Problem 105 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Given \(f(x)=1+x, \quad-1 \leq x \leq 0\) \(=-x, \quad 00\) ix. \(f(x)\) is injective function.

Step-by-Step Solution

Answer

i. \(f_1(x) = -x, \, -1 \leq x < 0; \, x-1, \, 0 \leq x < 1; \, 1, \, x=1\) ii. \(f_2(x) = -x, \, -1 \leq x \leq 0; \, x-1, \, 0 < x \leq 1\) iii. \(f_3(x) = 1, \, x=-1; \, -x-1, \, -10\) ix. \(f(x)\) is injective function.

Step 1: Find g(x) for the intervals

For the interval \(-1 \leq x < 0\), \(g(x) = -1 - x\). For the interval \(0 \leq x \leq 1\), \(g(x) = 1 - x\).

Step 2: Determine f_1(x) for both intervals

For the interval \(-1 \leq x < 0\), \(f_1(x) = f(g(x)) = f(-1 - x) = -(-1 - x) = x + 1\). For the interval \(0 \leq x \leq 1\), \(f_1(x) = f(g(x)) = f(1 - x) = -(1 - x) = x - 1\).

Step 3: Domain and Range of f_1(x)

The domain of \(f_1(x) = f(g(x))\) is \(-1 \leq x \leq 1\). The range of \(f_1(x)\) is \(-1 \leq f_1(x) \leq 1\). ii. Function: \(f_2(x) = g(f(x))\)

Step 4: Find f(x) for the intervals

For the interval \(-1 \leq x \leq 0\), \(f(x) = 1 + x\). For the interval \(0 < x \leq 1\), \(f(x) = -x\).

Step 5: Determine f_2(x) for both intervals

For the interval \(-1 \leq x \leq 0\), \(f_2(x) = g(f(x)) = g(1 + x) = -(1 + x) = -1 - x\). For the interval \(0 < x \leq1\), \(f_2(x) = g(f(x)) = g(-x) = 1 - (-x) = x + 1\).

Step 6: Domain and Range of f_2(x)

The domain of \(f_2(x) = g(f(x))\) is \(-1 \leq x \leq 1\). The range of \(f_2(x)\) is \(-1 \leq f_2(x) \leq 1\). iii. Function: \(f_3(x) = f_{1}(f_{2}(x))\)

Step 7: Find f_1(f_2(x)) for the intervals

For the interval \(-1 \leq x \leq 0\), \(f_1(f_2(x)) = f_1(-1-x) = -(-1-x) = x+1\). For the interval \(0 < x \leq1\), \(f_1(f_2(x)) = f_1(x+1) = -(x+1) = -x-1\).

Step 8: Domain and Range of f_3(x)

The domain of \(f_3(x) = f_{1}(f_{2}(x))\) is \(-1 \leq x \leq 1\). The range of \(f_3(x)\) is \(-1 \leq f_3(x) \leq 1\). iv. Function: \(f_4(x) = f_{2}(f_{1}(x))\)

Step 9: Find f_2(f_1(x)) for the intervals

For the interval \(-1 \leq x < 0\), \(f_2(f_1(x)) = f_2(x+1) = -x\). For the interval \(0 \leq x \leq 1\), \(f_2(f_1(x)) = f_2(x-1) = -x\).

Step 10: Domain and Range of f_4(x)

The domain of \(f_4(x) = f_{2}(f_{1}(x))\) is \(-1 \leq x \leq 1\). The range of \(f_4(x)\) is \(-1 \leq f_4(x) \leq 1\). v. Function: \(f_5(x) = f(|x|)\)

Step 11: Determine f_5(x)

For the interval \(-1 \leq x < 0\), \(f_5(x) = f(|x|)= f(-x) = -(-x) = x\). For the interval \(0 \leq x \leq 1\), \(f_5(x) = f(|x|) = f(x) = 1\).

Step 12: Domain and Range of f_5(x)

The domain of \(f_5(x) = f(|x|)\) is \(-1 \leq x \leq 1\). The range of \(f_5(x)\) is \(0 \leq f_5(x) \leq 1\). vi. Function: \(f_6(x) = |f(x)|\)

Step 13: Determine f_6(x)

For the interval \(-1 \leq x \leq 0\), \(f_6(x) = |f(x)| = |1 + x| = 1 + x\). For the interval \(0 < x \leq1\), \(f_6(x) = |f(x)| = |-x| = x\).

Step 14: Domain and Range of f_6(x)

The domain of \(f_6(x) = |f(x)|\) is \(-1 \leq x \leq 1\). The range of \(f_6(x)\) is \(0 \leq f_6(x) \leq 1\). vii. Function: \(f_7(x) = \operatorname{sgn}(f(x))\)

Step 15: Determine f_7(x)

For the interval \(-1 \leq x \leq 0\), \(f_7(x) = \operatorname{sgn}(1 + x) = 1\).

Step 16: Domain and Range of f_7(x)

The domain of \(f_7(x) = \operatorname{sgn}(f(x))\) is \(-1 \leq x \leq 1\). The range of \(f_7(x)\) is \(-1 \leq f_7(x) \leq 1\). viii. Function: \(f_8(x) = f(\operatorname{sgn}(x))\)

Step 17: Determine f_8(x)

For the interval \(-1 \leq x < 0\), \(f_8(x) = f(\operatorname{sgn}(x)) = f(-1) = 1 - 1 = 0\). For the interval \(0 \leq x \leq 1\), \(f_8(x) = f(\operatorname{sgn}(x)) = f(1) = -1\).

Step 18: Domain and Range of f_8(x)

The domain of \(f_8(x) = f(\operatorname{sgn}(x))\) is \(-1 \leq x \leq 1\). The range of \(f_8(x)\) is \(-1 \leq f_8(x) \leq 0\). ix. Which of the functions are injective?

Step 19: Determine injective functions

The only injective function among these is \(f(x)\) since it has a unique output for every input.