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Please solve RD Sharma class 12 chapter 16 Increasing and Decreasing Function Excercise Fill in the blanks Question 9: Maths Textbook Solution.

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Answer: The set of values of \lambda for which the function is strictly increasing is (4,\infty )

Hint: The function f(x)  is strictly increasing if f'(x)>0 .

Given: f(x)=\frac{\lambda sinx+6cosx}{2sinx+3cosx}

Explanation: We have,

f(x)=\frac{\lambda \sin x+6\cos x}{2\sin x+3\cos x}

This can be written as

f(x)=\frac{4\sin x+6\cos x+(\lambda -4)\sin x}{2\sin x+3\cos x}

f(x)=2+\frac{(\lambda -4)sinx}{2sinx+3cosx}

Differentiate (i)  with respect to x

f'(x)=(\lambda -4)\left [ \frac{(2\sin x+3\cos x)\cos x-\sin x(2\cos x-3\sin x)}{(2\sin x+3\cos x)^{2}} \right ]

 

=(\lambda -4)\frac{3(\cos ^{2}x+\sin ^{2}x)}{(2\sin x+3\cos x)^{2}}

=\frac{3(\lambda -4)}{(2\sin x+3\cos x)^{2}}

Now, as f(x)  is increasing,

                f'(x)>0

\Rightarrow \lambda -4>0

\Rightarrow \lambda >4

\Rightarrow \lambda \epsilon (4,\infty )

Thus, the values of \lambda for the increasing function is (4,\infty )

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