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If f(x) is an even function, then the graph y = f(x) will be symmetrical about
y — axis by definition.
In the previous question, what will be the minimum value of the function?
Required value = (5)^{2} + 10(5) + 11 = 2550 + 11 =14.
Find the minimum value of the function fix) = log_{2} (x^{2}  2x + 5).
The minimum value of the function would occur at the minimum value of (x^{2}  2x + 5) as this quadratic function has imaginary roots.
Thus, minimum value of the argument of the log is 4. So minimum value of the function is log_{2} 4 = 2.
If f(x) is an even function, then the graph y = f(x) will be symmetrical about
y – axis by definition.
x^{–8} is even since f(x) = f(–x) in this case.
For what value of x, x^{2} + 10x + 11 will give the minimum value?
dy/dx = 2x + 10 = 0 fi x = –5.
Since the denominator x^{2} – 3x + 2 has real roots, the maximum value would be infinity.
Read the instructions below and solve.
f(x) = f(x – 2) – f(x – 1), x is a natural number
f(1) = 0, f(2) = 1
The value of f(8) is
f(1) = 0, f(2) = 1,
f(3) = f(1) – f(2) = –1
f(4) = f(2) – f(3) = 2
f(5) = f(3) – f(4) = –3
f(6) = f(4) – f(5) = 5
f(7) = f(5) – f(6) = –8
f(8) = f(6) – f(7) = 13
f(9) = f(7) – f(8) = –21
13
Read the instructions below and solve.
f(x) = f(x – 2) – f(x – 1), x is a natural number
f(1) = 0, f(2) = 1
What will be the domain of the definition of the function f(x) = 8–^{x}C_{ 5–x} for positive values of x?
f(1) = 0, f(2) = 1,
f(3) = f(1) – f(2) = –1
f(4) = f(2) – f(3) = 2
f(5) = f(3) – f(4) = –3
f(6) = f(4) – f(5) = 5
f(7) = f(5) – f(6) = –8
f(8) = f(6) – f(7) = 13
f(9) = f(7) – f(8) = –21
For any ^{n}C_{r}, n should be positive and r ≥ 0.
Thus, for positive x, 5 – x ≥ 0
fi x = 1, 2, 3, 4, 5.
Ajesh saves Rs 50,000 every year and deposits the money in a bank at compound interest of 10%(compunded annually).What would be his total saving at the end of the 5th year?
At the end of the 1st year, he will get Rs 50000, it will give him interest for 4 years compounded annually
Hence at the end of 5 years, this amount will become 50000(1.1)^{4}
Similarly, the amount deposited in the 2nd year will give interest for 3 years. Hence it will become 50000(1.1)^{3}
Similarly, we can calculate for the remaining years.
The total saving at the end of the 5th year would be a GP, given by
Net saving = 50000(1.1)^{4 }+ 50000(1.1)^{3} ..... 50000
Thus net saving = = Rs 3,05,255
Define the following functions:
(a) (a M b) = a – b (b) (a D b) = a + b
(c) (a H b) = (ab) (d) (a P b) = a/b
Q.
Which of the following functions will represent a^{2} – b^{2}?
Option a = (a – b) (a + b) = a^{2} – b^{2}
Define the following functions:
(a) (a M b) = a – b (b) (a D b) = a + b
(c) (a H b) = (ab) (d) (a P b) = a/b
Q.
What is the value of (3M4H2D4P8M2)?
3 – 4 × 2 + 4/8 – 2 = 3 – 8 + 0.5 – 2 = – 6.5
(using BODMAS rule)
Define the following functions:
(a) (a M b) = a – b (b) (a D b) = a + b
(c) (a H b) = (ab) (d) (a P b) = a/b
Q.
Which of the four functions defined has the minimum value?
The minimum would depend on the values of a and b. Thus, cannot be determined.
If 0 < a <1 and 0 < b < 1 and if a < b, which of the following expressions will have the highestvalue?
Again (a + b) or a/b can both be greater than each other depending on the values we take for a and b.
E.g. for a = 0.9 and b = 0.91, a + b > a/b.
For a = 0.1 and b = 0.11, a + b < a/b
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