Maths

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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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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2 months ago

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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Maths Inverse Trigonometric Functions

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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2 months ago

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:

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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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2 months 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…………….

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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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2 months 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…………….

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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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2 months ago

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

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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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2 months ago

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:

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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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2 months ago

Maths Continuity and Differentiability

Let the coefficients of x^-1 and x^-3 in the expansion of ${(2 x^{\frac{1}{5}} - \frac{1}{x^{\frac{1}{5}}})}^{15}, x > 0,$ be m and n respectively. If r is a positive integer such that mn² = ¹⁵C_r . 2^r, then the value of r is equal to….

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New answer posted

2 months ago

Let f and g be twice differentiable even functions on (-2, 2) such that $f (\frac{1}{4}) = 0, f (\frac{1}{2}) = 0, f (1) = 1$ and $f (\frac{3}{4}) = 0, g (1) = 2$

Then, the minimum number of solutions of $f (x) g " (x) + f' (x) g' (x) = 0 i n (- 2, 2)$ is equal to……….

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Contributor-Level 10

f (x) is an even function

$f (- \frac{1}{4}) = f (- \frac{1}{2}) = f (\frac{1}{2}) = f (\frac{1}{4}) = 0$

So, f (x) has at least four roots in (-2, 2)

$g (\frac{- 3}{4}) = g (\frac{3}{4}) = 0$

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g " (x) = f' (x) \cdot g' (x) = 0$

It is same as number of roots of $\frac{d}{d x} (f (x) \cdot g' (x)) = 0$ will have atleast 4 roots in (-2, 2)

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New answer posted

2 months ago

Maths Class 12th

Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is:

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