JEE Math Previous Year Questions and Solutions

Let \(PQRS\) be a quadrilateral in a plane, where \(QR = 1\), \(\angle{PQR} = \angle{QRS} = 70^{\circ}\), \(\angle{PQS} = 15^{\circ}\) and \(\angle{PRS} = 40^{\circ}\). If \(\angle{RPS} = \theta^{\circ}\), \(PQ = \alpha\) and \(PS = \beta\), then the interval(s) that contain(s) the value of \(4\alpha \beta \sin{\theta}\) is/are A) \((0, \sqrt{2})\) B) \((1, 2)\) C) \((\sqrt{2}, 3)\) D) \((2\sqrt{2}, 3\sqrt{2})\) Sol :  The given information is coded in the following figure : \(\angle{SQR} = 70^{\circ} - 15^{\circ} = 55^{\circ}\) \(\angle{PRQ} = 70^{\circ} - 40^{\circ} = 30^{\circ}\) In triangle \(QSR\), \(\angle{QSR} = 180^{\circ} - (55^{\circ} + (40^{\circ} + 30^{\circ}))\)    \(= 55^{\circ}= \angle{SQR}\) \(\implies QR = SR = 1\)….{in a triangle, sides opposite to equal angles are equal} In triangle \(PQR\), \(\angle{QPR} = 180^{\circ} - (30^{\circ} + (15^{\circ} + 55^{\circ}))\)    \(= 80^{\circ}\) Using Sine rule of triangles, \(\frac{\alpha}{\sin{30^{\ci...

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 \(x\)           \( = 5^{2} \times 5^{-2} \times 5^{\frac{1}{2}} \times 5^{-\frac{1}{2}}\)         \( = \frac{5^{2}}{5^{2}} \times \frac{5^{\frac{1}{2}}}{5^{\frac{1}{2}}}\)         ...

Considering only the principal values of the inverse trigonometric function, the value of \(\frac{3}{2} \cos^{-1}{\sqrt{\frac{2}{2+\pi^{2}}}} + \frac{1}{4}\sin^{-1}{\frac{2\sqrt{2}\pi}{2 + \pi^{2}}} + \tan^{-1}{\frac{\sqrt{2}}{\pi}}\) is ____. Sol : Convert \(\cos^{-1}\) and \(\sin^{-1}\) terms into \(\tan^{-1}\). To Convert \(\cos^{-1}\) term into \(\tan^{-1}\) : Let \(t = \cos^{-1}{\sqrt{\frac{2}{2+\pi^{2}}}} = \cos^{-1}{\frac{\sqrt{2}}{\sqrt{2+\pi^{2}}}}\) …(1) \(\implies\)\(\cos{t} = \frac{\sqrt{2}}{\sqrt{2+\pi^{2}}}\) Use right angle triangle to determine  ‘\(\tan{t}\)’. Consider a right triangle as shown above, with side adjacent to angle ‘t’ equal to \(\sqrt{2}\), and hypotenuse equal to \(\sqrt{2+\pi^{2}}\).  Using Pythagoras’ theorem,  side opposite to ‘t’ \(= \sqrt{2+\pi^{2} - 2} = \pi\) Hence,  \(\sin{t} = \frac{\pi}{\sqrt{2+\pi^{2}}}\) And,  \(\tan{t} = \frac{\sin{t}}{\cos{t}}\)         \(  = \fra...

Let \(P_{1}\) and \(P_{2}\) be two planes given by  \(P_{1} : 10x + 15y + 12z - 60 = 0\). \(P_{2} : -2x + 5y + 4z - 20 = 0\). Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on \(P_{1}\) and \(P_{2}\)? A) \(\frac{x - 1}{0} = \frac{y - 1}{0} = \frac{z - 1}{5}\) B) \(\frac{x - 6}{-5} = \frac{y}{2} = \frac{z}{3}\) C) \(\frac{x}{-2} = \frac{y - 4}{5} = \frac{z}{4}\) D) \(\frac{x}{1} = \frac{y - 4}{-2} = \frac{z}{3}\) Sol :  The two faces of a tetrahedron lie on the planes \(P_{1}\) and \(P_{2}\). Let L be the line of intersection of the planes. L is one of the six edges of the tetrahedron. In the figure above, the other five edges of the tetrahedron are \(l_{1}\), \(l_{2}\), \(l_{3}\), \(l_{4}\) and \(l_{5}\), out of which, \(l_{1}\), \(l_{2}\), \(l_{3}\) and \(l_{4}\), all four of them completely lie in either plane \(P_{1}\) or \(P_{2}\) …AND… they each intersect the line L at a single point(A and B in the figure).  So the li...

