Home Maths Differentiation and Applications of Derivatives General The following questions given below consist …
Maths Differentiation and Applications of Derivatives General Single Correct MCQ
Published on: August 14, 2026

The following questions given below consist of an " Statement-I " and " Statement-II " Type questions. Use the following Key to choose the appropriate answer.

A
If both Statement-I and Statement-II are true, and Statement-II is the correct explanation of Statement-I.
B
If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I.
C
If Statement-I is true but Statement-II is false.
D
If Statement-I is false but Statement-II is true. (i) Statement-I : Let the coefficients in the cubic equation ax 3 + bx 2 + cx + d = 0 be related as – a + b – c + d = 3 and 8a + 4b + 2c + d = 6. Then the equation 3ax 2 + 2bx + c – 1 = 0 has at least one root in (–1, 2). Statement-II : If ƒ(x) be a continuous function in [a, b] such that ƒ'(x) exists, then between two consecutive roots of ƒ'(x) = 0 there will be at least one root of ƒ(x) = 0. (ii) Statement-I : The slope of normal at the point with abscissa x = –2 of the graph of the function f(x) = |x 2 – |x|| is 1/3. Statement-II : at x = –2, the slope of tangent of the curve is (–3) and normal perpendicular to tangent. (iii) Statement-I : If S = t 3 – 6t, then the acceleration at the time when velocity vanishes. Statement-II : The acceleration of the particle is given by a =

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Ans.

(i)

Sol.

(ii)

Sol. f(x) = |x 2 – |x|| for x = – 2

∴ f(x) = x 2 + x x = – ve ⇒ |x| = – x

= 2x + 1 |x 2 – |x|| = |x 2 + x|

Slope of tangent at x = – 2

= – 4 + 1 = – 3 x 2 + x = +ve

Slope of normal

= = . ⇒ |x 2 + x| = x 2 + x

(iii)

Sol. Statement-I : S = t 3 – 6t

= t 2 – 6 = 0 ⇒ t 2 = 4 ⇒ t = 2 sec.

Now = 3t = 3.2 = 6 unit/sec 2 Statement-II : It is also correct.

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