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Answer:
3√5 sq.cm ; 2/√5 ; (0 , 2)
Step-by-step explanation:
Question: The area of a triangle formed by vertices O, A and B, where vector OA = i + 2j + 3k and vector AB = -3i - 2j + k.
Solution: Area = 1/2 | OA x AB |
⇒ 1/2 * | (i + 2j + 3k) x (-3i - 2j + k) |
⇒ 1/2 * | [-2k - j + 6k + 2i - 9j + 6i ] |
⇒ 1/2 * | (8i - 10j + 4k) |
⇒ 1/2 * √8² + 10² + 4²
⇒ 1/2 * 6√5
⇒ 3√5 cm²
Based on:
i x i = j x j = k x k = 0
i x j = k ; j x k = i ; k x i = j
j x i = -k ; k x j = -i ; i x k = -j
Question: If cos(arcsin(2/√5) + arccosx) = 0, then x is equal to
Solution:
⇒ cos(arcsin(2/√5) + arcosx) = cos(π/2)
⇒ arcsin(2/√5) + arccosx = π/2
⇒ arcsin(2/√5) = π/2 - arccosx
⇒ arcsin(2/√5) = arcsinx
⇒ 2/√5 = x
Using:
arcsinx = inverse of sinx , so with cosx.
π/2 - arccosx = arcsinx
Question: The interval in which the function F given by f(x) = x²e^(-x) is strictly increasing.
Solution: If f(x) is increasing, then f'(x) > 0
⇒ d(x²e^(-x) )/dx > 0
⇒ 2xe^(-x) - x²e^(-x) > 0
⇒ 2x - x² > 0
⇒ x ∈ (0 , 2)
- (a) 3√5 square units , or, units²
- (c) x = 2/√5
- (d) (0,2)
for further explanation , refer the above attachment
