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Question 13 - Jee advanced Math 2022 P1 Questions with Solutions

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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...

Question 12 - Jee advanced Math 2022 P1 Questions with Solutions

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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\...

Question 8 - Jee advanced Math 2022 P1 Questions with Solutions

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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...

Question 10 - Jee advanced Math 2022 P1 Questions with Solutions

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}) ...

Question 9 - Jee advanced Math 2022 P1 Questions with Solutions

 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)^...

Question 7 - Jee advanced Math 2022 P1 Questions with Solutions

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...

Question 6 - Jee advanced Math 2022 P1 Questions with Solutions

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 ...

Question 5 - Jee advanced Math 2022 P1 Questions with Solutions

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...

Question 4 - Jee advanced Math 2022 P1 Questions with Solutions

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}...

Question 3 - Jee advanced Math 2022 P1 Questions with Solutions

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...

Question 2 - Jee advanced Math 2022 P1 Questions with Solutions

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...