Ncert Solutions Maths class 12th

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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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 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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P (H) = x . P (T) = 1 – x

P (4H. 1T) = P (5H)

6x = 5 = 0 $x = \frac{5}{6}$

P (atmost 2H)

$P (O H, 5 T) + P (1 H, 4 T) + P (2 H, 3 T)$

$= \frac{1}{6^{5}} (1 + 25 + 250) = \frac{276}{6^{5}} = \frac{4^{6}}{6^{4}}$

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2 months ago

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x – 3y + 5z = 8. If the mirror image of the point $(2, - \frac{1}{2}, 2)$ in the rotated plane in B(a, b, c), then

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Consider the equation of plane,

$P : (2 x + 3 y + z + 20) + λ (x - 3 y + 5 z - 8) = 0$

$?$ Plane P is perpendicular to 2x + 3y + z + 20 = 0

So, $4 + 2 λ + 9 - 9 λ + 1 + 5 λ = 0$

0 $λ = 7$

P : 9x – 18y + 36z – 36 = 0

Or P : x – 2y + 4z = 4

If image of

$(2, - \frac{1}{2}, 2)$

In plane P is (a, b, c) then

$\frac{a - 2}{1} = \frac{b + \frac{1}{2}}{- 2} = \frac{c - 2}{4}$

and $(\frac{a + 2}{2}) - 2 (\frac{b - \frac{1}{2}}{2}) + 4 (\frac{c + 2}{2}) = 4$

clearly

$a = \frac{4}{3}, b = \frac{5}{6} a n d c = - \frac{2}{3}$

So, a : b : c = 8 : 5 : 4

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2 months ago

Let the normal at the point P on the parabola y² = 6x pass through the point (5, 8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is

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Contributor-Level 10

Let P (at², 2 at) where

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

T : yt = x + at² so point Q is

$(- a, a t - \frac{a}{t})$

N : y = -tx + 2at + at³ passes through (5, -8)

$- 8 = - 5 t + 3 t + \frac{3}{2} t^{3}$

⇒ $3 t^{3} - 4 t + 16 = 0$

⇒ t = -2

So ordinate of point Q is $- \frac{9}{4}$

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2 months ago

If the probability that a randomly chose 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96 p is equal to……….

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Total number of numbers from given condition = n (s) = 26

Every required number is of the form

$A = 7 . (1 0^{a_{1}} + 1 0^{a_{2}} + 1 0^{a_{3}} + . . . .)$ + 111111

Here 111111 is always divisible by 21.

$∴$ Required probability = $\frac{22}{2^{5}} = p$ $\frac{}{^{}}$ $\frac{}{}$ $∴ \frac{11}{32}$ = p $∴$ 96p = 33

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2 months ago

Let a line L₁ be tangent to the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{4} = 1$ and let L₂ be the line passing through the origin and perpendicular to L₁. If the locus of the point of intersection of L₁ and L₂ is ${(x^{2} + y^{2})}^{2}$ = αx² + βy², then α + β is equal to……….

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Equation of L₁ = is

$\frac{x s e c θ}{4} - \frac{y t a n θ}{2} = 1$ ….(i)

Equation of line L₂ is

$\frac{x t a n θ}{2} + \frac{y s e c θ}{4} = 0$ ….(ii)

$?$ Required point of intersection of L₁ and L₂ is (x₁, y₁) then

$\frac{x_{1} s e c θ}{4} - \frac{y_{1} t a n θ}{2} - 1 = 0$ ….(iii)

$a n d \frac{y_{1} s e c θ}{4} + \frac{x_{1} t a n θ}{2} = 0$ ……(iv)

From equations (iii) and (iv)

$s e c θ = \frac{4 x_{1}}{x_{1}^{2} + y_{1}^{2}} a n d t a n θ = \frac{- 2 y_{1}}{x_{1}^{2} + y_{1}^{2}}$

$∴$ Required locus of (x₁, y₁) is

${(x^{2} + y^{2})}^{2} = 16 x^{2} - 4 y^{2}$

$\begin{array}{l} ∴ α = 16, β = - 4 \\ ∴ α = β = 12 \end{array}$

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2 months ago

Let C be a circle passing through the points A(2, 1) and B(3, 4). The line segment AB is not a diameter of C. If r is the radius of C and its centre lies on the circle ${(x - 5)}^{5} + {(y - 1)}^{2} = \frac{13}{2}$ , then r² is equal to

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Equation of perpendicular bisector of AB is

$y - \frac{3}{2} = - \frac{1}{5} (x - \frac{5}{2}) \Rightarrow x + 5 y = 10$

Solving it with equation of given circle,

${(x - 5)}^{2} + {(\frac{10 - x}{5} - 1)}^{2} = \frac{13}{2}$

$\Rightarrow x - 5 = \pm \frac{5}{2} \Rightarrow x = \frac{5}{2} o r \frac{15}{2}$

But

$x \neq \frac{5}{2}$

because AB is not the diameter.

So, centre will be

$(\frac{15}{2}, \frac{1}{2})$

Now,

$r^{2} = {(\frac{15}{2} - 2)}^{2} + {(\frac{1}{2} + 1)}^{2} = \frac{65}{2}$

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2 months ago

Let $r \in {p, q \sim p, \sim q}$ be such that the logical statement $r \lor (\sim p) \Rightarrow (p \land q) \lor r$ is a tautology. Then r is equal to:

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Clearly r must be equal to $\sim$ p

$? \sim p \lor \sim p = \sim p$

$a n d (p \land q) \lor \sim p = p$

$∴ \sim p \Rightarrow p =$ tautology.

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2 months ago

The mean and standard of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to:

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Contributor-Level 10

$\bar{x} = 15, σ = 2 \Rightarrow σ^{2} = 4$

$∴ x_{1} + x_{2} + . . . . + x_{50} = 15 * 50 = 750$

$4 = \frac{x_{1}^{2} + x_{2}^{2} + . . . . . + x_{50}^{2}}{50} - 225$

Let a be the correct observation and b is the incorrect observation then a + b = 70 and

$16 = \frac{75 - b + a}{50}$

$= \frac{50 * 229 + 6 0^{2} - 1 0^{2}}{50} - 256 = 43$

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2 months ago

Let $\vec{a} = \hat{i} + \hat{j} + 2 \hat{k}, \vec{b} = 2 \hat{i} - 3 \hat{j} + \hat{k} a n d \vec{c} = \hat{i} - \hat{j} + \hat{k}$ be three given vectors. let $\vec{v}$ be a vector in the plane of $\vec{a} a n d \vec{b}$ whose projection on $\vec{c}$ is $\frac{2}{\sqrt[]{3}} .$ $\frac{}{}$ If $\vec{v} . \hat{j} = 7, t h e n \vec{v} . (\hat{i} + \hat{k})$ is equal to:

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