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Maths Class 12th

Let f(x) = $m a x {| x + 1 |, | x + 2 |, . . . . . ., | x + 5 |} . T h e n \int_{- 6}^{0} f (x) d x$ is equal to………

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Raj Pandey

Contributor-Level 9

$∴ \int_{- 6}^{0} f (x) d x = 2 * \frac{1}{2} (2 + 5) * 3 = 21$

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New answer posted

a year ago

Maths Class 11th

LE the common tangents to the curves $4 (x^{2} + y^{2}) = 9$ and y² = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and $l$ respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\frac{l}{e^{2}}$ is equal to…….

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Raj Pandey

Contributor-Level 9

Let y = mx + c is the common tangent

$s o c = \frac{1}{m} = \pm \frac{3}{2} \sqrt[]{1 + m^{2}} \Rightarrow m^{2} = \frac{1}{3}$

so equation of common tangents will be

$y = \pm \frac{1}{\sqrt[]{3}} x \pm \sqrt[]{3}$

which intersects at Q (-3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$

...more

Let y = mx + c is the common tangent

$s o c = \frac{1}{m} = \pm \frac{3}{2} \sqrt[]{1 + m^{2}} \Rightarrow m^{2} = \frac{1}{3}$

so equation of common tangents will be

$y = \pm \frac{1}{\sqrt[]{3}} x \pm \sqrt[]{3}$

which intersects at Q (-3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$

and length of latus rectum

$= \frac{2 b^{2}}{a} = 3$

Hence $\frac{l}{e^{2}} = \frac{3}{3 / 4} = 4$

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New answer posted

a year ago

Maths Class 11th

There are ten boys B₁, B₂,…., B₁₀ and five girls G₁, G₂,…., G₅ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both B₁ and B₂ together should not be the members of a group, is………….

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Raj Pandey

Contributor-Level 9

Boys (10) Girls (5)

(3) (3)

B₁ & B₂ should not be selected together

Total number of ways

= (56 + 56) * 10 = 1120

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New answer posted

a year ago

Maths Class 11th

The sum of the cubes of all the roots of the equation $x^{4} - 3 x^{3} - 2 x^{2} + 3 x + 1 = 0$ is…………

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Raj Pandey

Contributor-Level 9

$x^{4} - 3 x^{3} - 2 x^{2} + 3 x + 1 = 0$

$\Rightarrow x = \pm 1$

and let a, b are roots of x² – 3x – 1 = 0

$∴ α + β = 3 α β = - 1$

$∴ 1^{3} + {(- 1)}^{3} + α^{3} + β^{3}$

= 27 + 3 (3) = 36

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New answer posted

a year ago

Maths Class 11th

Let $Δ, \nabla \in {\land, \lor}$ be such that $p \nabla q \Rightarrow ((p Δ q) \nabla r)$ is a tautology. Then $(p \nabla q) Δ r$ is logically equivalent to:

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Raj Pandey

Contributor-Level 9

Case – I $Δ \equiv \nabla \equiv \land$

$(p \land q) \to ((p \land q) \land r)$

it can be false if r is false,

so not a tautology

Case – II If $Δ \equiv \nabla \equiv \lor$

$(p \lor q) ? \to ((p \lor q) \lor r) \equiv$ tautology

then $(p \lor q) \lor r \equiv (p Δ r) \lor q$

Case – III If $Δ \equiv v, \nabla \equiv \land$ &nbs

...more

Case – I $Δ \equiv \nabla \equiv \land$

$(p \land q) \to ((p \land q) \land r)$

it can be false if r is false,

so not a tautology

Case – II If $Δ \equiv \nabla \equiv \lor$

$(p \lor q) ? \to ((p \lor q) \lor r) \equiv$ tautology

then $(p \lor q) \lor r \equiv (p Δ r) \lor q$

Case – III If $Δ \equiv v, \nabla \equiv \land$

then $(p \land q) \to {(p \lor q) \land r}$

Not a tautology

Case – IV If $Δ \equiv \land, \nabla \equiv \lor$

$(p \land q) \to {(p \land q) \lor r}$

Not a tautology

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New answer posted

a year ago

Maths Inverse Trigonometric Functions

Let f(x) = $2 c o s^{- 1} x + 4 c o t^{- 1} x - 3 x^{2} - 2 x + 10, x \in [- 1, 1] . I f [a, b]$ is the range of the function f, then 4a – b is equal to:

0 Follower 4 Views

Raj Pandey

Contributor-Level 9

$f (x) = 2 c o s^{- 1} x + 4 c o t^{- 1} x - 3 x^{2} - 2 x + 10 \forall x \in [- 1, 1]$

$\Rightarrow f' (x) = - \frac{2}{\sqrt[]{1 - x^{2}}} - \frac{4}{1 + x^{2}} - 6 x - 2 < 0 \forall x \in [- 1, 1]$

So, f (x) is decreasing function and range of f (x) is

$[f (1), f (- 1)],$ which is $[π + 5, 5 π + 9]$

Now 4a – b = 4 (p + 5) - (5p + 9) = 11 - “π”

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New answer posted

a year ago

Maths Polymers

Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If $f (g (x)) = 8 x^{2} - 2 x,$ and $g (f (x)) = 4 x^{2} + 6 x + 1,$ then the value of f(2) + g(2) is…………….

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

g (x) = px + q

$f (g (x)) = a {(p x + q)}^{2} + b (p) (+ q) + c$

$? 8 x^{2} ? 2 x = a {(p x + q)}^{2} + b (p x + q) + c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

$9 (f (x)) = p (a x^{2} + b x + c) + q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

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New answer posted

a year ago

Maths Matrices

Let $M = [\begin{matrix} 0 & - α \\ α & 0 \end{matrix}],$ where a is a non-zero real number an $N = \sum_{k = 1}^{49} M^{2 k} .$ If (I – M²)N = -2I, then the positive integral value of α is…………….

0 Follower 18 Views

Vishal Baghel

Contributor-Level 10

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

g (x) = px + q

$f (g (x)) = a {(p x + q)}^{2} + b (p) (+ q) + c$

$\Rightarrow 8 x^{2} - 2 x = a {(p x + q)}^{2} + b (p x + q) + c$

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

$9 (f (x)) = p (a x^{2} + b x + c) + q$

=>4x² + 6x + 1 = apx² + bpx + cp + q

=> Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

=> b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

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New answer posted

a year ago

Maths Class 11th

The total number of four digit numbers such that each of first three digits is divisible by the last digit, is equal to……………….

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

Fix the unit place, find the chances for the first three digits

unit digit as 1, total ways = 9.10²

unit digit as 2, total ways = 4.5²

unit digit as 3 total ways = 3.4²

unit digit as 4 total ways = 2.3²

unit digit as 5 total ways = 1.2²

unit digit as 6 total ways = 1.2²

unit digit as 7 total ways = 1.2²

unit digit as 8 total ways = 1.2²

unit digit as 9 total ways = 1.2²

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New answer posted

a year ago

Maths Class 11th

The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to:

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

$\bar{x} = 6 = \frac{a + b + 8 + 5 + 10}{5} \Rightarrow a + b = 7$ …… (i)

and

$σ^{2} = \frac{a^{2} + b^{2} + 8^{2} + 5^{2} + 1 0^{2}}{5} - 6^{2} = 6.8$

⇒a² + b² = 25 ……. (ii)

From (i) and (ii) (a, b) = (3, 4) or (4, 3)

Now mean deviation about mean

$M = \frac{1}{5} (3 + 2 + 2 + 1 + 4) = \frac{12}{5}$

⇒ 25M = 60

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