Ncert Solutions Maths class 11th

Let m₁, m₂ be the slopes of two adjacent sides of a square of side a such that a² + 11a + $3 (m_{1}^{2} + m_{2}^{2}) = 220 .$ If one vertex of the square is $(10 (c o s α - s i n α), 10 (s i n α + c o s α)),$ where $α \in (0, \frac{π}{2})$ and the equation of one diagonal is $(c o s α - s i n α) x + (s i n α + c o s α) y = 10,$ then $72 (s i n^{4} α + c o s^{4} α) + a^{2} - 3 a + 13$ is equal to:

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$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cos a - sina)x + (sina + cosa)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a² + 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

-> 210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

->m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2a - 10cos2a = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

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$m_{1} m_{2} = - 1,$ for square a,b,c,d let

$A (10 (c o s α - s i n α), 10 (s i n α + c o s α))$

Diagonal : (cos a - sina)x + (sina + cosa)y = 10

BD (diagonal)

Dist. Of BD from A is

$\frac{| 10 {(c o s α - s i n α)}^{2} + 10 {(s i n α + c o s α)}^{2} - 10 |}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}}$

$\frac{10}{\sqrt[]{2}} = \frac{a}{\sqrt[]{2}} \Rightarrow a = 10$

Also, a² + 11a + 3 $(m_{1}^{2} + m_{2}^{2}) = 220$

-> 210 + 3 $(c m_{1}^{2} + m_{2}^{2}) = 220$

$m_{1}^{2} + m_{2}^{2} = \frac{10}{3}$

Also, m₁ m₂ = -1

->m² + $\frac{1}{m^{2}} = \frac{10}{3}$

or $- \sqrt[]{3}, \frac{1}{\sqrt[]{3}}$

m = $\sqrt[]{3}, \frac{- 1}{\sqrt[]{3}}$

$m^{4} - \frac{10}{3} m^{2} + 1 = 0 \Rightarrow m^{2} = \frac{\frac{10}{3} \pm \sqrt[]{\frac{100}{9} - 4}}{2} - \frac{\frac{10}{3} \pm \frac{8}{3}}{2} = 3, \frac{1}{3}$

m = $\pm \sqrt[]{3}, \pm \frac{1}{\sqrt[]{3}}$

Diagonal AC:

$(s i n α + c o s α) x - (c o s α - s i n α) y$

=10 cos2a - 10cos2a = 0

Slope of AC = $\frac{s i n α + c o s α}{c o s α - s i n α} = \frac{t a n α + 1}{1 - t a n α} = t a n (α + \frac{π}{4}) α = 30 °$

FIGURE

? = $72 (\frac{1}{16} + \frac{9}{16}) + 100 - 30 + 13 = \frac{720}{16} + 83 = 128$

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Let C be the centre of the circle $x^{2} + y^{2} - x + 2 y = \frac{11}{4}$ $^{}^{} \frac{}{}$ and P be a point on the circle. A line passes through the point C, makes an angle of $\frac{π}{4}$ $\frac{}{}$ with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit2) is:

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Contributor-Level 10

$x^{2} + y^{2} - x + 2 y = \frac{11}{4}$

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

or $Δ P Q R P R = Q R s i n 2 \geq \frac{1}{3}$

$= 4.6 s i n \frac{π}{8}$

$a s Δ P Q R = \frac{1}{2} P R * P Q$

$= 4 s i n \frac{π}{4} = \frac{4}{\sqrt[]{2}} = 2 \sqrt[]{2}$

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If the tangents drawn at the points P and Q on the parabola y² = 2x – 3 intersect at the point R(0, 1), then the orthocenter of the triangle PQR is:

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Contributor-Level 10

$y^{2} = 2 x - 3$ ……. (i)

Equation of chord of contact

PQ : r = O

(y * 1) = (x + 0) – 3

y = x – 3 ……… (ii)

From (i) and (ii)

y = 1 or 3

$M P Q = \frac{4}{4} = 1$

$M Q R = \frac{2}{6} = \frac{1}{3}$

$M_{P Q} * M_{P R} = - 1 \Rightarrow P Q ⊥ P R$

Orthocentre = P (2, -1)

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For $t \in (0, 2 π),$ if ABC is an equilateral triangle with vertices $A (s i n t, - c o s t)$ , $B (c o s t, s i n t)$ and C(a, b) such that its orthocenter lies on a circle with centre $(1, \frac{1}{3})$ $\frac{}{}$ , then $(a^{2} - b^{2})$ $^{}^{}$ is equal to:

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Contributor-Level 10

S $\equiv s i n t, C \equiv c o s t$

Let orthocenter be (h, k)

Since it if an equilateral triangle hence orthocenter coincides with centroid.

$∴ a + s + c = 3 h, b + s - c = 3 k$

$\frac{a}{3} = 1, \frac{b}{3} = \frac{1}{3} \Rightarrow a = 3, b = 1$

$∴ a^{2} - b^{2} - 8$

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$\sum_{r = 1}^{20} (r^{2} + 1) (r!)$ is equal to

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Contributor-Level 10

$\sum_{r = 1}^{20} (r^{2} + 1) r!$

t_r = (r² + 1)r!

= r²r! + r!

