end{array}) = (\begin{array}{l} 1 \\ 1 \\ 2 \end{array})$

$\Rightarrow A [\begin{matrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{matrix}] = [\begin{matrix} 1 & - 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{matrix}]$

$A B = C \Rightarrow A = C B^{- 1}$

$| A - 2 l | = (- 4) (- 1) + 3 (- 1) + 1 (- 1)$

4 – 3 – 1 = 0

$l (- 1, 0, 1) + m (- 1, 1, 0) = (- 4, 3, 1) \Rightarrow - l - m = - 4$

$m = 3, l = 1$

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New answer posted

a year ago

Maths Functions

Let f : N -> R be a function such that f(x + y) = 2f(x) f(y) for natural number x and y. If f(1) = 2, then the value of a for which $\sum_{k = 1}^{10} f (α + k) = \frac{512}{3} (2^{20} - 1)$ holds is:

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