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Ncert Solutions Maths class 11th

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Ncert Solutions Maths class 11th Class 11th

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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Payal Gupta

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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Ncert Solutions Maths class 11th Class 11th

The number of solutions of the equation sin x = cos² x in the interval (0, 10) is

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Payal Gupta

Contributor-Level 10

sin x = 1 – sin² x

sin x = $\frac{- 1 + \sqrt[]{5}}{2}, \frac{- 1 - \sqrt[]{5}}{2} (r e j e c t e d)$

draw y = sin x

y = $\frac{\sqrt[]{5} - 1}{2},$ find their pt. of intersection.

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Ncert Solutions Maths class 11th Class 11th

Let 3, 6, 9, 12,…. upto 78 terms and 5, 9, 13, 17,…. upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to

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Payal Gupta

Contributor-Level 10

First common term to both AP's is 9

t₇₈ of $(3, 6, 9, . . . . . .) = 78 * 3 = 234$

t₅₉ of $(5, 9, 13, . . . . . . . .) = 5 + (5 - 1) 4 = 237$

n^th common term $\leq 234$

9 + (n – 1) 12 $\leq$ 234

n < $\frac{237}{12} \Rightarrow n = 19$

Now sum of 19 terms with a = 9, d = 12

$= \frac{19}{2} (2.9 + (19 - 1) \cdot 12) = 2223$

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Ncert Solutions Maths class 11th Class 11th

Negation of the Boolean statement $(p \lor q) \Rightarrow ((\sim r) \lor p)$ is equivalent to

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Payal Gupta

Contributor-Level 10

pva ⇒ $(\sim r v p)$

$\sim (p \lor q) \lor (\sim r v p)$

its negation as asked in question

$(p \lor q) \land (\sim p \land r)$

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

= $(p \land r \land \sim p) [a s p \land \sim p i s f a l s e]$

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Ncert Solutions Maths class 11th Class 11th

From the base of a pole of height 20 meter, the angle of elevation of the top of tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is

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Payal Gupta

Contributor-Level 10

Let base = b

$t a n 60 ° = \frac{h}{b}$

$t a n 30 ° = \frac{h ? 20}{b}$

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Ncert Solutions Maths class 11th Class 11th

From the base of a pole of height 20 meter, the angle of elevation of the top of tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is

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New answer posted

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Ncert Solutions Maths class 11th Class 11th

Let a triangle ABC be inscribed in the circle $x^{2} - \sqrt[]{2} (x + y) + y^{2} = 0$ such that $∠ B A C = \frac{π}{2} .$ If the length of side AB is $\sqrt[]{2},$ then the area of the $Δ A B C$ is equal to

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Payal Gupta

Contributor-Level 10

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

here AB = $\sqrt[]{2}$ , BC = 2, AC = $\sqrt[]{2}$

area = $\frac{1}{2} * \sqrt[]{2} * \sqrt[]{2} = 1$

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Ncert Solutions Maths class 11th Class 11th

Let P : y² = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is

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Payal Gupta

Contributor-Level 10

Tangents making angle $\frac{π}{4}$ with y = 3x + 5.

$t a n \frac{π}{4} = | \frac{m - 3}{1 + 3 m} | \Rightarrow m = - 2, \frac{1}{2}$

So, these tangents are $⊥$ . So ASB is a focal chord.

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Ncert Solutions Maths class 11th Class 11th

The sum of the infinite series $1 + \frac{5}{6} + \frac{12}{6^{2}} + \frac{22}{6^{3}} + \frac{35}{6^{4}} + \frac{51}{6^{5}} + \frac{70}{6^{6}} + . . . .$ is equal to

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Payal Gupta

Contributor-Level 10

Let S = 1 $+ \frac{5}{6} + \frac{12}{6^{2}} + \frac{22}{6^{3}} + . . . .$

$\underline{- \frac{S}{6} = \frac{1}{6} + \frac{5}{6^{2}} + . . . .}$

$\frac{5 S}{6} = 1 + \frac{4}{6} + \frac{7}{6^{2}} + \frac{10}{6^{3}} + . . . .$

$\underline{- \frac{5 S}{36} = \frac{1}{6} + \frac{4}{6^{2}} + \frac{7}{6^{3}} + . . . . .}$

$\frac{25 S}{36} = 1 + \frac{3}{6} + \frac{3}{6^{2}} + \frac{3}{6^{3}} + . . . . . .$

$\frac{25 S}{36} = 1 + \frac{3 / 6}{1 - 1 / 6}$

$\frac{25 S}{36} = 1 + \frac{3 / 5}{1}$

$S = \frac{288}{125}$

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Ncert Solutions Maths class 11th Class 11th

Let arg(z) represent the principal argument of the complex number z. Then $| z | = 3$ and art(z – 1) – arg(z + 1) = $\frac{π}{4}$ intersect

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Payal Gupta

Contributor-Level 10

$| z | = 3 \Rightarrow$ circle with radius = 3

arg $(\frac{z - 1}{z + 1}) = \frac{π}{4},$ part of a circle (with radius $\sqrt[]{2}$ ). no common points

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