TS EAMCET · Maths · Inverse Trigonometric Functions
For \(a>0\), if \(f(x)=a x+b\) is an onto function from \([-1,1]\) to \([0,2]\), then \(\cot \left[\tan ^1 \frac{1}{7}+\tan ^1 \frac{1}{8}+\tan ^1 \frac{1}{5}\right]=\)
- A \(f(-1)\)
- B \(f(1)\)
- C \(f(0)\)
- D \(f(2)\)
Answer & Solution
Correct Answer
(B) \(f(1)\)
Step-by-step Solution
Detailed explanation
We have, \(f(x)=a x+b\) is onto from \([-1,1]\) to \([0,2]\) \(\begin{aligned} & \because \quad f(-1)=-a+b=0 \\ & f(\mathbf{l})=a+b=2 \\ & a=1, b=1 \\ & \therefore \quad f(x)=x+1 \end{aligned}\) From Eqs. (i) and (ii), we get…
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