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Question: Considering only the principal values of the inverse trigonometric functions, let $P$ be the set of ...

Considering only the principal values of the inverse trigonometric functions, let PP be the set of all integral values of the parameter 'p' for which the equation 16((sec−1x)2+(cosec−1x)2)=pπ216\left( \left(sec^{-1}x\right)^2 + \left(cosec^{-1}x\right)^2 \right) = p\pi^2 has real solution and QQ be the set of all integral values of the parameter 'q' for which the equation (sin−1x)3+(cos−1x)3=q32π3\left(sin^{-1}x\right)^3 + \left(cos^{-1}x\right)^3 = \frac{q}{32}\pi^3 has real solution. Which of the following statement(s) is(are) INCORRECT?

A

n(P)=19n(P)= 19

B

n(Q)=28n(Q)= 28

C

n(P∩Q)=19n(P \cap Q)= 19

D

n(P−Q)=1n(P-Q)= 1

Answer

A, C, D

Explanation

Solution

The problem asks us to find the number of integral values for parameters 'p' and 'q' for which two given equations have real solutions. We then need to check the correctness of four statements regarding the cardinalities and intersections of the sets P and Q.

Part 1: Analyze the equation for parameter 'p'

The given equation is 16((sec−1x)2+(cosec−1x)2)=pπ216\left( \left(sec^{-1}x\right)^2 + \left(cosec^{-1}x\right)^2 \right) = p\pi^2.

For real solutions, xx must be in the domain of both sec−1xsec^{-1}x and cosec−1xcosec^{-1}x, which is (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty).

We use the identity sec−1x+cosec−1x=π2sec^{-1}x + cosec^{-1}x = \frac{\pi}{2}.

Let y=sec−1xy = sec^{-1}x. Then cosec−1x=π2−ycosec^{-1}x = \frac{\pi}{2} - y.

The equation becomes:

16(y2+(π2−y)2)=pπ216\left( y^2 + \left(\frac{\pi}{2} - y\right)^2 \right) = p\pi^2

16(y2+π24−πy+y2)=pπ216\left( y^2 + \frac{\pi^2}{4} - \pi y + y^2 \right) = p\pi^2

16(2y2−πy+π24)=pπ216\left( 2y^2 - \pi y + \frac{\pi^2}{4} \right) = p\pi^2

32y2−16πy+4π2=pπ232y^2 - 16\pi y + 4\pi^2 = p\pi^2

Let f(y)=32y2−16πy+4π2f(y) = 32y^2 - 16\pi y + 4\pi^2. This is a parabola opening upwards, with its vertex at y=−−16π2×32=16π64=π4y = -\frac{-16\pi}{2 \times 32} = \frac{16\pi}{64} = \frac{\pi}{4}.

The value of f(y)f(y) at the vertex is f(π4)=32(π4)2−16π(π4)+4π2=32π216−4π2+4π2=2π2f\left(\frac{\pi}{4}\right) = 32\left(\frac{\pi}{4}\right)^2 - 16\pi\left(\frac{\pi}{4}\right) + 4\pi^2 = 32\frac{\pi^2}{16} - 4\pi^2 + 4\pi^2 = 2\pi^2.

Now, we need to consider the range of y=sec−1xy = sec^{-1}x.

The principal value branch of sec−1xsec^{-1}x is [0,π]−{π2}[0, \pi] - \{\frac{\pi}{2}\}.

However, sec−1xsec^{-1}x and cosec−1xcosec^{-1}x must be defined for the same xx.

Case 1: x∈[1,∞)x \in [1, \infty).

In this case, y=sec−1x∈[0,π2)y = sec^{-1}x \in [0, \frac{\pi}{2}).

The function f(y)f(y) is decreasing in [0,π4][0, \frac{\pi}{4}] and increasing in [π4,π2)[\frac{\pi}{4}, \frac{\pi}{2}).

f(0)=4π2f(0) = 4\pi^2.

As y→π2−y \to \frac{\pi}{2}^-, f(y)→32(π2)2−16π(π2)+4π2=8π2−8π2+4π2=4π2f(y) \to 32\left(\frac{\pi}{2}\right)^2 - 16\pi\left(\frac{\pi}{2}\right) + 4\pi^2 = 8\pi^2 - 8\pi^2 + 4\pi^2 = 4\pi^2.

