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The range of the function
Calculation:
Concept:
The points where the f'(x) = 0 are known as critical values.
At critical point, if f ''(x) > 0 then function has minima.
If f ''(x) < 0 then function has maxima.
For x ≥ 0
f(x) = − 3sin x
⇒ f '(x) = − 3cos x
⇒ f ′(0) = − 3
For x < 0, f (x) = x3 + x2 + 10x
⇒ f ′(x) = 3x2 + 2x + 10
⇒ f ′(0) = 10
⇒ f ′(x) > 0 for x < 0 and f ′(x) < 0 for x ≥ 0
⇒ f (x) has maxima at x = 0.
∴ At x = 0, there is a point of maximum.
The correct answer is Option 1.
where α ∈ Q is equal to
Four numbers are multiplied together. Then the probability that the product will be divisible by 5 or 10 is
The divisibility of the product of four numbers depends upon the value of the last digit of each number.
The last digit of a number can be any of the 10 digits
⇒ The total number of ways of selecting last digits of four numbers is 10 × 10 × 10 × 10 = 104
Now, if the product of the 4 numbers is not divisible by 5 or 10
⇒ Then the number of choices for the last digit of each number is 8 (excluding 0 or 5).
⇒ Favourable number of ways = 8 × 8 × 8 × 8 = 84
∴ P(product is divisible by 5 or 10)
= 1 - P(product is divisible by 5 or 10)
The differential equation of all circles which pass through the origin and whose centres lie on y-axis is
Given, the circles pass through the origin.
They have their centres at (0, a) The circles have radius a.
∴ The equation of the family of circles in given by x2 + (y - a)2 = a2
⇒ x2 + y2 + a2 - 2ya = a2
⇒ x2 + y2 = 2ay ⋯ (i)
Differentiating wrt x, we get:
∴ The differential equation of all circles which pass through the origin and whose centres lie on y-axis is
The coefficient of x49 in the product (x – 1) (x – 3) ... (x – 99) is
The sum of first n terms of an AP with first term and common difference d is given by:
We can observe, (x − a)(x − b) = x2 − (a + b)x + ab
(x − a)(x − b)(x − c) = x3 − (a + b + c)x2 + (ab + bc + ca) x − abc
(x − a)(x − b)(x − c)(x − d) = x4 − (a + b + c + d)x3 + (ab + bc + cd + da)x2 − (abc + bcd + cda + abd)x + abcd
Now, (x − 1)(x − 3)(x − 5) ⋯ (x − 99)
= x50 − (1 + 3 + 5 + ⋯ + 99)x49 +(1⋅3 + 1⋅5 + 1⋅7 + ⋯ + 1⋅99 + 3⋅5 + 3⋅7 + ⋯ + 3⋅99 + 5⋅7 + ⋯ + 97⋅99)x48 + ⋯
∴ Coefficient of x48 =
= − [1 + 3 + 5 + ⋯ + 99]
= − 25[2 + 49 × 2]
= − 25 × 100
= − 2500
∴ The coefficient of x49 in the product (x – 1) (x – 3) ... (x – 99) is − 2500.
The correct answer is Option 3.
Area bounded by the curves y = |x| - 2 and y = 1 - |x - 1| is equal to
∴ Required area = ar(ABCD)
= ar(ΔOAB) + ar(ΔODC) + ar(ΔOBC)
= 1 + 1 + 2
= 4 sq. units
∴ The area bounded by the curves y = |x| - 2 and y = 1 - |x - 1| is equal to 4 sq. units.
If two events A and B are such that P(A') = 0.3, P(B) = 0.4 and P (A ∩ B') = 0.5, then
If the point (1, a) lines in between the line x + y = 1 and 2(x + y) = 3, then a lies in
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