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The area enclosed between the ellipse x^{2} + 9y^{2} = 9 and the straight line x + 3y = 3, in the first quadrant is
The intersection points of x^{2} + 9y^{2} = 9 and x + 3y = 3 are (0,1) and (3,0).
∴ A = 6π − 3π = 3π.
The coefficient of x^{9} in the expansion of ((x^{2}/2)  (2/x))^{9} is
The term x^{9} will be T(10). Therefore T(10) will be
⇒ (2/x)^{9}
⇒ 512/x^{9}
⇒ 512x^{9}
The differential equation of the family of lines through the origin is
Let y = mx be the family of lines through origin. Therefore, dy/dx = m
Eliminating m, we get y = (dy/dx).x or x(dy/dx) – y = 0.
The circles x^{2} + y^{2}  12x 12y = 0 and x^{2} + y^{2} + 6x + 6y = 0
Length of tangent drawn from (5,1) to the circle x^{2 }+ y^{2 }+ 6x  4y  3 = 0 is
Given circle is x^{2 }+ y^{2 }+ 6x  4y  3 = 0 .....(i)
Given point is (5, 1) Let P = (5,1)
Now length of the tangent from P(x, y) to circle (i) = √ x^{2 }+ y^{2 }+ 2gx + 2fy + c.
Now length of the tangent from P(5, 1) to circle (i) = √ 5^{2}+1^{2}+ 6.5 − 4.1 − 3 = 7
The solution of differential equation (dy/dx) = [(x(2logx + 1)) / (siny + ycosy)] is
(siny+ y.cosy)dy = [x(2logx + 1)]dx
Integrating both sides, we get
ydy = (c−x)dx
Integrating both sides,
∫ydy = ∫(c−x)dx
y^{2}/2 = cx − x^{2}/2 + k
x^{2}/2 + y^{2}/2 − cx = k
x^{2} + y^{2} − 2cx = 2k
(x−c)^{2} + y^{2} = 2k + c^{2}
This is equation of a circle with center on xaxis.
In an Argand plane the inequality π/4 < arg (z) < π/3 represent the region bounded by
We have π/4 < arg (z) < π/3
Let z = x + iy
∴ arg (z) = tan^{1}(y/x)
The given inequality can be written as
π/4 < tan^{1}(y/x) < π/3
⇒ tan(π/4) < y/x < tan(π/3)
⇒ 1 < y/x < √3
⇒ x < y < √3x
This inequality represents the region between the lines
y = x and y = √3x
The product of the perpendicular, drawn from any point on a hyperbola to its asymptotes is
From the top of the house 15 metres high the angle of depression of a point which is at a distance 15 m from the base of the house is
Given, Perpendicular = 15 m
Base = 15 m
Angle of depression = tanθ = 15/15 = 1
θ = 45^{°}
The parabolas y^{2} = 4x and x^{2} = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If S₁, S₂, S₃ are respectively the areas of these parts numbered from top to bottom; then S₁ : S₂ : S₃ is
The solution of the differential equation cos x sin y dx + sin x cos y dy = 0 is
A+B = 0
⇒ A + B = O, where O is a zero matrix.
⇒ A = O − B = −B
Take determinant on both sides
⇒ A = −B
⇒ A + B = 0
A and B are square matrices of order n x n, then (A  B)^{2} is equal to
(A  B)^{2}
⇒ (A  B) x (A  B)
⇒ (A  B) x A  (A  B) x B
⇒ A^{2}  AB  BA + B^{2}
f(x) = (x+1)^{3} + 1
∴ f'(x) = 3(x+1)^{2}
f'(x) = 0 ⇒ x = 1
Now, f" (1  ∈) = 3(∈)^{2} > 0, f'(1 + ∈)^{2} = 3∈^{2} > 0
∴ f(x) has neither a maximum nor a minimum at x = 1
Let f'(x) = φ ′ (x) = 3(x+1)^{2}
∴ φ′(x) = 6(x+1).