= r (r + 1 – 1)r! + r!

= r (r + 1)! – (r – 1)r!

= V_r – V_r-1

_{$\sum_{r = 1}^{20} (V_{r} - V_{r - 1})$}

= V₁ – V₀

_{$+ V_{2} - V_{1}$}

_{$+ V_{3} - V_{2}$}

_{$+ V_{20} - V_{19}$}

_{$= V_{20} - V_{0} = 20 (21!) - 0$}

_{$(22 - 2) (21!) = 22! - 2 (21!)$}

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Let the operations $*, ⊙ \in {\land, \lor} . I f (p * q) ⊙ (p ⊙ \sim q)$ is a tautology, then the ordered pair $(*, ⊙)$ is:

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(A) $(\land, \lor) : (P \lor Q) \land (P \land \sim Q)$ not a tautology

(B) $(\lor, \lor) : (P \lor Q) \lor (P \lor \sim Q) = (P \lor T)$ in a tautology

$(p \lor q) \lor (p \lor \sim q) =$ T using venn diagrams

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If [t] denotes the greatest integer $\leq t,$ then the value of $\int_{0}^{1} [2 x - | 3 x^{2} - 5 x + 2 | + 1] d x i s :$

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Contributor-Level 10

y = 2x - $| 3 x^{2} - 5 x + 2 |, 0 \leq x \leq 1$

$= {\begin{matrix} - 3 x^{2} + 7 x - 2, & 3 x^{2} - 3 x + 2, \end{matrix} \begin{matrix} 0 \leq x \leq \frac{2}{3} & \frac{2}{3} < x \leq 1 \end{matrix}$

3x² – 5x + 2 = 0

$x = \frac{+ 5 \pm \sqrt[]{25 - 24}}{6}$

= $\frac{5 + 1}{6} = 1, \frac{2}{3}$

3x² – 7x + 3 = 0

x = $\frac{7 \pm \sqrt[]{49 - 39}}{6} = \frac{7 \pm \sqrt[]{13}}{6}$

$- 3 x^{2} + 7 x - 2$

$\frac{- 7 \pm \sqrt[]{49 - 24}}{- 6} = \frac{7 + 5}{6} = 2, \frac{1}{3}$

3x² + 7x– 2 = -1

->3x² – 7x + 1 = 0

x = $\frac{7 \pm \sqrt[]{49 - 12}}{6} = \frac{7 \pm \sqrt[]{37}}{6}$

I = $\int_{0}^{6} (\begin{matrix} (- 2) & + 1 \end{matrix}) d x + \int_{\frac{7 - \sqrt[]{37}}{6}}^{\frac{1}{3}} (\begin{matrix} (- 1) & + 1 \end{matrix}) d x + \int_{- \frac{1}{3}}^{\frac{7 - \sqrt[]{13}}{6}} (\begin{matrix} 0 & + 1 \end{matrix}) d x + \int_{\frac{7 - \sqrt[]{13}}{6}}^{\frac{2}{3}} (1 + 1) d x + \int_{\frac{2}{3}}^{1} (1 + 1) d x$

$= \frac{\sqrt[]{37} + \sqrt[]{13} - 4}{6}$

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Section-A

If $z \neq 0$ be a complex number such that $| z - \frac{1}{z} | = 2,$ then the maximum value of $| z |$ is:

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Contributor-Level 10

$| z - \frac{1}{z} | = 2$

${| z |}_{m a x} = ?$

$| z - \frac{1}{2} | \geq | | z | - \frac{1}{| z |} |$

$2 \geq | r - \frac{1}{r} |$

0 $\leq r^{2} + 2 r - 1 & r^{2} - 2 r - 1 \leq 0$

$r = \frac{- 2 \pm \sqrt[]{8}}{2}$ $r = \frac{2 \pm \sqrt[]{8}}{2}$

$= - 1 \pm \sqrt[]{2}$ $= 1 \pm \sqrt[]{2}$

r $\geq \sqrt[]{2} - 1 & 0 \leq r \leq 1 + \sqrt[]{2}$

$\sqrt[]{2} - 1 \leq r \leq \sqrt[]{2} + 1$

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Let the point P(α,β) be at a unit distance from each of the two lines L₁ : 3x – 4y + 12 = 0, and L₂ : 8x + 6y + 11 = 0. If P lies below L₁ and above L₂, then 100(α + β) is equal to

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Payal Gupta

Contributor-Level 10

L₁ : 3x – 4y + 12 = 0

L₂ : 8x + 6y + 11 = 0

$(α, β)$ lies on that angle which contain origin

$∴$ Equation of angle bisector of that angle which contain origin is

$\frac{3 x - 4 y + 12}{5} = \frac{8 x + 6 y + 11}{10}$

$\Rightarrow 2 x + 14 y - 13 = 0$

$(α, β)$ lies on it

$\Rightarrow 2 α + 14 β - 13 = 0$ …… (i)

$\Rightarrow 3 α - 4 β + 7 = 0$ ……. (ii)

Solving (i) & (ii)

$α = - \frac{23}{25} & β = \frac{53}{50}$

$∴ α + β = \frac{7}{50}$

$\Rightarrow 100 (α + β) = 14$

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New answer posted

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