Class 12th
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New answer posted
2 months agoContributor-Level 9
Given f (g (x) is defined piecewise:
f (g (x) =
x³ + 2 ; x < 0
x? ; 0? x < 1
(3x - 2)² ; x? 1
fog (x) is discontinuous at x = 0.? non differentiable.
fog (x) is not differentiable at x = 0.
New answer posted
2 months agoContributor-Level 10
Using Lagrange's Mean Value Theorem (LMVT) for x ∈ [−7, -1].
[f (-1) - f (-7)] / [-1 - (-7)] ≤ 2
[f (-1) - (-3)] / 6 ≤ 2
f (-1) + 3 ≤ 12
f (-1) ≤ 9
Using LMVT for x ∈ [−7, 0].
[f (0) - f (-7)] / [0 - (-7)] ≤ 2
[f (0) - (-3)] / 7 ≤ 2
f (0) + 3 ≤ 14
f (0) ≤ 11
Therefore, f (0) + f (-1) ≤ 11 + 9 = 20.
New answer posted
2 months agoContributor-Level 10
Answer given by NTS is (1) which is wrong.
I = 1/ (a+b) ∫? x [f (x) + f (x+1)]dx . (1)
Using the property x → a + b - x
I = 1/ (a+b) ∫? (a+b-x) [f (a+b-x) + f (a+b+1-x)]dx
Given f (a+b+1-x) = f (x)
I = 1/ (a+b) ∫? (a+b-x) [f (x+1) + f (x)]dx . (2)
Adding (1) and (2):
2I = 1/ (a+b) ∫? (a+b) [f (x) + f (x+1)]dx
2I = ∫? [f (x) + f (x+1)]dx
2I = ∫? f (x)dx + ∫? f (x+1)dx
Let x+1 = t in the second integral, so dx = dt.
When x=a, t=a+1. When x=b, t=b+1.
∫? f (x+1)dx = ∫? ¹ f (t)dt = ∫? ¹ f (x)dx
New answer posted
2 months agoContributor-Level 10
c = λ (a x b).
a = I + j - k
b = I + 2j + k
a x b = | I j k |
| 1 -1 |
| 1 2 1 |
= I (1 - (-2) - j (1 - (-1) + k (2-1) = 3i - 2j + k.
c = λ (3i - 2j + k).
Given c ⋅ (i + j + 3k) = 8.
λ (3i - 2j + k) ⋅ (i + j + 3k) = 8
λ (3 - 2 + 3) = 8 => 4λ = 8 => λ = 2.
c = 2 (a x b).
We need to find c ⋅ (a x b).
c ⋅ (a x b) = 2 (a x b) ⋅ (a x b) = 2|a x b|².
|a x b|² = 3² + (-2)² + 1² = 9 + 4 + 1 = 14.
So, c ⋅ (a x b) = 2 * 14 = 28.
New answer posted
2 months agoContributor-Level 10
f (x) = {x+a, if x<0; |x-1|, if x0}
g (x) = {x+1, if x<0; (x-1)+b, if x0}
g (f (x) must be continuous. The potential points of discontinuity are where the definitions of f (x) and g (f (x) change. This is at x=0 and where f (x)=0.
f (x)=0 when x=-a (if a>0) or when x=1.
Continuity at x = 0:
lim (x→0? ) g (f (x) = lim (x→0? ) g (x+a). Since a could be anything, let's analyze f (0? )=a. So, lim is g (a).
lim (x→0? ) g (f (x) = g (f (0? ) = g (|0-1|) = g (1). Since 1≥0, g (1) = (1-1)²+b = b.
g (f (0) = g (|0-1|) = g (1) = b.
So we need g (a) = b.
Case 1: a < 0. g (a) = a+1. So a+1=b.
Case 2: a ≥ 0. g (a) = (a-1)²+b. So (a-1)²+b=b => (a-1)²=0 => a=1.
Now consider continuity at x=-a (assuming a>0).
l
New answer posted
2 months agoContributor-Level 10
[a? ] = [ 1 ] [-1 b? ]
[a? ] [√3 k] [ k b? ] (This is likely incorrect OCR, should be a 2x1 result)
The solution seems to derive from a matrix multiplication:
[√3a? ] = [ 1 ] [b? ]
[√3a? ] [√3 k] [b? ]
This leads to:
b? - b? = √3a?
b? + kb? = √3a?
Also given: a? ² + a? ² = (2/3) (b? ² + b? ²).
Squaring and adding the two derived equations and comparing with the given condition leads to k=1.
New answer posted
2 months agoContributor-Level 10
The integral is I = ∫ [ (x²-1) + tan? ¹ (x + 1/x)] / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
This is a complex integral. The provided solution splits it into two parts:
I? = ∫ (x²-1) / [ (x? +3x²+1)tan? ¹ (x+1/x)] dx
I? = ∫ 1 / (x? +3x²+1) dx
The solution proceeds with substitutions which are hard to follow due to OCR quality, but it seems to compare the final result with a given form to find coefficients α, β, γ, δ. The final expression shown is:
10 (α + βγ + δ) = 10 (1 + (1/2√5)*√5 + 1/2) seems incorrect.
The calculation is shown as 10 (1 + 1/10 - 1/2) = 10 (11/10 - 5/10) = 10 (6/10) = 6.
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