Ncert Solutions Maths class 11th

Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

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Kindly go through the solution

(3!)³ (4!)

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The shortest distance between the lines (x-1)/0 = (y+1)/-1 = z/1 and x+y+z+1=0, 2x-y+z+3=0 is:

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Family of planes: x+y+z+1+? (2x-y+z+3)=0. Parallel to line means dot product of normals is zero.

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The position of a moving car at time t is given by f(t)=at²+bt+c, t>0, where a,b and c are real numbers greater than 1. Then the average speed of the car over the time interval [t?,t?] is attained at the point:

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Average speed = (f (t? )-f (t? )/ (t? -t? ) = a (t? +t? )+b.
Instantaneous speed = f' (t)=2at+b.
2at+b=a (t? +t? ) ⇒ t= (t? +t? )/2.

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If α, β are roots of the equation x² + 5√2x + 10 = 0, α > β and P? = α? - β? for each positive integer n, then the value of (P?P? + 5√2P?P?)/(P?P? + 5√2P?²) is equal to……….

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α, β are roots of x²+5√2x+10=0. α+β=-5√2, αβ=10.
P? + 5√2P? + 10P? = 0.
So P? +5√2P? =-10P? P? +5√2P? =-10P?
Expression = P? (-10P? )/P? (-10P? ) = 1.

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A ray of light coming from the point (2,2√3) is incident at an angle 30° on the line x=1 at the point A. The ray gets reflected on the line x=1 and meets x-axis at the point B. Then, the line AB passes through the point:

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m (AP)=tan60°=√3. y-2√3=√3 (x-2). At x=1, y=√3. A= (1, √3).
m (AB)=tan120°=-√3. y-√3=-√3 (x-1) ⇒ √3x+y=2√3. (3, -√3) satisfies

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There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100k, then k is equal to…….

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| Class | No. of students | Number of possible cases |
| :-: | :-: | :- |
| 10 | 5 | (I) 2 | (II) 3 | (III) 2 |
| 11 | 6 | (I) 2 | (II) 2 | (III) 3 |
| 12 | 8 | (I) 6 | (II) 5 | (III) 5 |
Total cases =? C? *? C? *? C? +? C? *? C? *? C? +? C? *? C? *? C?
= 23,800 = 100K
∴ K = 238

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Let L₁ be a tangent to the parabola y²=4(x+1) and L₂ be a tangent to the parabola y²=8(x+2) such that L₁ and L₂ intersect at right angles. Then L₁ and L₂ meet on the straight line:

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Equation of tangent to y²=4 (x+1) is y=m (x+1)+1/m.
Equation of tangent to y²=8 (x+2) is y=m' (x+2)+2/m'.
m'=-1/m.
Solving for intersection point, x+3=0.

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The sum of all values of x in [0, 2π], for which sinx + sin2x + sin3x + sin4x = 0, is equal to:

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sinx+sin4x + sin2x+sin3x = 0
2sin (5x/2)cos (3x/2) + 2sin (5x/2)cos (x/2) = 0
2sin (5x/2) [cos (3x/2)+cos (x/2)] = 0
4sin (5x/2)cosxcos (x/2)=0.
sin (5x/2)=0 ⇒ 5x/2=kπ ⇒ x=2kπ/5. x=0, 2π/5, 4π/5, 6π/5, 8π/5, 2π.
cosx=0 ⇒ x=π/2, 3π/2.
cos (x/2)=0 ⇒ x=π.
Sum = 9π.

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Let S? be the sum of the first n terms of an arithmetic progression. If S? = 3S?, then the value of S?/S? is:

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S? = 3n/2 [2a+ (3n-1)d]. S? = 2n/2 [2a+ (2n-1)d].
S? =3S? ⇒ 3n/2 [2a+ (3n-1)d] = 3 (2n/2) [2a+ (2n-1)d].
2a+ (3n-1)d = 2 [2a+ (2n-1)d] ⇒ 2a+ (n-1)d=0.
S? /S? = (4n/2 [2a+ (4n-1)d]) / (2n/2 [2a+ (2n-1)d]) = 2 [- (n-1)d+ (4n-1)d]/ [- (n-1)d+ (2n-1)d] = 2 (3n)/ (n)=6.

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The locus of the centroid of the triangle formed by any point P on this hyperbola 16x² - 9y² + 32x + 36y - 164 = 0, and its foci is:

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