So, for y∈[0,π2)y \in [0, \frac{\pi}{2}), the range of f(y)f(y) is [2π2,4π2)[2\pi^2, 4\pi^2).

Case 2: x∈(−∞,−1]x \in (-\infty, -1].

In this case, y=sec−1x∈(π2,π]y = sec^{-1}x \in (\frac{\pi}{2}, \pi].

The function f(y)f(y) is increasing for y>π4y > \frac{\pi}{4}. Since (π2,π](\frac{\pi}{2}, \pi] is to the right of π4\frac{\pi}{4}, f(y)f(y) is increasing in this interval.

As y→π2+y \to \frac{\pi}{2}^+, f(y)→4π2f(y) \to 4\pi^2.

f(π)=32π2−16π(π)+4π2=32π2−16π2+4π2=20π2f(\pi) = 32\pi^2 - 16\pi(\pi) + 4\pi^2 = 32\pi^2 - 16\pi^2 + 4\pi^2 = 20\pi^2.

So, for y∈(π2,π]y \in (\frac{\pi}{2}, \pi], the range of f(y)f(y) is (4π2,20π2](4\pi^2, 20\pi^2].

Combining both cases, the range of f(y)f(y) is [2π2,4π2)∪(4π2,20π2][2\pi^2, 4\pi^2) \cup (4\pi^2, 20\pi^2].

Therefore, pπ2∈[2π2,4π2)∪(4π2,20π2]p\pi^2 \in [2\pi^2, 4\pi^2) \cup (4\pi^2, 20\pi^2].

Dividing by π2\pi^2, we get p∈[2,4)∪(4,20]p \in [2, 4) \cup (4, 20].

Since pp must be an integer, P={2,3}∪{5,6,…,20}P = \{2, 3\} \cup \{5, 6, \dots, 20\}.

The number of elements in PP is n(P)=2+(20−5+1)=2+16=18n(P) = 2 + (20 - 5 + 1) = 2 + 16 = 18.

Part 2: Analyze the equation for parameter 'q'

The given equation is (sin−1x)3+(cos−1x)3=q32π3\left(sin^{-1}x\right)^3 + \left(cos^{-1}x\right)^3 = \frac{q}{32}\pi^3.

For real solutions, xx must be in the domain of both sin−1xsin^{-1}x and cos−1xcos^{-1}x, which is [−1,1][-1, 1].

We use the identity sin−1x+cos−1x=π2sin^{-1}x + cos^{-1}x = \frac{\pi}{2}.

Let z=sin−1xz = sin^{-1}x. Then cos−1x=π2−zcos^{-1}x = \frac{\pi}{2} - z.

The equation becomes:

z3+(π2−z)3=q32π3z^3 + \left(\frac{\pi}{2} - z\right)^3 = \frac{q}{32}\pi^3

Using the identity a3+b3=(a+b)(a2−ab+b2)a^3+b^3 = (a+b)(a^2-ab+b^2):

(z+π2−z)(z2−z(π2−z)+(π2−z)2)=q32π3\left(z + \frac{\pi}{2} - z\right)\left(z^2 - z\left(\frac{\pi}{2} - z\right) + \left(\frac{\pi}{2} - z\right)^2\right) = \frac{q}{32}\pi^3

π2(z2−π2z+z2+π24−πz+z2)=q32π3\frac{\pi}{2}\left(z^2 - \frac{\pi}{2}z + z^2 + \frac{\pi^2}{4} - \pi z + z^2\right) = \frac{q}{32}\pi^3

π2(3z2−3π2z+π24)=q32π3\frac{\pi}{2}\left(3z^2 - \frac{3\pi}{2}z + \frac{\pi^2}{4}\right) = \frac{q}{32}\pi^3

Multiplying by 2π\frac{2}{\pi}:

3z2−3π2z+π24=q16π23z^2 - \frac{3\pi}{2}z + \frac{\pi^2}{4} = \frac{q}{16}\pi^2

Let g(z)=3z2−3π2z+π24g(z) = 3z^2 - \frac{3\pi}{2}z + \frac{\pi^2}{4}. This is a parabola opening upwards, with its vertex at z=−−3π/22×3=3π/26=π4z = -\frac{-3\pi/2}{2 \times 3} = \frac{3\pi/2}{6} = \frac{\pi}{4}.