φ′(x) = 0 ⇒ x = 1
φ ′ (1∈) = 6(∈) < 0, φ ′ (1∈) = 6∈ > 0
∴ φ (x) has a minimum at x = 1
f(n) = n^{3} + 2n
put n=1, to obtain f(1) = 1^{3} + 2.1 = 3
Therefore, f(1) is divisible by 3
Assume that for n=k, f(k) = k^{3} + 2k is divisible by 3
Now, f(k+1) = (k+1)^{3} + 2(k+1)
⇒ k^{3} + 2k + 3(k^{2} + k + 1)
⇒ f(k) + 3(k^{2 }+ k + 1)
Since, f(k) is divisible by 3
Therefore, f(k+1) is divisible by 3
and from the principle of mathematical induction f(n) is divisible by 3 for all n∈N
If equation λx^{2 }+ 2y^{2 } 5xy + 5x  7y + 3 = 0, represents two straight lines, the value of λ is
If α + i β = tan^{⁻1} z, z = x + yi and α is constant, then locus of \'z\' is
The locus of the point of intersection of two normals to the parabola x^{2}=8y, which are at right angles to each other,is
Let P(4t_{1}, 2t_{1}^{2}) and Q(4t_{2}, 2t_{2}^{2}) be the points on x^{2} = 8y. Equation of the normals at P and Q are
y  2t_{1}^{2} = 1/t_{1}(x  4t_{1}^{2}) ....(1)
y  2t_{2}^{2} = 1/t_{2} (x  4t_{2}^{2}) ....(2)
Since the normals are at right angles, we have
(1/t_{1})(1/t_{2}) = 1
⇒ t_{1}t_{2} = 1 ....(3)
Solving Eqs. (1) and (2) and from Eq. (3), we have
⇒ 2y = x^{2} + 12 which is the required locus.
The equation of the parabola with its vertex at (1,1) and focus at (3,1) is
Out of 6 boys and 4 girls a group of 7 is to be formed. In how many ways can this be done if the group is to have a majority of boys?
The boys are in majority, if groups formed are 4B 3G, 5B 2G, 6B 1G.
Total number of such combinations
⇒ ^{6}C_{4} x ^{4}C_{3} + ^{6}C_{5} x ^{4}C_{2} + ^{6}C_{6} x ^{4}C_{1}
⇒ 15 x 4 + 6 x 6 + 1 x 4
⇒ 60 + 36 + 4
⇒ 100
How many even numbers can be formed by using all the digits 2, 3, 4, 5, 6?
For a number to be even its last digit should be divisible by 2.
Therefore, only 2,4,6 can be used as last digits.
Number of such combinations,
⇒ 4!×3 = 24×3 = 72
A committee of five is to be chosen from a group of 9 people. The probability that a certain married couple will either serve together or not at all is
The total number of ways in which 5 person can be chosen out of 9 person is ^{9}C_{5} = 126
The couple serves the committee in ^{7}C_{3} x ^{2}C_{2} = 35 ways
The couple does not serve the committee in
^{7}C_{5}  ^{7}C_{2} = 21 ways
Since the couple will either be together or not at all
∴ favourable number of cases = 35 + 21 = 56
∴ required porbability = 56 126 = 4/9
A sixfaced dice is so biased that it is twice as likely to show an even number as an odd number when thrown. It is thrown twice. The probability that the sum of two numbers thrown is even is
∵ probability for odd = p
∴ probability for even = 2p
∵ p + 2p = 1
⇒ 3p = 1
⇒ p = 1/3
∴ probability for odd = 1/3 , probability for even = 2/3
Sum of two no. is even means either both are odd or both are even
∴ required probability = (1/3 × 1/3) + (2/3 × 2/3) = 1/9 + 4/9 = 5/9
The sum to infinity of (1/7) + (2/7^{2}) + (1/7^{3}) + (2/7^{4}) + ... is
If a, b are the roots of x^{2} + x + 1 = 0 then a^{2} + b^{2} =
Sum of the roots ⇒ a+b = −1 ...(1)
Product of the roots ⇒ ab = 1 ...(2)
We know that,
(a+b)^{2 }= a^{2 }+ b^{2 }+ 2ab
⇒ a^{2 }+ b^{2 }= (a+b)^{2 }− 2ab
Substituting (1) and (2) in the above equation, we get
⇒ a^{2 }+ b^{2 }= (−1)^{2 }− 2(1)
∴ a^{2 }+ b^{2 }= −1
The third term of a G.P. is 3. The product of its first five terms is
Third term of G.P. = ar^{2 }= 3
General term = ar^{n−1}
Product of first five terms = a^{5}⋅r^{0+1+2+3+4}
= a^{5}r^{10}=(ar^{2})^{5}
= (3)^{5}
= 243.
