JEE Math Previous Year Questions and Solutions

Let \(\alpha = \sum_{k = 1}^{\infty} \sin^{2k}{(\frac{\pi}{6})}\). Let \(g :[0, 1] \rightarrow R\) be the function defined by  \(g(x) = 2^{\alpha x} + 2^{\alpha (1 - x)}\). Then, which of the following statements is/are TRUE ? A) The minimum value of \(g(x)\) is \(2^{\frac{7}{6}}\). B) The maximum value of \(g(x)\) is \(1 + 2^{\frac{1}{3}}\). C) The function \(g(x)\) attains its maximum at more than one point. D) The function \(g(x)\) attains its minimum at more than one point. Sol : \(\alpha = \sum_{k = 1}^{\infty} \sin^{2k}{(\frac{\pi}{6})} = \sum_{k = 1}^{\infty} (\sin{(\frac{\pi}{6})})^{2k}\) \( =  \sum_{k = 1}^{\infty} (\frac{1}{2})^{2k} =  \sum_{k = 1}^{\infty} ((\frac{1}{2})^{2})^{k}\) \( =  \sum_{k = 1}^{\infty} (\frac{1}{4})^{k}\) \( = \sum_{k = 1}^{\infty} (\frac{1}{4} + (\frac{1}{4})^{2} + (\frac{1}{4})^{3} + ……)\) This is an infinite geometric series with the \(\text{first term}(a) = \frac{1}{4}\) and \(\text{common difference}(r) = \frac{1}{4}\). Thi...

 Consider the hyperbola \(\frac{x^{2}}{100} - \frac{y^{2}}{64} = 1\) with foci at \(S\) and \(S_{1}\),  where \(S\) lies on positive x-axis. Let \(P\) be a point on the hyperbola, in the first quadrant. Let \(\angle{SPS_{1}} = \alpha\), with \(\alpha < \frac{\pi}{2}\). The straight line passing through the point \(S\) and having the same slope as that of the tangent at \(P\) to the hyperbola, intersects the straight line \(S_{1}P\) at \(P_{1}\). Let \(\delta\) be the distance of \(P\) from the straight line \(SP_{1}\), and \(\beta = S_{1}P\). Then the greatest integer less than or equal to \(\frac{\beta \delta}{9}\sin{\frac{\alpha}{2}}\) is ______. Sol :  There are a couple of properties of hyperbola that will help us here. (1) The tangent line at a point bisects the angle between the lines connecting the two foci with the point of tangency.  So, in the figure above, tangent at \(P\) bisects the angle \(\alpha\) between the lines joining the foci(\(S\) an...

The greatest integer less than or equal to  \(\int_{1}^{2}\log_{2}{(x^{3} + 1)}dx + \int_{1}^{\log_{2}{9}}(2^{x} - 1)^{\frac{1}{3}}dx\) is ______. Sol :  Trick in this problem is realising that the integrands \(\log_{2}{(x^{3} + 1)}\) and \((2^{x} - 1)^{\frac{1}{3}}\) are inverse functions of each other. Here’s how. Let \(y(x) = \log_{2}{(x^{3} + 1)}\) \(y(x)\) exists for all \(x > -1\). It is one-one as well as onto function. Hence, its inverse exists.  \(\implies 2^{y} = x^{3} + 1\) \(\implies x = (2^{y} - 1)^{\frac{1}{3}}\) which is the integrand in the second term. Rewriting the same in more familiar notations, \(y^{-1}(x) = (2^{x} - 1)^{\frac{1}{3}}\) So the integrands are inverse functions of each other. Now think about the definite integrals in terms of areas.   For any non-zero invertible function \(y(x)\), the area under the curve between x = c to x = d is highlighted in red. And the blue area is equivalent to the area under the curve \(y^{-1}(x)\) be...

