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तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएसएसएलसी (अंग्रेजी माध्यम) कक्षा १०

A function f: [– 5, 9] → R is defined as follow : f(x) = {6x+1;-5≤x<25x2-1;2≤x<63x-4;6≤x≤9 Find 2f(-2)-f(6)f(4)+f(-2) - Mathematics

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प्रश्न

A function f: [– 5, 9] → R is defined as follow :

f(x) = `{{:(6x + 1";", -5 ≤ x < 2),(5x^2 - 1";", 2 ≤ x < 6),(3x - 4";", 6 ≤ x ≤ 9):}` Find `(2"f"(- 2) - "f"(6))/("f"(4) + "f"( -2))`

योग

उत्तर

f(x) = 6x + 1; x = {– 5, – 4, – 3, – 2, – 1, 0, 1}

f(x) = 5x2 – 1; x = {2, 3, 4, 5}

f(x) = 3x – 4; x = {6, 7, 8, 9}

`(2"f"(- 2) - "f"(6))/("f"(4) + "f"( -2))`

f(x) = 6x + 1

f(– 2) = 6(– 2) + 1 = – 12 + 1 = – 11

f(x) = 3x – 4

f(6) = 3(6) – 4 = 18 – 4 = 14

f(x) = 5x2 – 1

f(4) = 5(4)2 – 1 = 5(16) – 1

= 80 – 1 = 79

f(x) = 6x + 1

f(– 2) = 6(– 2) + 1 = – 12 + 1 = – 11

`(2"f"(- 2) - "f"(6))/("f"(4) + "f"( -2)) = (2(- 11) - 14)/(79 - 11)`

= `(- 22 - 14)/(68)`

= `(-36)/(68)`

= `(-9)/(17)`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Relations and Functions - Exercise 1.4 [पृष्ठ २५]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
अध्याय 1 Relations and Functions
Exercise 1.4 | Q 10. (iv) | पृष्ठ २५
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