Consider the parabola \(y^{2} = 4x\). Let \(S\) be the focus of the parabola. A pair of tangents drawn to the parabola from the point \(P(-2, 1)\) meet the parabola at \(P_{1}\) and \(P_{2}\). Let \(Q_{1}\) and \(Q_{2}\) be points on the lines \(SP_{1}\) and \(SP_{2}\) respectively such that \(PQ_{1}\) is perpendicular to \(SP_{1}\) and \(PQ_{2}\) is perpendicular to \(SP_{2}\). Then, which of the following is/are TRUE? A) \(SQ_{1} = 2\) B) \(Q_{2}Q_{1} = \frac{3\sqrt{10}}{5}\) C)  \(PQ_{1} = 3\) D) \(SQ_{2} = 1\) Sol :  \(y^{2} = 4x\) is a standard parabola \(y^{2} = 4ax\) with the vertex at the origin. \(\implies 4a = 4 \implies a = 1(> 0)\) So this is a parabola which opens to the right(since \(a > 0\)) in the Cartesian plane. \(\implies\) Focus of the parabola is S(1, 0).  \(PP_{1}\) and \(PP_{2}\) are tangents to the parabola from \(P(-2, 1)\). And \(PQ_{1}\), \(PQ_{2}\) are perpendiculars to \(SP_{1}\) and \(SP_{2}\).  We are looking for the coordinate...

Let S be the reflection of point Q with respect to the plane given by \(\vec{r} = -(t + p)\hat{i} + t\hat{j} + (1 + p)\hat{k}\) where t, p are real parameters and \(\hat{i}, \hat{j}, \hat{k}\) are the unit vectors along the three positive coordinate axes. If the position vectors of Q and S are \(10\hat{i}+ 15\hat{j} + 20\hat{k}\) and \(\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}\) respectively, then which of the following is/are TRUE? A) \(3(\alpha + \beta) = -101\) B) \(3(\beta + \gamma) = -71\) C) \(3(\gamma + \alpha) = -86\) D) \(3(\alpha + \beta + \gamma) = -121\) Sol :  S is the reflection of Q. \(\implies\) Both are same distance(perpendicular) from the given plane. Drop perpendiculars from S and Q onto the plane; let A be the point on the plane where they meet. Let \((x_{1}, y_{1}, z_{1})\) be the coordinates of A. \(\implies\) \(\overrightarrow{OA} = x_{1}\hat{i} + y_{1}\hat{j} + z_{1}\hat{k}\) is position vector of A. Also, let \(\overrightarrow{OQ} = 10\hat{i}+ 15\...

Let ABC be the triangle with \(AB = 1, AC = 3\) and \(\angle{BAC} = \frac{\pi}{2}\). If a circle of radius \(r > 0\) touches the sides AB, AC and also touches internally the circumcircle of the triangle ABC, then the value of \(r\) is _____.  Sol : Let \(C_{1}\) and \(r_{1}\) be the centre and the radius of the circumcircle(circle through the three vertices) of triangle ABC. Let \(C_{2}\) be the centre of the circle which is touching sides AB and AC of triangle ABC and touching the circumcircle at P(say). From the question, letter \(r\) denotes the radius of this circle.  The first result that we will use here is : If two circles touch each other internally(or externally) then their centres and the point of contact of circles are aligned, i.e., points \(C_{1}, C_{2}\) and P are collinear.  \(\implies r = r_{1} - d(C_{1}, C_{2})\) …..(1) where \(d(C_{1}, C_{2})\) is the distance between \(C_{1}\) and \(C_{2}\). The second result that we require here is related to the...