Suppose that g (x) = 1 + √x and f (g(x)) = 3 + 2 √x + x, then f (x) is
Given, g(x) = 1 + √x
And f(g(x)) = 3 + 2√x + x
Now, f(g(x)) = 3 + 2√x + x [ terms splitting]
f(g(x)) = 2 + (1 + 2√x + x)
Also, (1 + √x)^{2 }= 1 + x + 2√x
⇒ f(g(x)) = 2 + (1 + √x)^{2}
⇒ f(x) = 2 + x^{2}
f(x) = x^{2} + 2
The slope of the tangent of the curve at the point where x = 1 is
If a line passes through points (4,3) and (2,λ) and perpendicular to y=2x+3, then λ=
Two lines are perpendicular then product of slope be 1
m_{1}m_{2 }= −1
(3 − λ)/2 × 2 = −1
3 − λ = −1
3 + 1 = λ.
λ = 4.
The ratio in which the line joining the points (a,b,c) and (a,c,b) is divided by the xyplane is
Let λ:1 be the ratio in which the line joining the points (a,b,c) and (a, c, b) is divided by xy plane.
The centre of sphere passing through four points (0, 0, 0), (0, 2, 0), (1, 0, 0) and (0, 0, 4) is
Since the sphere passes through origin, its equation is of the form
x^{2 }+ y^{2 }+ z^{2 }+ 2ux + 2vy + 2wz = 0. ...(1)
Substituting the co  ordinates of the other given points, we get
u = 1, v = 2, w = 4
Centre of sphere = (u/2, v/2, w/2)
⇒ (1/2, 1, 2)
The graph of y = log_{a}x is reflection of the graph of y = a^{x} in the line
The graph of the logarithmic function y = log_{a}x is the reflection about the line y = x of the graph of the exponential function y = a^{x}, as shown in Figure.
The number of roots of the equation 2sin^{2}θ + 3sinθ + 1 = 0 in 0,2π is
If A+B+C=180^{o}, then [(tanA+tanB+tanC)/(tanA tanB tanC)]=
Given, A + B + C = 180
So, A + B = 180  C
Taking tan on both sides we get,
⇒ tan(A+B) = tan(180C)
⇒
⇒ tanA + tanB = tanC(1  tanA tanB)
⇒ tanA + tanB =  tanC + tanA tanB tanC
⇒ tanA + tanB + tanC = tanA tanB tanC
The points with position vectors 10î + 3ĵ, 12î  5ĵ and aî + 11ĵ are collinear if a =
Given, A = (10i + 3j)
B = (12i  5j)
C = (ai + 11j)
AB = 2i  8j
AC = (a  10)i + 8j
AB and AC are collinear
⇒ 2/(a  10) = 8/8
⇒ 2 = 10  a
⇒ a = 8
If a.b = 0
⇒ ab(cosθ) = 0
or cosθ = 0
⇒ θ = 90°
Therefore, a⊥b
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