alleryThe slope of the tangent to the curve $y = f(x)$ at the point $x = a$ is:
- $\dfrac{1}{f'(a)}$
- $f(a)$
- $-\dfrac{1}{f'(a)}$
- $f'(a)$
Show answer and solution
Correct answer: D. $f'(a)$
The derivative f'(a) is the slope of the tangent at x = a.
JEE Main Unit 4-5: Applications of Derivatives & Curve Sketching — practice questions — page 2
Explore 39 free Unit 4-5: Applications of Derivatives & Curve Sketching questions in JEE Main Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
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The slope of the tangent to the curve $y = f(x)$ at the point $x = a$ is:Read the answer and worked solution →The slope of the normal to the curve $y = f(x)$ at $x = a$ is:Read the answer and worked solution →At a critical point (a candidate for a local maximum or minimum) of a differentiable function $f$:Read the answer and worked solution →The function $f(x) = x^2$ is strictly increasing for:Read the answer and worked solution →The rate of change of the volume of a sphere $V = \tfrac{4}{3}\pi r^3$ with respect to its radius $r$ is:Read the answer and worked solution →For $f(x) = x^2 - 4x + 3$, the critical point occurs at:Read the answer and worked solution →The slope of the tangent to $y = x^2$ at the point $(1,1)$ is:Read the answer and worked solution →By the second derivative test, if $f'(c) = 0$ and $f''(c) > 0$, then $c$ is a point of:Read the answer and worked solution →By the second derivative test, at a point of local maximum the value of $f''(x)$ is:Read the answer and worked solution →
Worked practice questions
The slope of the normal to the curve $y = f(x)$ at $x = a$ is:
- $f(a)$
- $\dfrac{1}{f(a)}$
- $f'(a)$
- $-\dfrac{1}{f'(a)}$
Show answer and solution
Correct answer: D. $-\dfrac{1}{f'(a)}$
The normal is perpendicular to the tangent, so its slope is −1/f'(a).
At a critical point (a candidate for a local maximum or minimum) of a differentiable function $f$:
- $f(x) = 0$
- $f''(x) = 0$
- $f'(x) = 1$
- $f'(x) = 0$
Show answer and solution
Correct answer: D. $f'(x) = 0$
Critical points of a differentiable function satisfy f'(x) = 0.
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