Ncert Solutions Maths class 11th

Let the abscissae of the two points P and Q be the roots of 2x² – rx + p = 0 and the ordinates of P and Q be the roots of x² – sx – q = 0. If the equation of the circle described on PQ as diameter is 2(x² + y²) – 11x – 14y – 22 = 0, then 2r + s – 2q + is equal to….

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Contributor-Level 9

$P (x_{1}, y_{1})$

$Q (x_{2}, y_{2})$

$x_{1} + x_{2} = \frac{r}{2}, x_{1} x_{2} = \frac{P}{2}$

$y_{1} + y_{2} = s, y_{1}, y_{2} = - 9$

$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$

$2 (x^{2} + y^{2}) - r x - 2 s y + p - 2 q = 0$

r = 11, s = 7, p – 2q = -22

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3 weeks ago

The number of 3-digit odd numbers, whose sum of digits is a multiple of 7, is……….

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Contributor-Level 9

$x_{1} + x_{2} = 6, 13 \to 6 + 6 = 12$

$\begin{array}{l} x_{1} + x_{2} = 4, 11, 18 \to 4 + 8 + 1 = 13 \\ x_{1} + x_{2} = 2, 9, 16 \to 2 + 9 + 3 = 14 \\ x_{1} + x_{2} = 7, 14 \to 7 + 5 = 12 \\ x_{1} + x_{2} = 5, 12 \to 5 + 7 = 12 \end{array}$

= 63

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Let x = 2t, $y = \frac{t^{2}}{3}$ be a conic. Let S be the focus and B the point on the axis of the conic such that $S A ⊥ B A,$ where A is any point on the conic. If k is the ordinate of the centroid of the $Δ S A B,$ then $\underset{t \to 1}{l i m}$ k is equal to:

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x = 2t $A (2 t, \frac{t^{2}}{3})$

$y = \frac{t^{2}}{3}$ S (0, 3)

$3 y = {(\frac{x}{2})}^{2}$ $B (0, λ)$

$3 k = \frac{t^{2}}{3} + 3 + λ = \frac{2 t^{2}}{3} + 3 - \frac{12 t^{2}}{9 - t^{2}}$

$\underset{t \to 1}{l i m} 3 k = \frac{2}{3} + 3 - \frac{12}{8} = 3 - \frac{5}{6} = \frac{13}{6}$

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3 weeks ago

If y = m₁x + c₁ and y = m₂x + c₂, m₁ $\neq$ m₂ are two common tangents of circle x² + y² = 2 and parabola y² = x, Then the value of $8 | m_{1} m_{2} |$ is equal to:

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Contributor-Level 9

$y^{2} = 4 (\frac{1}{4}) x$

$x - t y + \frac{1}{4} t^{2} = 0$

$\frac{1}{t} = m, m_{2} \Rightarrow (8 m_{1} m_{2}) = 3 \sqrt[]{2} - 4$

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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 9

$p_{1} : \sim (p \leftrightarrow \sim q)$ $p_{2} : (p \land \sim q) \land ((\sim p) \lor q)$

$= \sim ((\sim p) \lor (\sim q))$

$= \sim (\sim (p \land q))$

$p \land q$

$(\sim p) \lor ((\sim p) \lor q)$ is false (given)

p q p₁ p₂ $(\sim p) \lor q$

T F T

...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

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3 weeks ago

If $e^{(c o s^{2} x + c o s^{4} x + c o s^{6} x + . . . \infty) l o g_{e} 2}$ satisfies the equation t² – 9t + 8 = 0, then the value of $\frac{2 s i n x}{s i n x + \sqrt[]{3} c o s x} (0 < x < \frac{π}{2}) i s$

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Contributor-Level 10

$e^{(c o s^{2} x + c o s^{4} x + c o s^{6} x + . . . . . \infty)} l o g_{e} 2$

$= e^{\frac{c o s^{2} x}{1 - c o s^{2} x} l o g_{e} 2} = e^{c o t^{2} x l o g_{e} 2} = 2^{c o t^{2} x}$

t² – 9t + 8 = 0

(t – 8) (t – 1) = 0

$t = 2^{c o t^{2} x} = 8 = 2^{3}$

$\Rightarrow c o t^{2} x = 3 = c o t^{2} \frac{π}{6}$

$\frac{2 s i n x}{s i n x + \sqrt[]{3} c o s x} = \frac{2 * \frac{1}{2}}{\frac{1}{2} + \sqrt[]{3} * \frac{\sqrt[]{3}}{2}} = \frac{1}{2}$

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3 weeks ago

The area (in sq. units) of the part of the circle $x^{2} + y^{2} = 36,$ which is outside the parabola y² = 9x is:

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Contributor-Level 10

x² + y² = 36 ……(i)

y² = 9x .(ii)

Solving (i) & (ii)

x² + 9x – 36 = 0

(x + 12) (x – 3) = 0

x = 3

Let $A = \int_{0}^{3} (\sqrt[]{36 - x^{2}} - 3 \sqrt[]{x}) d x$

$= {[\frac{x \sqrt[]{36 - x^{2}}}{2} + 18 s i n^{- 1} \frac{x}{6}]}_{0}^{3} - 3 . {\frac{x^{3 / 2}}{3 / 2}]}_{0}^{3}$

$= 3 π - \frac{3 \sqrt[]{3}}{2}$

$∴$ Required area = $= \frac{1}{2} π {(6)}^{2} + 2 (3 π - \frac{3 \sqrt[]{3}}{2}) = (24 π - 3 \sqrt[]{3}) s q . u n i t$

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3 weeks ago

Two vertical poles are 150m apart and the height of one is three times that of other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

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Contributor-Level 10

BC = CN = x = 75

$c o t θ = \frac{3 h}{75}$

$t a n θ = \frac{h}{75}$

$\frac{3 h^{2}}{75 * 75} = 1 \Rightarrow h = \frac{75}{\sqrt[]{3}} = 25 \sqrt[]{3}$

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3 weeks ago

Let p and q be two positive numbers such that p + q = 2 and p⁴ + q⁴ = 272. Then p and q are roots of the equation:

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Contributor-Level 10

p + q = 2

p⁴ + q⁴ = 272

${(p^{2} + q^{2})}^{2} - 2 p^{2} q^{2} = 272$

${[{(p + q)}^{2} - 2 p q]}^{2} - 2 p^{2} q^{2} = 272$

Let pq = t -> (4 – 2t)² – 2t² = 272

2t² – 16 t – 256 = 0

->t = pq = 16

Required equation x² – 2x + 16 = 0

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3 weeks ago

The statement among the following that is a tautology is:

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