Question 10 - Jee advanced Math 2022 P2 Questions with 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 \(\text{common difference}(r) = \frac{1}{4}\).

This series will converge if the limit 

\(\lim_{n \rightarrow \infty} \sum_{k = 1}^{n} (\frac{1}{4})^{k}\) 

exists and is finite.

Consider, \(\lim_{n \rightarrow \infty} \sum_{k = 1}^{n} (\frac{1}{4})^{k}\)

       \( = \lim_{n \rightarrow \infty} (\frac{1}{4} + (\frac{1}{4})^{2} + (\frac{1}{4})^{3} + …+(\frac{1}{4})^{n})\)

     \( = \lim_{n \rightarrow \infty} \frac{a( 1 - r^n)}{1 - r}\)

      \( = \lim_{n \rightarrow \infty} \frac{\frac{1}{4}(1 - (\frac{1}{4})^{n})}{1 - \frac{1}{4}}\)

       \( = \lim_{n \rightarrow \infty} \frac{1}{3} (1 - \frac{1}{4^{n}})\)

       \(= \frac{1}{3} [\lim_{n \rightarrow \infty} 1 - \lim_{n \rightarrow \infty} \frac{1}{4^{n}}]\)

       \(= \frac{1}{3}[1 - 0] = \frac{1}{3}\)

Substituting \(\alpha = \frac{1}{3}\) in the given function:

\(g(x) = 2^{\frac{1}{3}x} + 2^{\frac{1}{3}(1 - x)}\)

Differentiating w.r.t \(x\),

\(g’(x) = (\frac{1}{3} \ln{2}) 2^{\frac{1}{3}x} + (-1) (\frac{1}{3} \ln{2}) 2^{\frac{1}{3}(1 - x)}\)

\(g’(x) = (\frac{1}{3} \ln{2}) (2^{\frac{1}{3}x} - 2^{\frac{1}{3}(1 - x)})\)

\(g’(x) = 0 \implies 2^{\frac{1}{3}x} - 2^{\frac{1}{3}(1 - x)} = 0\)

\(\implies 2^{\frac{1}{3}x} = 2^{\frac{1}{3}(1 - x)}\)

\(\implies \frac{1}{3}x = \frac{1}{3}(1 - x)\)

\(\implies x = \frac{1}{2}\)

The domain of \(g(x)\) is \([0, 1]\). So the critical point \(x = \frac{1}{2}\) is in the domain of the function. 

To check if the function attains maximum value or minimum value or neither at the critical point: 

Differentiating \(g(x)\) the second time,

\(g’’(x) = (\frac{1}{3} \ln{2}) [(\frac{1}{3} \ln{2}) 2^{\frac{1}{3}x} - ((-1) \frac{1}{3} \ln{2}) 2^{\frac{1}{3}(1 - x)}]\)

\(   = (\frac{1}{3} \ln{2})^{2} (2^{\frac{1}{3}x} + 2^{\frac{1}{3}(1 - x)})\)

          \(> 0\) for every \(x \in [0,1]\)

\(\implies g(x)\) attains minimum value at \(x = \frac{1}{2}\) by second derivative test.

\(\implies\) the minimum value of \(g(x)\) is 

\(g(\frac{1}{2}) = 2^{\frac{1}{3} \frac{1}{2}} + 2^{\frac{1}{3}(1 - \frac{1}{2})}\)

              \( = 2^{\frac{7}{6}}\)

The pattern followed here by \(g(x)\) is: decreasing initially in the interval \(0 \leq x \leq \frac{1}{2}\) and then increasing in the interval \(\frac{1}{2} \leq x \leq 1\). So the maximum value will be attained at any of the two end points.

\(g(0) = 1 + 2^{\frac{1}{3}}\) and

\(g(1) = 1 + 2^{\frac{1}{3}}\)

Hence, the maximum value of \(g(x)\) is \(1 + 2^{\frac{1}{3}}\). Also, the function attains the same maximum value at two points(0 & 1), but the minimum value at only one point(\(\frac{1}{2}\)).

Therefore, options A, B and C are True. 


Comments

Post a Comment

Popular posts from this blog

Question 7 - Jee advanced Math 2022 P2 Questions with Solutions

Image
 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...
Read more

Question 6 - Jee advanced Math 2022 P2 Questions with Solutions

 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...
Read more

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...
Read more