If $ f(x) = (sin^2 x- 1)^n (2 + cos^2 x)$ , then x = π/2 is a point of

Q: If $\large f(x) = (sin^2 x- 1)^n (2 + cos^2 x)$ , then x = π/2 is a point of

(A) local maximum, if n is odd

(B) local minimum, if n is odd

(C) local maximum, if n is even

(D) local minimum, if n is even

Sol. If x = a is the point of local extremum of

y = f(x), then f( a –h).f(a +h) > 0

⇒ f( π/2 –h) . f( π/2+h) > 0

f( π/2- h) = (- ve)n ….(1)

f( π/2 +h) =( -ve)n ….(2)

f(π/2) = 0 ….(3)

⇒ f( π/2 – h).f(π/2 +h) = ( -ve )2n > 0

⇒ n can be odd or even.

So from (1),(2) and (3),if n is odd or even maxima or mimima occurs accordingly .

Hence (A), (D) are correct.

If f (x) = |4 x – x^2 – 3| when x ∈ [0 , 4] then

Q: If f (x) = |4 x – x2 – 3| when x ∈ [0 , 4] then

(A) x = 0 is a global maximum

(B) x = 4 is a global maximum

(C) x = 2 is a local maximum

(D) x = 1 and 3 are global minimum

Sol. Clearly, x = 1, 3 are the points of global minimum (values being equal) and x = 0, 4 are the points of global maximum (values being equal) and x = 2 is a local maximum.

Hence (A), (B), (C) and (D) are correct.

f(x) is cubic polynomial which has local maximum at x = – 1. If f(2) = 18, f(1) = -1 and f'(x) has local minima at x = 0

Q: f(x) is cubic polynomial which has local maximum at x = – 1. If f(2) = 18, f(1) = -1 and f'(x) has local minima at x = 0, then

(A) the distance between (-1, 2) and (a, f(a)), where x = a is the point of local minima is 2

(B) f(x) is increasing for x ∈ [1, 2√5 ]

(C) f(x) has local minima at x = 1

(D) the value of f(0) = 5

Sol: The required polynomial which satisfy the condition is $\large f(x) = \frac{1}{4}(19x^3 – 57x + 34)$

f(x) has local maximum at x = -1 and local minimum at x = 1

Hence f(x) is increasing for x ∈ [1, 2√5 ]

Hence (B) and (C) are correct.

$ Let \; f(x) = \left\{\begin{array}{ll} |x-1| + a \; , x < 1 \\ 2x + 3 \; , x \geq 1 \end{array} \right. $

Q: $ \large Let \; f(x) = \left\{\begin{array}{ll} |x-1| + a \; , x < 1 \\ 2x + 3 \; , x \geq 1 \end{array} \right. $ If f(x) has a local minima at x = 1 , then

(A) a ≥ 5

(B) a > 5

(C) a > 0

(D) none of these

Sol. f(x) = 1- x + a , x < 1

= 2x + 3 , x  ≥ 1

Local minimum value of f(x) at x = 1, will be 5

i.e. 1- x + a  ≥ 5 at x = 1 or, a  ≥ 5.

Hence (A) is correct.