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5 (a) diberi garis lurus ( y = x + p ) menyilang lengkung ( x ^ { 2 } +…

Question

5 (a) diberi garis lurus ( y = x + p ) menyilang lengkung ( x ^ { 2 } + y ^ { 2 } - 2 = 0 ) pada titik m dan titik n. cari julat nilai p. given the straight line ( y = x + p ) intersects with the curve ( x ^ { 2 } + y ^ { 2 } - 2 = 0 ) at point m and point n. find the range of values of p. 4 markah 4 marks (b) diberi fungsi kuadratik ( f ( x ) = 3 + 5 x - 2 x ^ { 2 } ). given the quadratic function ( f ( x ) = 3 + 5 x - 2 x ^ { 2 } ). (i) dengan menggunakan kaedah penyempurnaan kuasa dua, ungkapkan ( f ( x ) ) dalam bentuk ( f ( x ) = a ( x + p ) ^ { 2 } + q ), dengan keadaan a, p dan q ialah pemalar. by using the method of completing the square, express ( f ( x ) ) in the form of ( f ( x ) = a ( x + p ) ^ { 2 } + q ), such that a, p and q are constants. (ii) seterusnya, lakar graf bagi ( f ( x ) ) untuk ( 0 leq x leq 5 ). hence, sketch the graph of ( f ( x ) ) for ( 0 leq x leq 5 ). 4 markah 4 marks

Explanation:

Step1: Substitute \(y=x + p\) into \(x^{2}+y^{2}-2 = 0\)

Substitute \(y\) in the curve equation:

$$\begin{align*} x^{2}+(x + p)^{2}-2&=0\\ x^{2}+x^{2}+2px+p^{2}-2&=0\\ 2x^{2}+2px+(p^{2}-2)&=0 \end{align*}$$

Since the line intersects the curve at two points \(M\) and \(N\), the discriminant \(\Delta=b^{2}-4ac>0\) for the quadratic equation \(ax^{2}+bx + c = 0\). Here \(a = 2\), \(b=2p\), \(c=p^{2}-2\).

Step2: Calculate the discriminant

$$\begin{align*} \Delta&=(2p)^{2}-4\times2\times(p^{2}-2)\\ &=4p^{2}-8(p^{2}-2)\\ &=4p^{2}-8p^{2}+16\\ &=-4p^{2}+16 \end{align*}$$

Step3: Solve the inequality \(\Delta>0\)

$$\begin{align*} -4p^{2}+16&>0\\ 4p^{2}-16&<0\\ p^{2}-4&<0\\ (p - 2)(p + 2)&<0 \end{align*}$$

The solution of the inequality \((p - 2)(p + 2)<0\) is \(-2

Answer:

The range of values of \(p\) is \(-2