The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. Practice Paper 1 Question 6

Which values of $$k$$ give a maximum at $$x=-1$$ for $$f(x)=(k+1)x^4-(3k+2)x^2-2kx$$?

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Hints

• Hint 1
How do you find points of maxima/minima of a given curve?
• Hint 2
What is the value of derivative of $$f(x)$$ at $$x=-1$$?
• Hint 3
How do you determine if a stationary point is a point of maxima?

Solution

First derivative of $$f(x)$$ with respect to $$x$$ is $$(4k+4)x^3-(6k+4)x-2k$$, and second derivative of $$f(x)$$ with respect to $$x$$ is $$(12k+12)x^2-(6k+4)$$. Notice that the first derivative is always $$0$$ at $$x=-1,$$ it does not depend on $$k$$. For maxima, the second derivative must be negative at $$x=-1$$ which gives us $$12k+12 -(6k+4)<0,$$ and hence $$k<-\frac{4}{3}$$.

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