JEE Math Previous Year Questions and Solutions

 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} \...

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...

Let \(l_{1}, l_{2},….,l_{100}\) be consecutive terms of an arithmetic progression with common difference \(d_{1}\), and let  \(w_{1}, w_{2},….,w_{100}\) be consecutive terms of another arithmetic progression with common difference \(d_{2}\), where \(d_{1}d_{2} = 10\). For each \(i = 1, 2,…,100\), let \(R_{i}\) be a rectangle with length \(l_{i}\), width \(w_{i}\) and area \(A_{i}\). If  \(A_{51}-A_{50}=1000\), then the value of  \(A_{100}-A_{90}\) is _____. Sol :  \(l_{1}, l_{2},….,l_{100}\) are consecutive terms of an arithmetic progression with common difference \(d_{1}\). \(\implies l_{2} = l_{1} + d_{1}; \: l_{3} = l_{1} + 2 d_{1}; \: l_{4} = l_{1} + 3d_{1}; ….l_{i}= l_{1} + (i-1)d_{1}…\) ….where \(1 \leq i \leq 100\) \(w_{1}, w_{2},….,w_{100}\) are consecutive terms of an arithmetic progression with common difference \(d_{2}\). \(\implies w_{2} = w_{1} + d_{2}; \: w_{3} = w_{1} + 2 d_{2}; \: w_{4} = w_{1} + 3d_{2}; ….w_{j}= w_{1} + (j-1)d_{2}…\) ….where ...

Let \(\bar{z}\) denote the complex conjugate of a complex number \(z\) and let \(i=\sqrt{-1}\). In the set of complex numbers, the number of distinct roots of the equation  \(\bar{z} - z^{2} = i(\bar{z} + z^{2})\) is ______. Sol: \(\bar{z} - z^{2} = i(\bar{z} + z^{2})\) \(\implies \bar{z}(1-i) = z^{2}(1+i)\) Clearly \(z = 0 + 0i\) is one of the solutions of the equation. To find other non-zero solutions, let \(z \neq 0 + 0i \implies \bar{z} \neq 0+0i\)  \(\implies \frac{z^{2}}{\bar{z}} = \frac{1-i}{1+i}\) \(\implies \frac{z^{2}}{\bar{z}} = \frac{(1-i)(1-i)}{(1+i)(1-i)}\) \(\implies \frac{z^{2}}{\bar{z}} = \frac{-2i}{2}\) \(\implies \frac{z^{2}}{\bar{z}} = -i\) Rewriting the above equation in exponential form, \(\implies \frac{(r e^{i \theta})^{2}}{r e^{i (-\theta)}} = 1 e^{i (- \frac{\pi}{2})}\) where \(r = |z|\) and \(\theta = arg(z)\) \(\implies r e^{i (3\theta)} = 1 e^{i (- \frac{\pi}{2})}\) Solutions to the equation are all the possible values of \(r\) and \(\theta\) for...

Let \(z\) be a complex number with non-zero imaginary part. If \(\frac{2+3z+4z^{2}}{2-3z+4z^{2}}\) is a real number, then the value of \(|z|^{2}\) is _____. Sol :  It is given that the fraction \(\frac{2+3z+4z^{2}}{2-3z+4z^{2}}\) is a real number. Let \(p\) be a real number such that  \(\frac{2+3z+4z^{2}}{2-3z+4z^{2}} = p\) \(\implies 2+3z+4z^{2} = p(2-3z+4z^{2})\) \(\implies 2(1-p) + 3z(1+p) + 4z^{2}(1-p) = 0\) \(\implies (1-p)(2+4z^{2}) + 3z(1+p) = 0\) \(\implies (1-p)(2+4z^{2}) = -3z(1+p)\) \(\implies \frac{2+4z^{2}}{3z} = -\frac{(1+p)}{(1-p)}\) (We are dividing both sides by \((1-p)\) because \(p \neq 1\). If \(p=1 \implies 2+3z+4z^{2} = 1(2-3z+4z^{2}) \implies z = 0 + 0 i\), which contradicts the given fact that \(z\) has non-zero imaginary part.) \(\implies \frac{2+4z^{2}}{3z} = \frac{(1+p)}{(p-1)}\) As \(p\) is a real number, \(\frac{1+p}{p-1}\) is also a real number. \(\implies \frac{2+4z^{2}}{3z}\) has imaginary part equal to zero. \(\frac{2+4z^{2}}{3z} = \frac{2}...

In a study about a pandemic, data of 900 persons was collected. It was found that 190 persons had symptom of fever, 220 persons had symptom of cough, 220 persons had symptom of breathing problem, 330 persons had symptom of fever or cough or both, 350 persons had symptom of cough or breathing problem or both, 340 persons had symptom of fever or breathing problem or both, 30 persons had all three symptoms(fever, cough and breathing problem). If a person is chosen randomly from these 900 persons, then the probability that the person has at-most one symptom is _____. Sol :  Let C be a letter for representing ‘Cough’; F for ‘Fever’; and B for ‘breathing problem’.  i) 190 = n(F) = n(F only) + n(F and C) + n(F and B) + n(all three) ii) 220 = n(C) = n(C only) + n(F and C) + n(C and B) + n(all three) iii) 220 = n(B) = n(B only) + n(F and B) + n(C and B) + n(all three) n(F or C or both) = n(F) + n(C) - n(F and C) - n(all three) (We are subtracting n(F and C) because it is already in...

2) Let \(\alpha\) be a positive real number. Let \(f:R \rightarrow R\) and \(g:(\alpha, \infty) \rightarrow R\) be the functions defined by \(f(x) = \frac{\sin{\pi x}}{12}\)  and \(g(x) = \frac{2\ln{(\sqrt{x}-\sqrt{\alpha}})}{\ln{(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}}\). Then the value of \(\lim_{x\rightarrow \alpha^{+}}{f(g(x))}\) is ______. Sol :  Limit of a composite function \(f(g(x))\) as \(x\) approaches \(a\) can be found using the following result: If,       i) \(\lim_{x\rightarrow a}{g(x)}\) exists and is equal to L and     ii) f(x) is continuous at L then \(\lim_{x\rightarrow a}{f(g(x))} = f(\lim_{x\rightarrow a}{g(x)}) = f(L)\) i)  Let’s check the first condition for our example.  \(\lim_{x\rightarrow \alpha^{+}}{g(x)} = \lim_{x\rightarrow \alpha^{+}}{\frac{2\ln{(\sqrt{x}-\sqrt{\alpha})}}{\ln{(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}}}\) On substituting \(x = \alpha\) the limit will result in indeterminate \(\frac{0}{0}\) form. So  L'Hôpi...