mathemerize

Find the domain and range of function f(x) = \(x-2\over 3-x\).

Solution : we have,  f(x) = \(x-2\over 3-x\) Domain of f : Clearly f(x) is defined for all x satisfying 3 – x \(\ne\) 0 i.e. x \(\ne\) 3 Hence, Domain of f is R – {3} Range of f : Let y = f(x), i.e.  y = \(x-2\over 3-x\) \(\implies\) 3y – xy = …

Find the domain and range of function f(x) = \(x-2\over 3-x\). Read More »

The foci of an ellipse are \((\pm 2, 0)\) and its eccentricity is 1/2, find its equation.

Solution : Let the equation of the ellipse be \(x^2\over a^2\) + \(y^2\over b^2\) = 1. Then, coordinates of the foci are \((\pm ae, 0)\). Therefore,  ae = 2 \(\implies\)  a = 4 We have \(b^2\) = \(a^2(1 – e^2)\) \(\implies\) \(b^2\) =12 Thus, the equation of the ellipse is \(x^2\over 16\) + \(y^2\over 12\) …

The foci of an ellipse are \((\pm 2, 0)\) and its eccentricity is 1/2, find its equation. Read More »

Find the equation of the ellipse whose axes are along the coordinate axes, vertices are \((0, \pm 10)\) and eccentricity e = 4/5.

Solution : Let the equation of the required ellipse be \(x^2\over a^2\) + \(y^2\over b^2\) = 1                 ……….(i) Since the vertices of the ellipse are on y-axis. So, the coordinates of the vertices are \((0, \pm b)\). \(\therefore\)    b = 10 Now, \(a^2\) = \(b^2(1 – e^2)\)  …

Find the equation of the ellipse whose axes are along the coordinate axes, vertices are \((0, \pm 10)\) and eccentricity e = 4/5. Read More »

If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) = 1, then prove that \(1\over p^2\) = \(1\over a^2\) + \(1\over b^2\)

Solution : The given line is bx + ay – ab = 0 ………….(i) It is given that p = Length of the perpendicular from the origin to line (i) \(\implies\) p = \(|b(0) + a(0) – ab|\over {\sqrt{b^2+a^2}}\) = \(ab\over \sqrt{a^2+b^2}\) \(\implies\) \(p^2\) = \(a^2b^2\over a^2+b^2\) \(\implies\) \(1\over p^2\) = \(a^2+b^2\over a^2b^2\) \(\implies\) \(1\over …

If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) = 1, then prove that \(1\over p^2\) = \(1\over a^2\) + \(1\over b^2\) Read More »

Find the distance between the line 12x – 5y + 9 = 0 and the point (2,1)

Solution : We have line 12x – 5y + 9 = 0 and the point (2,1) Required distance = |\(12*2 – 5*1 + 9\over {\sqrt{12^2 + (-5)^2}}\)| = \(|24-5+9|\over 13\) = \(28\over 13\) Similar Questions If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) = …

Find the distance between the line 12x – 5y + 9 = 0 and the point (2,1) Read More »

Let \(a_n\) be the nth term of an AP. If \(\sum_{r=1}^{100}\) \(a_{2r}\) = \(\alpha\) and \(\sum_{r=1}^{100}\) \(a_{2r-1}\) = \(\beta\), then the common difference of the AP is

Solution : Given, \(a_2 + a_4 + a_6 + …… + a_{200}\) = \(\alpha\)      ………(i) and \(a_1 + a_3 + a_5 + ….. + a_{199}\) = \(\beta\)           ………(ii) On subtracting equation (ii) from equation (i), we get (\(a_2 – a_1\)) + (\(a_4 – a_3\)) + ……… + (\(a_{200} – …

Let \(a_n\) be the nth term of an AP. If \(\sum_{r=1}^{100}\) \(a_{2r}\) = \(\alpha\) and \(\sum_{r=1}^{100}\) \(a_{2r-1}\) = \(\beta\), then the common difference of the AP is Read More »

A man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs 11040 after

Solution : Let the time taken to save Rs 11040 be (n + 3) months. for first three months, he saves Rs 200 each month. In (n + 3) months, 3 \(\times\) 200 + \(n\over 2\) { 2(240) + (n – 1) \(\times\) 40 } = 11040 \(\implies\)  600 + \(n\over 2\) {40(12+ n – …

A man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs 11040 after Read More »

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is

Solution : Let a be the first term and d (d \(\ne\) 0) be the common difference of the given AP, then \(T_{100}\) = a + (100 – 1)d = a + 99d \(T_{50}\) = a + (50 – 1)d = a + 49d \(T_{150}\) = a + (150 – 1)d = a + 149d …

If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is Read More »

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