Let \(a_{1}, a_{2}, a_{3}, ….\) be an arithmetic progression with \(a_{1}=7\) and common difference \(8\). Let \( T_{1}, T_{2}, T_{3},…\) be such that \(T_{1} = 3\) and \(T_{n+1} - T_{n} = a_{n}\) for \(n \geq 1\). Then, which of the following is/are TRUE? A) \(T_{20} =1604\) B) \(\sum_{k=1}^{20} T_{k} = 10510\) C) \(T_{30} = 3454\) D)  \(\sum_{k=1}^{30} T_{k} = 35610\) Sol :  \(a_{1}, a_{2}, a_{3}, ….\) is an arithmetic progression. \(\implies a_{n} = a_{1} + (n - 1)d\) Also, \(T_{n+1} - T_{n} = a_{n}\) \(\implies T_{n+1} = T_{n} + a_{n}\)  i.e., \(T_{2} = T_{1} + a_{1}; \:  T_{3} = T_{2} + a_{2}; \: T_{4} = T_{3} + a_{3}; \:\) and so on….. Let’s investigate the four options. A)  \(T_{20} = 1604\) \(\implies T_{19} + a_{19} = 1604\) \(\implies T_{18} + a_{18} + a_{19} = 1604\) …… \(\implies T_{1} + a_{1} + a_{2} + a_{3} + ….+ a_{19} = 1604\)  \(\implies 3 + a_{1} + (a_{1} + d) + (a_{1} + 2d) + ….+ (a_{1} + 18d) = 1604\) \(\implies 3 + (19 \times a_{1}) ...

 Consider the equation \(\int_{1}^{e} \frac{(\log_{e}{x})^{\frac{1}{2}}}{x(a - (\log_{e}{x})^\frac{3}{2})^{2}}dx = 1, \: \: a \in (-\infty, 0) U (1, \infty)\). Which of the following statements is/are true? (A) No \(a\) satisfies the above equation (B) An integer \(a\) satisfies the above equation (C) An irrational number \(a\) satisfies the above equation (D) More than one \(a\) satisfy the above equation Sol :  \(\int_{1}^{e} \frac{(\log_{e}{x})^{\frac{1}{2}}}{x(a - (\log_{e}{x})^\frac{3}{2})^{2}}dx = 1\) \(\implies \int_{1}^{e} \frac{(\log_{e}{x})^{\frac{1}{2}}\times \frac{1}{x}}{(a - (\log_{e}{x})^\frac{3}{2})^{2}}dx = 1\) Let \((\log_{e}{x})^\frac{3}{2} = t\) \(\implies (\frac{3}{2}(\log_{e}{x})^\frac{1}{2} \times \frac{1}{x})dx = dt\) \(\implies ((\log_{e}{x})^\frac{1}{2} \times \frac{1}{x})dx = \frac{2}{3}dt\) \(x = 1 \implies (\log_{e}{1})^\frac{3}{2} = 0 = t\) \(x = e \implies (\log_{e}{e})^\frac{3}{2} = 1 = t\) \(\implies \int_{0}^{1} \frac{\frac{2}{3}dt}{(a - t)^...

The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7 is _____. Sol : Method 1 (long) Let a, b, c and d be the four digits of a number from left to right in the given interval [2022, 4482]. \(\underline{a} \; \underline{b} \; \underline{c} \; \underline{d}\) For a four digit number in [2022, 4482], the first digit ‘a’ is always 2 or 3 or 4. All three of them are present in the given set of digits. So ‘a’ can be chosen in 3 ways.  But the choice for the second digit ‘b’ will depend on the choice of ‘a’. For instance, if ‘a’ is 3, ‘b’ can be any of the 6 given numbers(6 ways), but if ‘a’ is 4, ‘b’ cannot be 6 and 7 as the numbers 46_ _  and 47_ _are not a part of the interval. So if ‘a’ is 4, possible values of b are 0, 2, 3, 4(4 ways).  So to make this a little easier, we divide [2022, 4482] into union of three smaller intervals(for 3 different choices of ‘a’) as follows : [2022, 2999] U [3000, 3999] U [4000, 4482] I...