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Algebra Level 3 Test 9
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Question 1 |
Let f(x) = ax2 – b|x| , where a and b are constants. Then at x = 0, f(x) is
maximized whenever a > 0, b > 0 | |
maximized whenever a > 0, b < 0 | |
minimized whenever a > 0, b > 0 | |
minimized whenever a > 0, b< 0 |
Question 1 Explanation:
When a > 0, b < 0,
ax2 and –b |x| are non negative for all x,
i.e. ax2 – b|x| ≥ 0
Therefore ax2 – b |x| is minimum at x = 0
when a > 0, b < 0.
ax2 and –b |x| are non negative for all x,
i.e. ax2 – b|x| ≥ 0
Therefore ax2 – b |x| is minimum at x = 0
when a > 0, b < 0.
Question 2 |
Consider the following two curves in the x-y plane:
y = x3 + x2 + 5
y = x2 + x + 5
Which of following statements is true for −2 ≤ x ≤ 2 ?
y = x3 + x2 + 5
y = x2 + x + 5
Which of following statements is true for −2 ≤ x ≤ 2 ?
The two curves intersect once. | |
The two curves intersect twice. | |
The two curves do not intersect. | |
The two curves intersect thrice. |
Question 2 Explanation:
When we substitute two values of x in the above curves, at x = –2 we get
y = –8 + 4 + 5 = 1
y = 4 – 2 + 5 = 7
Hence at x = –2 the curves do not intersect.
At x = 2, y1 = 17 and y2 = 11
At x = –1, y1 = 5 and 2 and y2 = 5
When x = 0, y1 = 5 and y2 = 5
And at x = 1, y1 = 7 and y2 = 7
Therefore, the two curves meet thrice when x = –1, 0 and 1.
y = –8 + 4 + 5 = 1
y = 4 – 2 + 5 = 7
Hence at x = –2 the curves do not intersect.
At x = 2, y1 = 17 and y2 = 11
At x = –1, y1 = 5 and 2 and y2 = 5
When x = 0, y1 = 5 and y2 = 5
And at x = 1, y1 = 7 and y2 = 7
Therefore, the two curves meet thrice when x = –1, 0 and 1.
Question 3 |
Suppose n is an integer such that the sum of digits on n is 2, and 1010 < n 10n. The number of different values of n is
11 | |
10 | |
9 | |
8 |
Question 3 Explanation:
We have
(1) 1010 < n < 1011
(2) Sum of the digits for 'n' = 2
Clearly- (n)min = 10000000001 (1 followed by 9 zeros and finally 1)
Obviously, we can form 10 such numbers by shifting
'1' by one place from right to left again and again.
Again, there is another possibility for 'n'
n = 20000000000
So finally : No. of different values for n = 10 + 1 = 11
(1) 1010 < n < 1011
(2) Sum of the digits for 'n' = 2
Clearly- (n)min = 10000000001 (1 followed by 9 zeros and finally 1)
Obviously, we can form 10 such numbers by shifting
'1' by one place from right to left again and again.
Again, there is another possibility for 'n'
n = 20000000000
So finally : No. of different values for n = 10 + 1 = 11
Question 4 |
If f(x) = x3 – 4x + p , and f(0) and f(1) are of opposite sings, then which of the following is necessarily true
–1 < p < 2 | |
0 < p < 3 | |
–2 < p < 1 | |
–3 < p < 0 |
Question 4 Explanation:
We are given by the statements
f(0) = 03 – 4(0) + p = p
f(1) = 13 – 4(1) + p = p – 3
If P and P – 3 are of opp. signs then p(p – 3) < 0
Hence 0 < p < 3.
f(0) = 03 – 4(0) + p = p
f(1) = 13 – 4(1) + p = p – 3
If P and P – 3 are of opp. signs then p(p – 3) < 0
Hence 0 < p < 3.
Question 5 |
If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?
0 | |
–1 | |
1 | |
Not unique |
Question 5 Explanation:
We are given by the statement that if the sum of the first 11
terms of an arithmetic progression equals that of the first 19 terms, thus
t1 + t2 + t3 + t4 + t5 …………..+t11= t1 + t2 + t3 + t4 ………. +t19
Therefore sum of series
= 11/2[2a +(11 -1)d] = 19/2[2a +(19 -1)d]
22a +110d = 28a + 342d
16a + 232d = 0
2a + 29d = 0
30/2 [2a + (30 – 1)d] = 0
So the sum of first terms are = 0
terms of an arithmetic progression equals that of the first 19 terms, thus
t1 + t2 + t3 + t4 + t5 …………..+t11= t1 + t2 + t3 + t4 ………. +t19
Therefore sum of series
= 11/2[2a +(11 -1)d] = 19/2[2a +(19 -1)d]
22a +110d = 28a + 342d
16a + 232d = 0
2a + 29d = 0
30/2 [2a + (30 – 1)d] = 0
So the sum of first terms are = 0
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