MHT CET · Maths · Quadratic Equation
The number of roots of the equation, \((81)^{\sin ^2 x}+(81)^{\cos ^2 x}=30\) in the interval \([0, \pi]\), is equal to
- A 4
- B 8
- C 3
- D 2
Answer & Solution
Correct Answer
(A) 4
Step-by-step Solution
Detailed explanation
\((81)^{\sin ^2 x}+(81)^{\cos ^2 x}=30 ...(i)\)
\( \text {Let } y=81^{\sin ^2 x} \)
\( \therefore81^{\cos ^2 x}=81^{\left(1-\sin ^2 x\right)}=\frac{81}{81^{\sin ^2 x}}=\frac{81}{y}\)
\(\therefore\) Equation (i) becomes
\(y+\frac{81}{y}=30 \)
\( \therefore y^2-30 y+81=0 \)
\( \therefore (y-27)(y-3)=0 \)
\( \therefore y=27 \text { or } 3 \)
\( \therefore 81^{\sin ^2 x}=27 \text { or } 81^{\sin ^2 x}=3 \)
\( \therefore 3^{4 \sin ^2 x}=3^3 \text { or } 3^{4 \sin ^2 x}=3^1\)
\(\therefore 4 \sin ^2 x=3\) or \(4 \sin ^2 x=1\)
\(\therefore \sin x=\frac{\sqrt{3}}{2}, \frac{1}{2}\) \(\ldots[\because x \in[0, \pi]]\)
\(\therefore x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{\pi}{6}, \frac{5 \pi}{6}\) \(\ldots[\because x \in[0, \pi]]\)
\(\therefore\) Required number of roots are 4 .
\( \text {Let } y=81^{\sin ^2 x} \)
\( \therefore81^{\cos ^2 x}=81^{\left(1-\sin ^2 x\right)}=\frac{81}{81^{\sin ^2 x}}=\frac{81}{y}\)
\(\therefore\) Equation (i) becomes
\(y+\frac{81}{y}=30 \)
\( \therefore y^2-30 y+81=0 \)
\( \therefore (y-27)(y-3)=0 \)
\( \therefore y=27 \text { or } 3 \)
\( \therefore 81^{\sin ^2 x}=27 \text { or } 81^{\sin ^2 x}=3 \)
\( \therefore 3^{4 \sin ^2 x}=3^3 \text { or } 3^{4 \sin ^2 x}=3^1\)
\(\therefore 4 \sin ^2 x=3\) or \(4 \sin ^2 x=1\)
\(\therefore \sin x=\frac{\sqrt{3}}{2}, \frac{1}{2}\) \(\ldots[\because x \in[0, \pi]]\)
\(\therefore x=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{\pi}{6}, \frac{5 \pi}{6}\) \(\ldots[\because x \in[0, \pi]]\)
\(\therefore\) Required number of roots are 4 .
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