The value of g(z)g(z) at the vertex is g(π4)=3(π4)2−3π2(π4)+π24=3π216−3π28+π24=3π2−6π2+4π216=π216g\left(\frac{\pi}{4}\right) = 3\left(\frac{\pi}{4}\right)^2 - \frac{3\pi}{2}\left(\frac{\pi}{4}\right) + \frac{\pi^2}{4} = \frac{3\pi^2}{16} - \frac{3\pi^2}{8} + \frac{\pi^2}{4} = \frac{3\pi^2 - 6\pi^2 + 4\pi^2}{16} = \frac{\pi^2}{16}.

The principal value branch of z=sin−1xz = sin^{-1}x is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

The vertex z=π4z = \frac{\pi}{4} lies within this interval.

So, the minimum value of g(z)g(z) is π216\frac{\pi^2}{16}.

Now, we evaluate g(z)g(z) at the endpoints of the interval [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}].

g(−π2)=3(−π2)2−3π2(−π2)+π24=3π24+3π24+π24=7π24g\left(-\frac{\pi}{2}\right) = 3\left(-\frac{\pi}{2}\right)^2 - \frac{3\pi}{2}\left(-\frac{\pi}{2}\right) + \frac{\pi^2}{4} = \frac{3\pi^2}{4} + \frac{3\pi^2}{4} + \frac{\pi^2}{4} = \frac{7\pi^2}{4}.

g(π2)=3(π2)2−3π2(π2)+π24=3π24−3π24+π24=π24g\left(\frac{\pi}{2}\right) = 3\left(\frac{\pi}{2}\right)^2 - \frac{3\pi}{2}\left(\frac{\pi}{2}\right) + \frac{\pi^2}{4} = \frac{3\pi^2}{4} - \frac{3\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{4}.

The maximum value of g(z)g(z) in the interval [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] is 7π24\frac{7\pi^2}{4}.

So, the range of g(z)g(z) is [π216,7π24]\left[\frac{\pi^2}{16}, \frac{7\pi^2}{4}\right].

Therefore, q16π2∈[π216,7π24]\frac{q}{16}\pi^2 \in \left[\frac{\pi^2}{16}, \frac{7\pi^2}{4}\right].

Dividing by π2\pi^2: q16∈[116,74]\frac{q}{16} \in \left[\frac{1}{16}, \frac{7}{4}\right].

Multiplying by 16: q∈[1,16×74]=[1,28]q \in \left[1, 16 \times \frac{7}{4}\right] = [1, 28].

Since qq must be an integer, Q={1,2,…,28}Q = \{1, 2, \dots, 28\}.

The number of elements in QQ is n(Q)=28−1+1=28n(Q) = 28 - 1 + 1 = 28.

Part 3: Evaluate the statements

A. n(P)=19n(P)=19

From our calculation, n(P)=18n(P)=18. So statement A is INCORRECT.

B. n(Q)=28n(Q)=28

From our calculation, n(Q)=28n(Q)=28. So statement B is CORRECT.

C. n(P∩Q)=19n(P \cap Q)=19

P={2,3}∪{5,6,…,20}P = \{2, 3\} \cup \{5, 6, \dots, 20\}

Q={1,2,…,28}Q = \{1, 2, \dots, 28\}

P∩Q=({2,3}∪{5,6,…,20})∩{1,2,…,28}P \cap Q = (\{2, 3\} \cup \{5, 6, \dots, 20\}) \cap \{1, 2, \dots, 28\}

P∩Q={2,3}∪{5,6,…,20}P \cap Q = \{2, 3\} \cup \{5, 6, \dots, 20\}

So n(P∩Q)=n(P)=18n(P \cap Q) = n(P) = 18.

Therefore, statement C is INCORRECT.

D. n(P−Q)=1n(P-Q)=1

P−Q=P∖(P∩Q)P-Q = P \setminus (P \cap Q).

Since P∩Q=PP \cap Q = P, P−Q=P∖P=∅P-Q = P \setminus P = \emptyset.

So n(P−Q)=0n(P-Q) = 0.

Therefore, statement D is INCORRECT.

The question asks for the INCORRECT statement(s).

Statements A, C, and D are incorrect.

The final answer is A,C,D\boxed{A, C, D}.