 Let \(\beta\) be a real number. Consider the matrix \(\Biggl(\matrix{\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2}\Biggl)\). If \(A^{7} - (\beta - 1)A^{6} - \beta A^{5}\) is a singular matrix, then the value of \(9\beta\) is _____. Sol:  \(A^{n}\) is a matrix A multiplied to itself n times. Consider \(A^{7}\)            \( = A^{6} \cdot A\)            \( = (A^{5} \cdot A) \cdot A\)            \( = A^{5} \cdot (A \cdot A)\)….. Associative law of matrix multiplication            \( = A^{5} \cdot A^{2}\) \(A^{7} - (\beta - 1)A^{6} - \beta A^{5}\)  \( = A^{5} \cdot A^{2} - (\beta - 1)[A^{5} \cdot A] - \beta A^{5}\) For any constant \(p\) and square matrices \(A\) and \(B\),  \(p (A \cdot B) = (pA) \cdot B = A \cdot (pB)\)  \(\implies = A^{5} \cdot A^{2} - A^{5} \cdot [(\beta - 1) A] - \beta A^{5}\) Also, for any square mat...

The product of all positive real values of \(x\) satisfying the equation \(x^{(16(\log_{5}{x})^{3} - 68\log_{5}{x})} = 5^{-16}\) is _____. Sol :  Let \(t = \log_{5}{x}\). \(\implies x = 5^{t}\) \(x^{(16(\log_{5}{x})^{3} - 68\log_{5}{x})} = 5^{-16}\) \( = 5^{t(16t^{3} - 68t)} = 5^{-16}\) \(\implies t(16t^{3} - 68t) = -16\) \(\implies 16t^{4} - 68t^{2} + 16 = 0\) Let \(p = t^{2}\). \(\implies 4p^{2} - 17p + 4 = 0\) \(\implies 4p^{2} - 16p - p + 4 = 0\) \(\implies 4p(p - 4) - 1(p - 4) = 0\) \(\implies p = 4\) and \(p = \frac{1}{4}\) \(\implies t = \pm 2\) and \(t = \pm \frac{1}{2}\) \(\implies x = 5^{2} ; 5^{-2} ; 5^{\frac{1}{2}} ; 5^{-\frac{1}{2}}\) All will yield positive value answers. \(\implies\) product of all positive values of \(x\)           \( = 5^{2} \times 5^{-2} \times 5^{\frac{1}{2}} \times 5^{-\frac{1}{2}}\)         \( = \frac{5^{2}}{5^{2}} \times \frac{5^{\frac{1}{2}}}{5^{\frac{1}{2}}}\)         ...

 Let  \(\alpha\) and \(\beta\) be real numbers such that \(-\frac{\pi}{4} < \beta < 0 < \alpha < \frac{\pi}{4}\). If \(\sin{(\alpha + \beta)} = \frac{1}{3}\) and \(\cos{(\alpha - \beta)} = \frac{2}{3}\), then the greatest integer less than or equal to  \((\frac{\sin{\alpha}}{\cos{\beta}} + \frac{\cos{\beta}}{\sin{\alpha}} + \frac{\cos{\alpha}}{\sin{\beta}} + \frac{\sin{\beta}}{\cos{\alpha}})^{2}\) is ______. Sol :  \(\sin{(\alpha + \beta)} = \frac{1}{3}\) \(\implies \cos^{2}{(\alpha + \beta)} = 1 - \frac{1}{9} = \frac{8}{9}\) \(\cos{(\alpha - \beta)} = \frac{2}{3}\) \(\implies \sin^{2}{(\alpha - \beta)} = 1 - \frac{4}{9} = \frac{5}{9}\) Re-grouping the terms(1 & 3 together and 2 & 4 together) in the given trigonometric expression:  \( = ((\frac{\sin{\alpha}}{\cos{\beta}} + \frac{\cos{\alpha}}{\sin{\beta}}) + (\frac{\cos{\beta}}{\sin{\alpha}} + \frac{\sin{\beta}}{\cos{\alpha}}))^{2}\) \( = ((\frac{\sin{\alpha} \sin{\beta} + \cos{\alpha} \...