Ncert Solutions Maths class 11th

Let A(1, 4) and B(1, -5) be two points. Let P be a point on the circle ${(x - 1)}^{2} + {(y - 1)}^{2} = 1$ such that ${(P A)}^{2} + {(P B)}^{2}$ have a maximum value, then the points P, A and B lie on:

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$P A^{2} + P B^{2} = c o s^{2} θ + {(s i n θ - 3)}^{2} + c o s^{2} θ + {(s i n θ + 6)}^{2}$

= 2 cos² θ + 2 sin² θ + 6 sin θ + 45

= 6 sin θ + 47

for maximum of PA² + PB², sin q = 1

then P (1, 2)

Hence P, A & B will lie on a straight line.

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A natural number has prime factorization given by n = $2^{x} 3^{y} 5^{z},$ where y and z are such that y + z = 5 and $y^{- 1} + z^{- 1} = \frac{5}{6}, y > z .$ Then the number of odd divisors of n, including 1, is:

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Given n = 2^x. 3^y. 5^z . (i)

$y + z = 5 & \frac{1}{y} + \frac{1}{z} = \frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

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The sum of the series $\sum_{n = 1}^{\infty} \frac{n^{2} + 6 n + 10}{(2 n + 1)!}$ is equal to:

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$\sum_{n = 1}^{\infty} \frac{n^{2} + 6 n + 10}{(2 n + 1)!}$

put 2n + 1 = r, r = 3, 5, 7,.

so n = $\frac{r - 1}{2}$

$N o w \sum_{r = (3, 5, 7, . . . . .)}^{\infty} \frac{n^{2} + 6 n + 10}{(2 n + 1)!} = \sum_{r = (3, 5, 7 . . . . .)}^{\infty} (\frac{\frac{{(r - 1)}^{2}}{4} + 3 r - 3 + 10}{r!}) = \sum_{r = (3, 5, 7, . . . . . .)}^{\infty} (\frac{r^{2} + 10 r + 29}{4 . r!})$

$= \frac{1}{8} {41 e - \frac{19}{e} - 80) = \frac{41}{8} e - \frac{19}{8} e^{- 1} - 10$

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A seven digit number is formed using digit 3,3,4,4,4,5,5. The probability, that number so formed is divisible by 2, is:

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3, 4, 5, 5

In remaining six places you have to arrange

3, 4, 5,5

So no. of ways $= \frac{6!}{2! 2! 2!}$

Total no. of seven digits nos. = $\frac{7!}{2! 3! 2!} * 1$

Hence Req. prob. $\frac{\frac{6!}{2! 2! 2!}}{\frac{7!}{2! 3! 2!}} = \frac{6! 3!}{2! 7!} = \frac{3}{7}$

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If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x² + y² = 1 is a circle of radius r, then r is equal to:

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Contributor-Level 10

$h = \frac{c o s θ + 3}{2}$

$k = \frac{s i n θ + 2}{2}$

-> $c o s θ = 2 h - 3 & s i n θ = 2 k - 2$

-> ${(h - 3 / 2)}^{2} + {(k - 1)}^{2} = \frac{1}{4}$

$∴$ circle of radius r = $\frac{1}{2}$

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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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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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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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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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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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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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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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