JEE Main 2022 (Online) 24th June Morning Shift | Functions Question 54 | Mathematics

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JEE Main 2022 (Online) 24th June Morning Shift

MCQ (Single Correct Answer)

-1

For the function

$$f(x) = 4{\log _e}(x - 1) - 2{x^2} + 4x + 5,\,x > 1$$, which one of the following is NOT correct?

f is increasing in (1, 2) and decreasing in (2, $$\infty$$)

f(x) = $$-$$1 has exactly two solutions

$$f'(e) - f''(2) < 0$$

f(x) = 0 has a root in the interval (e, e + 1)

JEE Main 2022 (Online) 24th June Morning Shift

MCQ (Single Correct Answer)

-1

The sum of absolute maximum and absolute minimum values of the function $$f(x) = |2{x^2} + 3x - 2| + \sin x\cos x$$ in the interval [0, 1] is :

$$3 + {{\sin (1){{\cos }^2}\left( {{1 \over 2}} \right)} \over 2}$$

$$3 + {1 \over 2}(1 + 2\cos (1))\sin (1)$$

$$5 + {1 \over 2}(\sin (1) + \sin (2))$$

$$2 + \sin \left( {{1 \over 2}} \right)\cos \left( {{1 \over 2}} \right)$$

JEE Main 2022 (Online) 24th June Morning Shift

MCQ (Single Correct Answer)

-1

The domain of the function

$$f(x) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}({x^2} - 3x + 2)}}$$ is :

$$( - \infty ,1) \cup (2,\infty )$$

$$(2,\infty )$$

$$\left[ { - {1 \over 2},1} \right) \cup (2,\infty )$$

$$\left[ { - {1 \over 2},1} \right) \cup (2,\infty ) - \left\{ 3,{{{3 + \sqrt 5 } \over 2},{{3 - \sqrt 5 } \over 2}} \right\}$$

JEE Main 2021 (Online) 1st September Evening Shift

MCQ (Single Correct Answer)

-1

The range of the function,

$$f(x) = {\log _{\sqrt 5 }}\left( {3 + \cos \left( {{{3\pi } \over 4} + x} \right) + \cos \left( {{\pi \over 4} + x} \right) + \cos \left( {{\pi \over 4} - x} \right) - \cos \left( {{{3\pi } \over 4} - x} \right)} \right)$$ is :

$$\left( {0,\sqrt 5 } \right)$$

[$$-$$2, 2]

$$\left[ {{1 \over {\sqrt 5 }},\sqrt 5 } \right]$$

[0, 2]

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