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**Q. 1. The values of f(x) lie in the interval**

**Ans. **

**Solution.** For the given function to be defined

and sine function increases on [0, π/4]

**Q. 2. ****For the function **

**the derivative from the right, f '(0+) = ......................., and the derivative from the left, f '(0–) = ...............**

**Ans.** 0, 1

**Solution.**

Thus f ' (0^{+}) = 0 and f ' (0^{–}) = 1

**Q. 3.** **The domain of the function given by .................**

**Ans. **[–2, –1] ∪ [1, 2]

**Solution.** To find domain of function

For f (x) to be defined we should have

**NOTE THIS STEP :**

**Q. 4. ****Let A be a set of n distinct elements. Then the total number of distinct functions from A to A is ................. and out of these ................. are onto functions.**

**Ans. **n^{n}, n^{!}

**Solution. **Set A has n distinct elements. Then to define a function from A to A, we need to associate each element of set A to any one the n elements of set A. So total number of functions from set A to set A is equal to the number of ways of doing n jobs where each job can be done in n ways. The total number such ways is n × n × n × .... × n (n - times).**Hence the total number of functions from** A to A is n^{n}.

Now for an onto function from A to A, we need to associate each element of A to one and only one element of A. So total number of onto functions from set A to A is equal to number of ways of arranging n elements in range (set A) keeping n elements fixed in domain (set A). n elements can be arranged in n! ways.

Hence, the total number of functions from A to A is n!.

**Q. 5. **If **, then domain of f(x) is .... and its range is .................**

**Ans. **(-2, 1), [-1, 1]

**Solution. **The given function is,

Combining these two inequalities, we get x ∈ (– 2, 1)

∴ Domain of f is (– 2, 1)

Also sin q always lies in [– 1, 1].

∴ Range of f is [– 1, 1]

**Q. 6. There are exactly two distinct linear functions, ................., and ........... which map [– 1, 1] onto [0, 2].**

**Ans. **x + 1 and - x+ 1

**Solution. **KEY CONCEPT : Every linear function is either strictly increasing or strictly decreasing. If f (x) = ax + b, Df = [p, q], Rf = [m, n]

Then f (p) = m and f (q) = n, if f (x) is strictly increasing and f (p) = n, f (q) = m, If f (x) is strictly decreasing function.

Let f (x) = ax + b be the linear function which maps [–1, 1] onto [0, 2]. then f (–1) = 0 and f (1) = 2 or f (–1) = 2 and f (1) = 0

Depending upon f (x) is increasing or decreasing respectively.

⇒ – a + b = 0 and a + b = 2 ....(1)

or – a + b = 2 and a + b = 0 ....(2)

Solving (1), we get a = 1, b = 1.

Solving (2), we get a = – 1, b = 1

Thus there are only two functions i.e., x + 1 and – x + 1.

**Q. 7. ****If f is an even function defined on the interval (-5, 5), then four real values of x satisfying the equation **

**are ................., ................., ................., and ...........**

**Ans. **

**Solution. **Given that and f is an even function

**Q. 8. If f(x) = sin ^{2} x + ** then (gof) (x) = .................

**Ans. **1

**Solution. **

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