Binayak Bhaiya

Conic Sections (Ex – 11.1) In each of the following Exercises 1 to 5, find the equation of the circle with Question 1.center (0, 2) and radius 2 Solution:Here h = 0,k = 2 and r = 2The equation of circle is,(x-h)2 + (y- k)2 = r2∴ (x – 0)2 + (y – 2)2 = (2)2⇒ x2 + y2 + 4 […]

Straight Lines (Ex – 10.3) Question 1.Reduce the following equations into slope- intercept form and find their slopes and the y-intercepts.(i) x + 7y = 0,(ii) 6x + 3y-5 = 0,(iii) y=0 Solution: Question 2.Reduce the following equations into intercept form and find their intercepts on the axes.(i) 3x + 2y – 12 = 0,(ii) 4x – 3y = 6,(iii) 3y

Straight Lines (Ex – 10.2) In Exercises 1 to 8, find the equation of the line which satisfy the given conditions: Question 1.Write the equations for the x-and y-axes. Solution:We know that the ordinate of each point on the x-axis is 0.If P(x, y) is any point on the x-axis, then y = 0.∴ Equation

Straight Lines (Ex – 10.1) Question 1.Draw a quadrilateral in the Cartesian plane, whose vertices are (- 4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its area. Solution:The figure of quadrilateral whose vertices are A(- 4, 5), B(0, 7), C(5, -5) and D(-4, -2) is shown in the below figure. Question 2.The

Sequences and Series (Ex – 9.4) Find the sunt to n terms of each of the series in Exercises 1 to 7. Question 1.1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ……… Solution:In the given series, there is a sum of multiple of corresponding terms of two

Sequences and Series (Ex – 9.3) Question 1.Find the 20th and nth terms of the G.P. 5/2,5/4,5/8, …. Solution: Question 2.Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2. Solution:We have, as = 192, r = 2 Question 3.The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively.

Sequences and Series (Ex – 9.2) Question 1.Find the sum of odd integers from 1 to 2001. Solution:We have to find 1 + 3 + 5 + ……….. + 2001This is an A.P. with first term a = 1, common difference d = 3-1 = 2 and last term l = 2001∴ l = a

Sequences and Series (Ex – 9.1) Question 1.an = n(n + 2) Solution:We haven an = n(n + 2)substituting n = 1, 2, 3, 4, 5, we geta1 = 9(1 + 2) = 1 x 3 = 3a2 = 2(2 + ) = 2 x 4 = 8a3 = 3(3 + 2) = 3 x 5 = 15a4 = 4(4 +

Binomial Theorem (Ex – 8.2) Question 1.Find the coefficient of x5 in (x + 3)8 Solution.Suppose x5 occurs in the (r + 1)th term of the expansion (x + 3)8 Question 2.a5 b7in (a-2b)12 Solution.Suppose a5 b7 occurs in the (r + 1)th term of the expansion (a – 2b)12. Write the general term in the expansion of Question 3.(x2 – y)6 Solution.

Binomial Theorem (Ex – 8.1) Expand each of the expressions in Exercises 1 to 5. Question 1.(1−2x)5 Solution. Question 2.(2/x−x/2)5 Solution. Question 3.(2x−3)6 Solution. Question 4.(x/3+1/x)5 Solution. Question 5.(x+1/x)6 Solution. Using binomial theorem, evaluate each of the following Question 6.(96)3 Solution. Question 7.(102)5 Solution. Question 8.(101)4 Solution. Question 9.(99)5 Solution. Question 10.Using Binomial Theorem, indicate