Ncert Solutions Maths class 12th
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Let y = (ex)?
ln y = x ln (ex) = x [1 + ln x]
(1/y) (dy/dx) = (1) (1 + ln x) + x (1/x) = 2 + ln x
⇒ dy = (ex)? (2 + ln x)dx
∫? ² (ex)? (2 + log? x)dx = [ (ex)? ]? ² = (2e)² - (1e)¹ = 4e² - e
Upvote 
In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is:
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In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is:
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sum 6 → (1,5), (5,1), (3,3), (2,4), (4,2)
sum 7 → (1,6), (6,1), (5,2), (2,5), (3,4), (4,3)
= P (A) + P (? ) · P (B) · P (A) + P (? )P (B)P (? ) · P (B) · P (A) + …
This is infinite G.P. with common ratio P (? ) × P (B)
Probability of A wins = P (A) / (1 - P (? )P (B? )
= (5/36) / (1 - (31/36)* (30/36) =
Upvote The distance of the point (1, −2,3) from the plane x − y + z = 5 measured parallel to the line x/2 = y/3 = z/-6 is:
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The distance of the point (1, −2,3) from the plane x − y + z = 5 measured parallel to the line x/2 = y/3 = z/-6 is:
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Equation PQ
(x-1)/2 = (y+2)/3 = (z-3)/ (-6) = λ
Let Q = (2λ + 1, 3λ − 2, −6λ + 3)
Q lies on x - y + z = 5
⇒ (2λ + 1) − (3λ − 2) + (−6λ + 3) = 5
⇒ λ = -1/7
Q = (5/7, -17/7, 15/7)
∴ PQ = √ (2/7)² + (3/7)² + (6/7)²)
Upvote If the system of equations
x - 2y + 3z = 9
2x + y + z = b
x - 7y + az = 24
has infinitely many solutions, then a − b is equal to
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If the system of equations
x - 2y + 3z = 9
2x + y + z = b
x - 7y + az = 24
has infinitely many solutions, then a − b is equal to
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D = |1 -2 3; 2 1; 1 -7 a| = 0 ⇒ a = 8
also, D? = |9 -2 3; b 1; 24 -7 8| = 0 ⇒ b = 3
hence, a-b = 8-3=5
Upvote Let [t] denote the greatest integer ≤ t. Then the equation in x, [x]² + 2[x + 2] − 7 = 0 has:
Let [t] denote the greatest integer ≤ t. Then the equation in x, [x]² + 2[x + 2] − 7 = 0 has:
[x]² + 2 [x+2] - 7 = 0
⇒ [x]² + 2 [x] + 4 - 7 = 0
⇒ [x] = 1, -3
⇒ x ∈ [1,2) U [-3, -2)
Upvote