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Question5 Q.13) If f(x)=sec⁡xx+3f(x)=\frac{\sec x}{x+3} and f−1(c)=0f^{-1}(c)=0, then c=c= A. 13\frac{1}{3} B. 0 C. 23\frac{2}{\sqrt{3}} D. 14\frac{1}{4} cibtifantstius orsil 2=2= Q.14) One of the following equations is symmetric about origin A. y=x+1xy=\frac{x+1}{x} B. −x5+3x-x^{5}+3 x C. y=x4−2x2+6y=x^{4}-2 x^{2}+6 D. None −2=−x5)3x−(x5\begin{aligned} -2= & \left.-x^{5}\right)^{3 x} \\ & -\left(x^{5}\right. \end{aligned} Q.15) One of the following functions is an even function A. f(x)=sec⁡4x5xf(x)=\frac{\sec 4 x}{5 x} B.None C. f(x)=cos⁡5x2xf(x)=\frac{\cos 5 x}{2 x} D. f(x)=sin⁡2x3xf(x)=\frac{\sin 2 x}{3 x} Q.16)The range of the function f(x)=14−x2f(x)=\frac{1}{\sqrt{4-x^{2}}} is A. (0,2)(0,2) B. [0,2][0,2] C. (12,∞)\left(\frac{1}{2}, \infty\right) D. [12,∞)\left[\frac{1}{2}, \infty\right) E.None Q.17) Given that f(x)=1x−3f(x)=\frac{1}{x-3} and g(x)=1xg(x)=\frac{1}{x} then the domain of the function f∘gf \circ g is A. R\{0,13}R \backslash\left\{0, \frac{1}{3}\right\} B.R\{0}B . R \backslash\{0\} C. R\{13}R \backslash\left\{\frac{1}{3}\right\} D. R\{0,3}R \backslash\{0,3\} E.None Q.18) (The greatest integer less than or equals xx ) The range of f(x)=2[x]f(x)=2[x] is A. {0,61,62,63\{0,61,62,63, \qquad B. R\{0,61,62,63R \backslash\{0,61,62,63 \qquad f∣x∣f|x| sec - C. {0,62,64,66\{0,62,64,66 \qquad D. (−∞,∞)(-\infty, \infty) Q.19) If the domain of the function y=f(x)y=f(x) is [2,3)[2,3) then the domain of g(x)=f(3−x)g(x)=f(3-x) is A. [2,3)[2,3) B. (0,1](0,1] C. [0,1)[0,1) D. (2,3](2,3] E.None Q.20) Given that f(x)=sec⁡−1xf(x)=\sec ^{-1} x then f(2)=f(2)= A. 1sin⁡2\frac{1}{\sin 2} B. 1cos⁡2x\frac{1}{\cos 2}^{x} C. cos⁡−1(12)\cos ^{-1}\left(\frac{1}{2}\right) D. sin⁡−1(12)\sin ^{-1}\left(\frac{1}{2}\right) E.None Q.21) Given that f(x)=x2f(x)=x^{2} and g(x)={2x,x+3,x≥4g(x)=\left\{\begin{array}{l}2 x, \\ x+3, x \geq 4\end{array}\right. then (f∘g)(x)=f(x))\left.(f \circ g)(x)=f(x)\right) A. {4x2,x<4(x+1)2,x≥4\left\{\begin{array}{cl}4 x^{2} & , x<4 \\ (x+1)^{2} & , x \geq 4\end{array}\right. B. {4x2,x≤4(x+1)2,x>4\left\{\begin{array}{cl}4 x^{2} & , x \leq 4 \\ (x+1)^{2} & , x>4\end{array}\right. C. {4x2,x<16(x+1)2,x≥16\left\{\begin{array}{cc}4 x^{2} & , \quad x<16 \\ (x+1)^{2} & , \quad x \geq 16\end{array}\right. D. {4x2,x≤16(x+1)2,x>16\left\{\begin{array}{cc}4 x^{2} & , x \leq 16 \\ (x+1)^{2} & , x>16\end{array}\right. Q.22) The domain of the function f(x)=Ln⁡(5−∣7x+3∣)f(x)=\operatorname{Ln}(5-|7 x+3|) is A. [−87,27]\left[-\frac{8}{7}, \frac{2}{7}\right] B. (−87,27)\left(-\frac{8}{7}, \frac{2}{7}\right) C. R\[−87,27]R \backslash\left[-\frac{8}{7}, \frac{2}{7}\right] D. R\(−87,27)\mathrm{R} \backslash\left(-\frac{8}{7}, \frac{2}{7}\right) Q.23) The domain of f(x)=cos⁡−1(3x+1)f(x)=\cos ^{-1}(3 x+1) is −1,1-1,1 −24−-24- ولا تعت:

Studdy Solution

STEP 1

1. We are given a function f(x)=sec⁡xx+3 f(x) = \frac{\sec x}{x+3} .
2. We know that f−1(c)=0 f^{-1}(c) = 0 , meaning that when the function is evaluated at 0, it equals c c .

STEP 2

1. Evaluate the function f(x) f(x) at x=0 x = 0 .
2. Solve for c c .

STEP 3

Evaluate the function f(x)=sec⁡xx+3 f(x) = \frac{\sec x}{x+3} at x=0 x = 0 :
f(0)=sec⁡(0)0+3 f(0) = \frac{\sec(0)}{0 + 3}
Since sec⁡(0)=1cos⁡(0)=1 \sec(0) = \frac{1}{\cos(0)} = 1 , we have:
f(0)=13 f(0) = \frac{1}{3}

STEP 4

Since f−1(c)=0 f^{-1}(c) = 0 implies f(0)=c f(0) = c , we have:
c=13 c = \frac{1}{3}
The value of c c is 13 \boxed{\frac{1}{3}} .

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